AP Inter 2nd Year

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4e Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4e

I.

Question 1.
Examine the consistency of the system of equations x + 2y – 2, 2x + 3y = 3
Solution:
The given system of equations is: x + 2y – 2, 2x + 3y = 3
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\), x = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Hence, |A| = 1(3) – 2(2) = 3 – 4 = -1 ≠ 0
So, A is non-singular. Hence, A^-1 exists.
Thus, the given system of equations is consistent.

Question 2.
Examine the consistency of the system of equations 2x – y = 5, x + y = 4
Solution:
The given system of equations is 2x – y = 5, x + y = 4
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ll}
2 & -1 \\
1 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
5 \\
4
\end{array}\right]\)
Hence, |A| = 2(1) – 1(-1) = 2 + 1 = 3 ≠ 0 .
So, A is non-singular. Hence, A^-1 exists.
Thus, the given system of equations is consistent.

Question 3.
Examine the consistency of the system of equations x + 3y = 5, 2x + 6y = 8
Solution:
The given system of equations is x + 3y = 5, 2x + 6y = 8
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 6
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
5 \\
8
\end{array}\right]\)
Hence, |A| = 1(6) – 2(3) = 6 – 6 = 0 .
So, A is a singular matrix.
Now, (adjA) = \(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\)
(adj A)B = \(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
5 \\
8
\end{array}\right]=\left[\begin{array}{c}
30-24 \\
-10+8
\end{array}\right]=\left[\begin{array}{c}
6 \\
-2
\end{array}\right]\) ≠ 0
Hence, A^-1 exists.
Thus, the solution of the given system of equations does not exist.
Thus, the given system of equations is inconsistent.

II.

Question 1.
Examine the consistency of the system of equations x + y + z = 1, 2x + 3y + 2z = 2, ax + ay + 2az = 4
Solution:
The given system of equations is x + y + z = 1, 2x + 3y + 2z = 2,ax + ay + 2az =4
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 2 \\
\mathrm{a} & \mathrm{a} & 2 \mathrm{a}
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
1 \\
2 \\
4
\end{array}\right]\)
Hence, |A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a) = 4a – 2a – a
= 4a – 3a = a ≠ 0
So, A is non-singular. Hence, A^–1 exists.
Thus, the given system of equations is consistent.

Question 2.
Examine the consistency of the system of equations 3x – y – 2z = 2, 2y – 2z = -1, 3x – 5y = 3
Solution:
The given system of equations is 3x – y – 2z = 2, 2y – 2z = -1, 3x – 5y = 3
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
3 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right],\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
2 \\
-1 \\
3
\end{array}\right]\)
Hence, |A| = 3(0 – 5) – 0 + 3(1 + 4) = -15 + 15 = 0
So, A is a singular matrix.
Now, (adjA) = \(\left[\begin{array}{rcr}
-5 & 10 & 5 \\
-3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right]\)
∴ (adjA)B = \(\left[\begin{array}{ccc}
-5 & 10 & 5 \\
-3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right]\left[\begin{array}{c}
2 \\
-1 \\
3
\end{array}\right]=\left[\begin{array}{c}
-10-10+15 \\
-6-6+9 \\
-12-12+18
\end{array}\right]=\left[\begin{array}{l}
-5 \\
-3 \\
-6
\end{array}\right]\) ≠ 0
Thus, the given system of equations does not exist.
Hence, the system of equations is inconsistent.

Question 3.
Examine the consistency of the system of equations 5x – y + 4z = 5, 2x + 3y + 5z = 2, 5x – 2y + 6z = -1
Solution:
The given system of equations is 5x – y + 4z = 5, 2x + 3y + 5z = 2, 5x – 2y + 6z = -1
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
5 \\
2 \\
-1
\end{array}\right]\)
Hence, |A| = 5(18 + 10) + 1(12 – 25) + 4(-4 – 15)
= 5(28) + 1(-13) + 4(-19)
= 140 – 13 – 76 = 51 ≠ 0
So, A is non-singular. Hence, A^-1 exists.
Thus, the given system of equations is consistent.

Question 4.
Examine the consistency of the system of equations 5x + 2y = 4, 7x + 3y = 5
Solution:
The given system of equations is 5x + 2y = 4, 7x + 3y = 5
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ll}
5 & 2 \\
7 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), B = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
Hence, |A| = 15 – 14 = 1 ≠ 0
So, A is non-singular. Hence, A^-1 exists.
Now, A^-1 = \(\frac{1}{|\mathrm{~A}|}\)(adjA) = \(\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\)
⇒ X = A^-1B \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\left[\begin{array}{l}
4 \\
5
\end{array}\right]=\left[\begin{array}{c}
12-10 \\
-28+25
\end{array}\right]=\left[\begin{array}{c}
2 \\
-3
\end{array}\right]\)
∴ x = 2 and y = -3

Question 5.
Solve system of linear equations, using matrix method 2x – y = -2, 3x + 4y = 3.
Solution:
The given system of equations is 2x – y = -23x + 4y = 3
The given system of equations can be wrEtten in the form of AX = B where
A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & 4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), B = \(\left[\begin{array}{l}
-2 \\
3
\end{array}\right]\)
Hence, |A| = 8 + 3 = 11 ≠ 0
So, A is non-singular. Hence, A^-1 exists.

Question 6.
Solve system of linear equations, using matrix method 4x – 3y = 3, 3x – 5y = 7.
Solution:
The given system of equations is 4x – 3y = 3, 3x – 5y = 7
The given system of equations can be wrEtten in the form of AX = B where
A = \(\left[\begin{array}{ll}
4 & -3 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), B = \(\left[\begin{array}{l}
3 \\
7
\end{array}\right]\)
Hence, |A| = -20 + 9 = -11 ≠ 0
So, A is non-singular. Hence, A^-1 exists.

Question 7.
Solve system of linear equations, using matrix method 5x + 2y = 3, 3x + 2y = 5.
Solution:
The given system of equations is 5x + 2y = 3, 3x + 2y = 5.
The given system of equations can be wrEtten in the form of AX = B where
A = \(\left[\begin{array}{ll}
5 & 2 \\
3 & 2
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), B = \(\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)
Hence, |A| = 10 – 6 = 4 ≠ 0
So, A is non-singular. Hence, A^-1 exists.
Now, A^-1 = \(\frac{1}{|\mathrm{~A}|}(\) (adjA) = \(\frac{1}{4}\left[\begin{array}{cc}
2 & -2 \\
-3 & 5
\end{array}\right]\)
⇒ X = A^-1B ⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\frac{1}{4}\left[\begin{array}{cc}
2 & -2 \\
-3 & 5
\end{array}\right]\)
= \(\frac{1}{4}\left[\begin{array}{c}
6-10 \\
-9+25
\end{array}\right]\)
= \(\frac{1}{4}\left[\begin{array}{c}
-4 \\
16
\end{array}\right]\) = \(\left[\begin{array}{c}
-1 \\
4
\end{array}\right]\)

III.

Question 1.
Solve the system of equations using matrix method 2x + y + z = 1, x – 2y – z = \(\frac{3}{2}\), 3y – 5z = 9
Solution:
The given system of equations is 2x + y + z = 1, x – 2y – z = \(\frac{3}{2}\), 3y – 5z = 9
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
2 & 1 & 1 \\
1 & -2 & -1 \\
0 & 3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
\frac{3}{2} \\
9
\end{array}\right]\)
|A| = 2(10 + 3) -1(-5 – 3) + 0 = 2(13) – 1(-x) = 26 + 8 = 34 ≠ 0
So, A is non-singular. Hence. A^-1 exists.
Now, A₁₁ = 13; A₁₂ = 5; A₁₃ = 3
A₂₁ = 8; A₂₂ = -10; A₂₃ = -6
A₃₁ = 1; A₃₂ = 3; A₃₃ = -5
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}(\) (adjA) = \(\frac{1}{34}\left[\begin{array}{ccc}
13 & 8 & 1 \\
5 & -10 & 3 \\
3 & -6 & -5
\end{array}\right]\)
Also AX = B ⇒ X = A^-1B

∴ x = 1, y = \(\frac{1}{2}\) and z = \(\frac{-3}{2}\)

Question 2.
Solve the system of equations using matrix method x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2
Solution:
The given system of equations is x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
⇒ |A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) = 4 + 5 + 1 = 10 ≠ 0
So, A is non-singular. Hence A^-1 exists.
Now, A₁₁ = 4; A₁₂ = -5; A₁₃ = 1
A₂₁ = 2; A₂₂ = 0; A₂₃ = -2
A₃₁ = 2; A₃₂ = 5; A₃₃ = 3

∴ x = 2, y = -1 and z = 1

Question 3.
Solve the system of equations using matrix method 2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3
Solution:
The given system of equations is 2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
5 \\
-4 \\
3
\end{array}\right]\)
|A| = 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6) = 10 + 15 + 15 = 40 ≠ 0
So, A is non-singular. Hence. A^-1 exists.
Now, A₁₁ = 5; A₁₂ = 5; A₁₃ = 5
A₂₁ = 3; A₂₂ = -13; A₂₃ = 11
A₃₁ = 9; A₃₂ = 1; A₃₃ = -7

Hence, x = 1, y = 2 and z = 1

Question 4.
Solve the system of equations using matrix method x – y + 2z = 7, 3x + 4y – 5z = -5, 2x – y + 3z = 12
Solution:
The given system of equations is x – y + 2z = 7, 3x + 4y – 5z = -5, 2x – y + 3z = 12
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
7 \\
-5 \\
12
\end{array}\right]\)
|A| = 1(12 – 5) + 1(9 + 10) + 2(-3 – 8) = 7 + 19 = 4 ≠ 0
So, A is non-singular. Hence. A^-1 exists.
Now, A₁₁ = 7; A₁₂ = -19; A₁₃ = -11
A₂₁ = 1; A₂₂ = -1; A₂₃ = -1
A₃₁ = -3; A₃₂ = 11; A₃₃ = 7
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}(\) (adjA) = \(\frac{1}{4}\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right]\)
∴ X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)=\(\frac{1}{4}\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right]\left[\begin{array}{c}
7 \\
-5 \\
12
\end{array}\right]\)=\(\frac{1}{4}\left[\begin{array}{c}
49-5-36 \\
-133+5+132 \\
-77+5+84
\end{array}\right]\)=\(\frac{1}{4}\left[\begin{array}{c}
8 \\
4 \\
12
\end{array}\right]\)=\(\left[\begin{array}{l}
2 \\
1 \\
3
\end{array}\right]\)
Hence, x = 2, y = 1 and z = 3

Question 5.
If A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\), find A^-1. Using A^-1 solve the system of equations
2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3
Solution:
Given that A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\)
⇒ |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 – 2) = 0 – 6 + 5 = -1 ≠ 0
Now, A₁₁ = 0; A₁₂ = 2; A₁₃ = 1
A₂₁ = -1; A₂₂ = -9; A₂₃ = -5
A₃₁ = 2; A₃₂ = 23; A₃₃ = 13
Thus, A^-1 = \(\frac{1}{|\mathrm{~A}|}(\) (adjA) = –\(\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\)
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\), X = \(\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]\) and B = \(\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]\)
The solution of the system of equations is given by X = A^-1B
⇒ X = A^-1B
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right]=\left[\begin{array}{c}
0-5+6 \\
-22-45+69 \\
-11-25+39
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Hence, x = 1, y = 2 and z = 3

Question 6.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70. Find cost of each item per kg b matrix method.
Solution:
Let the cost of onions, wheat, and rice per kg in ₹ be x, y and z respectively.
Then, the given situation can be represented by a system of equations as
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
The given system of equations can be written in the form of AX = B where
A = \(\left[\begin{array}{lll}
4 & 3 & 2 \\
2 & 4 & 6 \\
6 & 2 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
60 \\
90 \\
70
\end{array}\right]\)
∴ |A| = 4(12 – 12) – 3(6 – 36) + 2(4 – 24) = 0 + 90 – 40 = 50 ≠ 0
Now, A₁₁ = 0; A₁₂ = 30; A₁₃ = -20
A₂₁ = -5; A₂₂ = 0; A₂₃ = 10
A₃₁ = 10; A₃₂ = -20; A₃₃ = 10

Thus, x = 5, y = 8 and z = 8
Hence, the cost of onions is ₹ 5 per kg, the cost of wheat is 8 per kg, and the cost of rice is ₹ 8 per kg

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4d

I.

Question 1.
Find the adjoint of the matrix \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) ⇒ A₁₁ = 4; A₁₂ = -3; A₂₁ = -2; A₂₂ = 1
∴ adjA = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]\)

Question 2.
Find the adjoint of the matrix \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 3 & 5 \\
-2 & 0 & 1
\end{array}\right]\)
Solution:

Question 3.
Find the inverse of the matrix \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\). Then|A| = (2 × 3) – (-2 × 4) = 6 – (-8) = 14
Now, A₁₁ = 3; A₁₂ = -4
A₂₁ = 2; A₂₂ = 2
Hence, adjA = \(\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|\mathrm{A}|}\) adjA = \(\frac{1}{14}\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)

Question 4.
Find the inverse of the matrix \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Then, |A| = (-1 × 2) – (5 × -3) = -2 + 15 = 13
Now, A₁₁ = 2; A₁₂ = 3
A₂₁ = -5; A₂₂ = -1
Hence, adjA = \(\left[\begin{array}{ll}
2 & -5 \\
3 & -1
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|\mathrm{A}|}\) adjA = \(\frac{1}{13}\left[\begin{array}{cc}
2 & -5 \\
3 & -1
\end{array}\right]\)

II.

Question 1.
If A = \(\left[\begin{array}{cc}
2 & 3 \\
-4 & -6
\end{array}\right]\), Verify A(adj A) = (adj A) A = |A| I
Solution:

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\), Verify A(adj A) = (adj A) A = |A| I
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\)
Then, |A| = 1(0 – 0) + 1(9 + 2) + 2(0 – 0) = 11
Also, |A|I = 11 \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]\)
A₁₁ = 0; A₁₂ = -11; A₁₃ = 0
A₂₁ = 3; A₂₂ = 1; A₂₃ = -1
A₃₁ = 2; A₃₂ = 8; A₃₃ = 3

Question 3.
Find the inverse of the matrix \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Then, |A| = 1(10 -0) – 2(0 – 0) + 3(0 – 0) = 10 ≠ 0
So, A is non singular. hence A exists.
A₁₁ = 10; A₁₂ = 0; A₁₃ = 0
A₂₁ = -10; A₂₂ = 5; A₂₃ = 0
A₃₁ = 2; A₃₂ = -4; A₃₃ = 2
Hence, adj A = \(\left[\begin{array}{ccc}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|A|}\) adjA = \(\frac{1}{10}\left[\begin{array}{ccc}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right]\)

Question 4.
Find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Then, |A| = 1(-3 – 0) – 0 + 0 = -3 ≠ 0
So, A is non singular. hence A^-1 exists.
A₁₁ = -3; A₁₂ = 3; A₁₃ = -9
A₂₁ = 0; A₂₂ = -1; A₂₃ = -2
A₃₁ = 0; A₃₂ = 0; A₃₃ = 3
Hence, adj A = \(\left[\begin{array}{ccc}
-3 & 0 & 0 \\
3 & -1 & 0 \\
-9 & -2 & 3
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|A|}\) adjA = \(\frac{-1}{3}\left[\begin{array}{ccc}
-3 & 0 & 0 \\
3 & -1 & 0 \\
-9 & -2 & 3
\end{array}\right]\)

Question 5.
Find the inverse of the matrix \(\left[\begin{array}{ccc}
2 & 1 & 3 \\
4 & -1 & 0 \\
-7 & 2 & 1
\end{array}\right]\) if it exists
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 1 & 3 \\
4 & -1 & 0 \\
-7 & 2 & 1
\end{array}\right]\)
Then, |A| = 2(-1 – 0) – 1(4 – 0) + 3(8 – 7)
= 2(-1) -1(4) + 3(1) ≠ 0
So, A is non singular. hence A^-1 exists.
A₁₁ = -1; A₁₂ = -4; A₁₃ = 1
A₂₁ = 5; A₂₂ = 23; A₂₃ = -11
A₃₁ = 3; A₃₂ = 12; A₃₃ = -6
Hence, adj A = \(\left[\begin{array}{ccc}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|A|}\) adjA = \(\frac{-1}{3}\left[\begin{array}{ccc}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right]\)

Question 6.
Find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\)
Then, |A| = 1(8 – 6) – 0 + 3(3 – 4)
= 2 – 3 = -1 ≠ 0
So, A is non singular. hence A^-1 exists.
A₁₁ = 2; A₁₂ = -9; A₁₃ = -6
A₂₁ = 0; A₂₂ = -2; A₂₃ = -1
A₃₁ = -1; A₃₂ = 3; A₃₃ = 2
Hence, adj A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|A|}\) adjA = \(-1\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)

Question 7.
Find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]\)
Then, |A| = 1(-cos²α – sin²α) = -(cos²α + sin²α) = -1
So, A is non singular. hence A^-1 exists.
A₁₁ = -cos²α – sin²α = -1; A₁₂ = 0; A₁₃ = 0
A₂₁ = 0; A₂₂ = -cos α; A₂₃ = -sin α
A₃₁ = 0; A₃₂ = -sin α; A₃₃ = cos α
Hence, adj A = \(\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|A|}\) adjA = \(-1\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]\)

Question 8.
Let A = \(\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right]\). Verify that (AB)^-1 = B^-1A^-1.
Solution:
Let A = \(\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\)
Then, |A| = 15 – 14 = 1
Now, A₁₁ = 5; A₁₂ = -2; A₂₁ = -7; A₂₂ = 3;
Hence, adj A = \(\left[\begin{array}{cc}
5 & -7 \\
-2 & 3
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|A|}\) adjA = \(\left[\begin{array}{cc}
5 & -7 \\
-2 & 3
\end{array}\right]\)
Now, Let B = \(\left[\begin{array}{cc}
6 & 8 \\
7 & 9
\end{array}\right]\), Then, |B| = 54 – 56 = -2
Now, Now, A₁₁ = 9; A₁₂ = -8; A₂₂ = 6
Hence, adj B = \(\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right]\)
∴ B^-1 = \(\frac{1}{|B|}\) adjB = \(-\frac{1}{2}\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right]=\left[\begin{array}{cc}
-\frac{9}{2} & 4 \\
\frac{7}{2} & -3
\end{array}\right]\)

Question 9.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), show that A² – 5A + 7I = 0. Hence find A^-1
Solution:

Thus A² – 5A + 7I = 0
⇒ A.A – 5A = -7I
⇒ AA(A^-1) – 5AA<sup-1 = -7IA^-1 [post-multip1ying by A^-1 as |A| ≠ 0]
⇒ A(AA^-1) – 5I = -7A^-1 AI – 5I = -7A^-1
⇒ A^-1 = –\(\frac{1}{7}\) (A – 5I) = A^-1 = \(\frac{1}{7}\) (5I – A) .
⇒ A^-1 = \(\frac{1}{7}\left[\left(\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right)-\left(\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right)\right]\)
⇒ A^-1 = \(\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)

Question 10.
For the matrix A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\), find the numbers a and b such that A² + aA + bI = 0
Solution:
Let A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\)
|A| = 3×1—2×1=1
A² = A.A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
9+2 & 6+2 \\
3+1 & 2+1
\end{array}\right]\) = \(\left[\begin{array}{cc}
11 & 8 \\
4 & 3
\end{array}\right]\)
Now A² + aA + bI = 0
⇒ (A.A)A^-1 + aA.A^-1 + bIA^-1 = 0 [post. multiplying by A^-1 as |A| ≠ o]
⇒ A(AA^-1) + aI + b(IA^-1) = 0
⇒ AI + aI + bA^-1 = 0 ⇒ A + aI = -bA^-1
⇒ A^-1 = –\(\frac{1}{b}\)(A + aI) …………… (1)
A^-1 = \(\frac{1}{|A|}\) adjA = \(\frac{1}{1}\left[\begin{array}{cc}
1 & -2 \\
-1 & 3
\end{array}\right]=\left[\begin{array}{cc}
1 & -2 \\
-1 & 3
\end{array}\right]\) …….. (2)
From (1) and(2), we have,
⇒ \(\left(\begin{array}{cc}
1 & -2 \\
-1 & 3
\end{array}\right)=\frac{1}{b}\left[\left(\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right)+\left(\begin{array}{cc}
a & 0 \\
0 & a
\end{array}\right)\right]\)
⇒ \(\left[\begin{array}{cc}
1 & -2 \\
-1 & 3
\end{array}\right]=-\frac{1}{b}\left[\begin{array}{cc}
3+a & 2 \\
1 & 1+a
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
1 & -2 \\
-1 & 3
\end{array}\right]=\left[\begin{array}{cc}
\frac{-3-a}{b} & -\frac{2}{b} \\
-\frac{1}{b} & \frac{-1-a}{b}
\end{array}\right]\)
Now, comparing the corresponding elements of the two matrices, we have:
–\(\frac{1}{b}\) = -1 ⇒ b = 1
Also, \(\frac{-3-a}{b}\) = 1⇒ -3 – a = 1 ⇒ a = -4
∴ a = -4, b = 1

III.

Question 1.
For the matrix A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\). Show that A³ – 6A² + 5A + 11 I = 0. Hence, find A^-1
Solution:


Thus, A³ – 6A² + 5A + 11 I = 0
Now A³ – 6A² + 5A + 11 I = 0
⇒ (AAA)A^-1 – 6(AA)A^-1 + 5AA^-1 + 11 IA^-1 = 0 [Post-multiplying by A^-1 as |A| ≠ 0]
⇒ AA(AA^-1) – 6A(AA^-1) + 5(AA^-1) = -11(IA^-1)
⇒ A² – 6A + 5I = -11A^-1
⇒ A^-1 = –\(\frac{1}{11}\)(A^-1 – 6A + 5I) ………….. (1)

Question 2.
If A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\), Verify that A³ – 6A² + 9A – 4I = 0 and hence find A^-1
Solution:


A³ – 6A² + 9A – 4I = 0
⇒ (AAA)A^-1– 6(AA)A^-1 + 9AA^-1 – 4IA^-1 = 0 [Post multipIying by A^-1 as |A| ≠ 0]
⇒ AA(AA^-1) – 6A(AA^-1) + 9(AA^-1) = 4(IA^-1)
⇒ AAI – 6AI + 9I = 4A^-1
⇒ A² – 6A + 9I = 4A^-1
⇒ A^-1 = \(\frac{1}{4}\) (A² – 6A + 9I) ………………. (1)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7b

I.

Question 1.
Find integral of \(\frac{2 x}{1+x^2}\)
Solution:
Put 1 + x² = t ⇒ 2xdx = dt
∴ ∫\(\frac{2 x}{1+x^2}\) dx = ∫\(\frac{d t}{t}\) = ∫\(\frac{1}{t}\) dt = log |t| + c = log |1 + x²| + c = log(1 + x²) + c. [∵ t = 1 + x²]

Question 2.
Find integral of \(\frac{(\log x)^2}{x}\)
Solution:
Put log x = t ⇒ \(\frac{1}{x}\) dx = dt ⇒ \(\frac{dx}{x}\) = dt
∴ ∫\(\frac{(\log x)^2}{x}\) dx = ∫(log x)²\(\left(\frac{d x}{x}\right)\) = ∫t² dt = \(\frac{t^3}{3}\) + c = \(\frac{1}{3}\)(log x)³ + c [∵ t = log x]

Question 3.
Find integral of \(\frac{1}{x+x \log x}\)
Solution:
Put 1 + log x = t ⇒ \(\frac{d x}{x}\) = dx
∴ \(\int \frac{1}{x+x \log x} d x=\int \frac{1}{1+\log x}\left(\frac{d x}{x}\right)\) = ∫\(\frac{1}{t}\) dt = log |t| + c = log |1 + log x| + c [∵ t = 1 + log x]

Question 4.
Find integral of sinx sin(cos x)
Solution:
We have to find ∫sin x sin(cos x) dx = -∫sin(cos x)(-sin x)dx
Put cos x = t ⇒ -sin x dx = dt
∴ ∫sinx sin(cos x)dx = -∫sin(cox x)(-sin x dx)
= -∫sint dt = -(-cos t) + c = cos t + c = cos(cos x) + c

Question 5.
Find integral of sin(ax + b) ocs(ax + b)
Solution:
∫sin(ax + b) cos(ax + b) dx = \(\frac{1}{2}\)∫2sin(ax + b)cos(ax + b)dx
= \(\frac{1}{2}\)∫sin2(ax + b) dx = \(\frac{1}{2}\)∫sin(2ax + 2b) dx [∵ 2 sin A cos A = sin 2A]
= \(\frac{1}{2} \frac{[-\cos (2 a x+2 b)]}{2 a}\) + c = \(\frac{-1}{4 a}\)cos2(ax + b) + c. [∵ sin(ax + b) dx = \(-\frac{1}{a}\)cos(ax + b) + c]

Question 6.
Find integral of \(\sqrt{a x}+b\)
Solution:

Question 7.
Find integral of \(x \sqrt{x+2}\)
Solution:
∫\(x \sqrt{x+2}\) dx = ∫\(x \sqrt{x+2}\)dx = ∫((x + 2) – 2)\(\sqrt{x+2}\) dx

Question 8.
Find integral of \(x \sqrt{1+2 x^2}\)
Solution:
Let I = ∫\(x \sqrt{1+2 x^2}\) dx = \(\frac{1}{4}\)∫\(\sqrt{1+2 x^2}\)(4xdx) ………….(i) [∵ \(\frac{d}{d x}\)(1 + 2x²) = 0 + 2.2x = 4x]
Put 1 + 2x² = t ⇒ 4xdx = dt
∴ From (i), I = \(\frac{1}{4}\)∫\(\sqrt{t}\)dt = \(\frac{1}{4}\)∫t^1/2 dt
[∵ t = 1 + 2x²]

Question 9.
Find integral of (4x + 2)\(\sqrt{x^2+x}+1\)
Solution:
Let I = ∫(4x + 2)\(\sqrt{x^2+x}+1\) dx = ∫2(2x + 1)\(\sqrt{x^2+x+1} d x\)
= ∫2\(\sqrt{x^2+x+1}\)(2x + 1) dx ……(i)
Put x² + x + 1 = t ⇒ (2x + 1)dx = dt
From (i), I = ∫2\(\sqrt{t}\) dt = 2∫t^1/2 dt
= \(2 \frac{t^{3 / 2}}{\frac{3}{2}}+c=\frac{4}{3} t^{3 / 2}+c=\frac{4}{3}\left(x^2+x+1\right)^{3 / 2}+c\) [∵ t = x² + x + 1]

Question 10.
Find integral of \(\frac{1}{x-\sqrt{x}}\)
Solution:
Let I = \(\int \frac{1}{x-\sqrt{x}} d x\) …………….(i)
Put \(\sqrt{\text { Linear }}\) = t, i.e., \(\sqrt{x}\) = t ⇒ x = t² ⇒ dx = 2t dt
∴ From (i), I = \(\int \frac{1}{t^2-t}\)2tdt = 2∫\(\frac{t}{t(t-1)}\) dt
= 2\(\int \frac{1}{t-1}\) dt = 2log |t – 1| + c = 2log \(|\sqrt{x}-1|\) + c [∵ \(\int \frac{1}{a x+b} d x=\frac{1}{a}\) log |ax + b|]

Question 11.
Find integral of \(\frac{x}{\sqrt{x+4}}\), x > 0
Solution:
Let I = \(\int \frac{x}{\sqrt{x+4}} d x\) ………….(i)

Question 12.
Find integral of (x³ – 1)^1/3x⁵
Solution:
Let I = ∫(x³ – 1)^1/3x⁵ dx = ∫(x³ – 1)^1/3x³x² dx
= \(\frac{1}{3}\)∫(x³ – 1)^1/3x³(3x² dx) …..(i) [∵ \(\frac{d}{dx}\)(x³ – 1) = 3x²]
Put x³ – 1 = t ⇒ x³ = t + 1 ⇒ 3x² = \(\frac{dt}{dx}\) ⇒ 3x² dx = dt

Question 13.
Find integral of \(\frac{x^2}{\left(2+3 x^3\right)^3}\)
Solution:
Let I = \(\) …….(i) [\(\frac{d}{d x}\int \frac{x^2}{\left(2+3 x^3\right)^3} d x=\frac{1}{9} \int \frac{9 x^2}{\left(2+3 x^3\right)^3} d x\)(2 + 3x³) = 9x²]
Put 2 + 3x³ = t ⇒ 9x² dx = dt
∴ Fron (i), I = \(\frac{1}{9} \int t^{-3} d t=\frac{1}{9}\left(\frac{t^{-2}}{-2}\right)+c=\frac{-1}{18 t^2}+c=\frac{-1}{18\left(2+3 x^3\right)^2}+c\) [∵ t = 2 + 3x³]

Question 14.
Find integral of \(\frac{1}{x(\log x)^m}\), x > 0, m ≠ 1
Solution:
Let I = \(\int \frac{1}{x(\log x)^m} d x(x>0) \Rightarrow I=\int \frac{\frac{I}{x} d x}{(\log x)^m}\) ……(i)
Put log x = t ⇒ \(\frac{d x}{x}\) = dt
From (i), I = \(\int \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{m}}}=\int \mathrm{t}^{-\mathrm{m}} \mathrm{dt}=\frac{\mathrm{t}^{-\mathrm{m}+1}}{-\mathrm{m}+1}+\mathrm{c}\) (Assuming m ≠ 1)
= \(\frac{(\log x)^{1-m}}{1-m}\) + c [∵ t = log x]

Question 15.
Find integral of \(\frac{x}{9-4 x^2}\)
Solution:
Let I = \(\int \frac{x}{9-4 x^2} d x=\frac{-1}{8} \int \frac{-8 x}{9-4 x^2} d x\) ……….(i) [∵ \(\frac{d}{d x}\)(9 – 4x²) = -8x]
Put 9 – 4x² = t ⇒ -8xdx = dt [∵ \(\) \int \frac{f^{\prime}(x)}{f(x)} d x= logf(x) + c]

Question 16.
Find integral of e^2x+3
Solution:
Put 2x + 3 = t ⇒ 2 dx = dt [∵ ∫e^ax+bdx = \(\frac{1}{a}\)e^{ax + b} + c]
∴ ∫e[sup]2x+3[/sup]dx = \(\frac{1}{2}\)∫e^t dt = \(\frac{1}{2}\)(e^t) + C = \(\frac{1}{2}\)e^(2x+3) + C

Question 17.
Find integral of \(\frac{x}{e^{x^2}}\)
Solution:
Put x² = t ⇒ 2xdx = dt
∴ \(\int \frac{x}{e^{x^2}} d x=\frac{1}{2} \int \frac{1}{e^t} d t=\frac{1}{2} \int e^{-t} d t=\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+C=-\frac{1}{2} e^{-x^2}+C=\frac{-1}{2 e^{x^2}}+C\)

Question 18.
Find integral of \(\frac{e^{\tan -x}}{1+x^2}\)
Solution:
Put tan^-1 x = t ⇒ \(\frac{1}{1+x^2}\)dx = dt ∴ \(\int \frac{e^{\tan ^{-1} x}}{1+x^2}\) dx = ∫e^t dt = e^t + C = e^tan-1x + C

Question 19.
Find integral of \(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Solution:
\(\frac{e^{2 x}-1}{e^{2 x}+1}\) Dividing Nr. and Dr. by e^x we get \(\frac{\frac{e^{2 x}-1}{e^x}}{\frac{e^{2 x}+1}{e^x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\) [\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]
Put e^x + e^-x = t ⇒ (e^x – e^-x)dx = dt
⇒ \(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{e^x-e^{-x}}{e^x+e^{-x}} d x=\int \frac{d t}{t}\) = log|t| + C = log |e^x + e^-x| + C

Question 20.
Find integral of \(\)
Solution:
Put e^2x + e^-2x = t ⇒ (2e^2x – 2e^-2x) dx = dt ⇒ (2e^2x – 2e^-2x) dx = dt

Question 21.
Find integral of tan²(2x – 3)
Solution:
We have tan²(2x – 3) = sec²(2x – 3) – 1
Put 2x – 3 = t ⇒ 2 dx = dt
⇒ \(\int \tan ^2(2 x-3) d x=\int\left[\sec ^2(2 x-3)-1\right] d x\)
= \(\frac{1}{2} \int \sec ^2 \mathrm{tdt}-\int 1 \mathrm{dx}=\frac{1}{2} \tan \mathrm{t}-\mathrm{x}+\mathrm{C}=\frac{1}{2} \tan (2 \mathrm{x}-3)-\mathrm{x}+\mathrm{C}\)

Question 22.
Find integral of sec²(7 – 4x)
Solution:
Put 7 – 4x = t ⇒ -4 dx = dt
∴ ∫sec²(7 – 4x)dx = \(\frac{-1}{4}\)∫sec² tdt = \(\frac{-1}{4}\)(tan t) + C = \(\frac{-1}{4}\)tan(7 – 4x) + C

Question 23.
Find integral of \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\)
Solution:

Question 24.
Find integral of \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Solution:

Question 25.
Find integral of \(\frac{1}{\cos ^2 x(1-\tan x)^2}\)
Solution:
We have \(\frac{1}{\cos ^2 x(1-\tan x)^2}=\frac{\sec ^2 x}{(1-\tan x)^2}\)
Put (1 – tan x) = t ⇒ -sec² xdx = dt
∴ \(\int \frac{\sec ^2 x}{(1-\tan x)^2} d x=\int \frac{-d t}{t^2}=-\int t^{-2} d t=\frac{1}{t}+C=\frac{1}{(1-\tan x)}+C\)

Question 26.
Find integral of \(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Solution:
Put \(\sqrt{x}\) = t ⇒ \(\frac{1}{2 \sqrt{x}}\)dx = dt ⇒ \(\int \frac{\cos \sqrt{x}}{\sqrt{x}}\) = 2∫costdt = 2 sin t + C = 2 sin\(\sqrt{x}\) + C

Question 27.
Find integral of \(\sqrt{\sin 2 x} \cos 2 x\)
Solution:
Put sin 2x = t ⇒ 2 cos 2x dx = dt
∴ \(\int \sqrt{\sin 2 x} \cos 2 x d x=\frac{1}{2} \int \sqrt{t} d t=\frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C=\frac{1}{3} t^{\frac{3}{2}}+C=\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C\)

Question 28.
Find integral of \(\frac{\cos x}{\sqrt{1+\sin x}}\)
Solution:
Put sin 2x = t ⇒ cosxdx = dt
∴ \(\int \frac{\cos x}{\sqrt{1+\sin x}} d x=\int \frac{d t}{\sqrt{t}}=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C=2 \sqrt{t}+C=2 \sqrt{1+\sin x}+C\)

Question 29.
Find integral of cotx logsin x
Solution:
Put logsin x = t ⇒ \(\frac{1}{\sin x}\) cos xdx = dt ∴ cot x dx = dt
⇒ ∫cotx log sin xdx = ∫ tdt = \(\frac{t^2}{2}\) + C = \(\)(log sin x)² + \(\frac{1}{2}\)

Question 30.
Find integral of \(\frac{\sin x}{1+\cos x}\)
Solution:
Put 1 + cosx = t ⇒ -sinx dx = dt
⇒ ∫\(\frac{\sin x}{1+\cos x}\) dx = ∫\(-\frac{\mathrm{dt}}{\mathrm{t}}\) = – log |t | + C = -log|1 + cos x| + C

Question 31.
Find integral of \(\frac{\sin x}{(1+\cos x)^2}\)
Solution:
Put 1 + cosx = t ⇒ -sinx dx = dt
∴ \(\int \frac{\sin x}{(1+\cos x)^2} d x=\int-\frac{d t}{t^2}=-\int t^{-2} d t=\frac{1}{t}+C=\frac{1}{(1+\cos x)}+C\)

Question 32.
Find integral of \(\frac{(1+\log x)^2}{x}\)
Solution:
Put 1 + log x = t ⇒ \(\frac{1}{x}\)dx = dt ∴ \(\int \frac{(1+\log x)^2}{x} d x=\int t^2 d t=\frac{t^3}{3}+C=\frac{(1+\log x)^3}{3}+C\)

Question 33.
Find integral of \(\frac{(x+1)(x+\log x)^2}{x}\)
Solution:

Question 34.
Find integral of \(\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}\)
Solution:
Put x³ = t ⇒ 4x³dx = dt

Question 35.
Find integral of \(\frac{x^3}{\sqrt{1-x^8}}\)
Solution:
Put x³ = t ⇒ 4x³dx = dt
∴ \(\int \frac{x^3}{\sqrt{1-x^8}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^2}}=\frac{1}{4} \sin ^{-1} t+C=\frac{1}{4} \sin ^{-1}\left(x^4\right)+C\)

Question 36.
Find integral of cos³x e^{logsin x}
Solution:
cos³ xe^logsinx = cos³ x sin x
Let cos x = t ⇒ -sin xdx = dt
∴ \(\int \cos ^3 x e^{\log \sin x} d x=\int \cos ^3 x \sin x d x=-\int t^3 d t=-\frac{t^4}{4}+C=-\frac{\cos ^4 x}{4}+C\)

Question 37.
Find integral of e^{3log x}(x⁴ + 1)^-1
Solution:
e^{3log x}(x⁴ + 1)^-1 = e^{log x3}(x⁴ + 1)^-1 = \(\frac{x^3}{\left(x^4+1\right)}\) [∵ \(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]
Let x⁴ + 1 = t ⇒ 4x³ dx = dt
⇒ \(\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=\int \frac{x^3}{\left(x^4+1\right)} d x=\frac{1}{4} \int \frac{d t}{t}=\frac{1}{4} \log |t|+C=\frac{1}{4} \log \left|x^4+1\right|+C\)

Question 38.
Find integral of f'(ax + b)[f(ax + b)]ⁿ
Solution:
Given integral is f'(ax + b)[f(ax + b)]ⁿ
Put f(ax + b) = t ⇒ af'(ax + b) dx = dt
⇒ f'(ax + b)[f(ax + b)]ⁿ dx = \(\frac{1}{a} \int t^n d t=\frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]=\frac{1}{a(n+1)}(f(a x+b))^{n+1}+C\)

II.

Question 1.
Find integral of \(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\)
Solution:
Given integrand is \(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\). Multiplying and dividing by x^-3, we get

Question 2.
Find integral of \(\frac{1}{1+\cot x}\)
Solution:

Question 3.
Find integral of \(\frac{1}{1-\tan x}\)
Solution:

Question 4.
Find integral of \(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4c

I.

Question 1.
Write Minors and Cofactors of the elements of \(\left|\begin{array}{cc}
2 & -4 \\
0 & 3
\end{array}\right|\)
Solution:
The given determinant is \(\left|\begin{array}{cc}
2 & -4 \\
0 & 3
\end{array}\right|\)
Minor means little determinant
Minor of the element a_jj is M_jj
Here a₁₁ = 1. So M₁₁ = Minor of a₁₁ = 3
M₁₁ = Minor of the element a₁₁ = 3; M₁₂ = Minor of the element a₁₂ = 0;
M₂₁ = Minor of the element a₂₁ = -4; M₂₂ = Minor of the element a₂₂ = 2;
Now, cofactor of a_ij is A_ij = (-1)^i+j M_ij
A₁₁ =(-1)^{1 + 1}(3) = 3;
A₁₂ =(-1)^{1 + 2} (0) = 0;
A₂₁ = (-1)^{2 + 1} (-4) = 4;
A₂₂ = (-1)^{2 + 2} (2) = 2

Question 2.
Write Minors and Cofactors of the elements of \(\left|\begin{array}{ll}
a & c \\
b & d
\end{array}\right|\)
Solution:
The given determinant is \(\left|\begin{array}{ll}
a & c \\
b & d
\end{array}\right|\)
Minor of the element a_jj is M_jj
M₁₁ = Minor of the element a₁₁ = d; M₁₂ = Minor of the element a₁₂ = b;
M₂₁ = Minor of the element a₂₁ = c; M₂₂ = Minor of the element a₂₂ = a;
Now, cofactor of a_jj is A_jj = (-1)^{i + j} M_jj
A₁₁ = (-1)^{1 +1} (d) = d; A₁₂ = (-1)¹⁺² (b) = -b
A₂₁ = (-1 )^{2 + 1} (c) = -c; A₂₂ = (-1)^{2 + 2} (a) = a

Question 3.
Using Cofactors of elements of second row, evaluate ∆ = \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
Solution:
The given determinant is \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
Minor of the element a_jj is M_jj
M₂₁ = Minor of the element a₂₁ = \(\left|\begin{array}{ll}
3 & 8 \\
2 & 3
\end{array}\right|\) = (3 × 3) – (8 × 2) = -7
A₂₁ = (-1)²⁺¹ (-7) = 7
M₂₂ = Minor of the element a₂₂ = \(\left|\begin{array}{ll}
5 & 8 \\
1 & 3
\end{array}\right|\) = (5 × 3) – (8 × 1) = 15 – 8 = 7
A₂₂ = (-1)²⁺² (7) = 7
M₂₃ Minor of the element a₂₃ = \(\left|\begin{array}{ll}
5 & 3 \\
1 & 2
\end{array}\right|\) = (5 × 2) – (3 × 1) = 10 – 3 = 7
A₂₃ = (-1)^{2 + 3} (7) = -7
We know that ∆ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
∆ = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃
= 2(7) + 0(7) + 1(-7) = 14 – 7 = 7

Question 4.
Using Cofactors of elements of third column, evaluate ∆ = \(\left|\begin{array}{ccc}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|\)
Solution:
The given determinant is \(\left|\begin{array}{ccc}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|\)
M₁₃ = \(\left|\begin{array}{ll}
1 & y \\
1 & z
\end{array}\right|\) (1 × z) – (1 × y) = z – y; A₁₃ = (-1)⁴(z – y) = zy
M₂₃ =\(\left|\begin{array}{ll}
1 & x \\
1 & z
\end{array}\right|\) = 1 × Z – x × 1 = z – x; A₂₃ = (-1)⁵(z – x) = -(z – x) = x – z
M₃₃ = \(\left|\begin{array}{ll}
1 & x \\
1 & y
\end{array}\right|\) =1 × y – x × 1 = y – x. A₃₃ = (-1)⁶(y – x) = y – x
We know that ∆ is equal to the sum of the product of the elements of the second row
with their corresponding cofactors.
∆ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃ .
= yz(z – y) + zx(x – z) + xy(y – x) = yz² – y²z + x²z – xz² + xy² – x²y
=(x²z – y²z) + (yz² – xz²) + (xy² – x²y) = z(x² – y²) + z²(y – x) + xy(y – x)
= z(x – y)(x + y) + z² (y – x) + xy(y – x) = (x – y)[zx – z² + zy – xy]
= (x – y)[z(x – z) + y(z – x)} =(x – y)(z – x)[-z + y]
= (x – y)(y – z)(z – x)
∴ ∆ = (x – y)(y – z)(z – x)

II.

Question 1.
Find minors and cofactors of the elements of the determinant \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
Solution:
The given determinant is \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\). Minor of the element a_ij is M_ij
Minor of the elements a₁₁ is M₁₁ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1; M₁₂ = \(\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|\) = 0; M₁₃ = \(\left|\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right|\) = 0
Similarly, M₂₁ = \(\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|\) = 0; M₂₂ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1; M₂₃ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0
M₃₁ = \(\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|\) = 0; M₃₁ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0; M₃₃ = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1
Now, cofactor of a_ij is A_ij = (-1)^{i + j} M_ij
A₁₁ = (-1)^{1 + 1} (1) = 1; A₁₂ =(-1)^{1 + 2}(0) = 0; A₁₃ =(-1)^{1 + 3}(0) = 0;
A₂₁ = (-1)^{2 + 1} (0) = 0; A₂₂ = (-1)^{2 + 2}(1) = 1; A₂₃ = (-1)^{2 + 3}(0) = 0
A₃₁ = (-1)^{3 + 1} (0) = 0; A₃₂ = (-1)^{3 + 2}(0) = 0; A₃₃ = (-1)^{3 + 3} (1) = 1

Question 2.
Find minors and cofactors of the elements of the determinant \(\left|\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right|\)
Solution:
The given determinant is \(\left|\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right|\). Minor of the element a_ij is M_ij
Minor of the element a₁₁ is M₁₁ = \(\left|\begin{array}{cc}
5 & -1 \\
1 & 2
\end{array}\right|\) = (5 × 2) – (1 × -1) = 10 + 1 = 11
M₁₂ = \(\left|\begin{array}{cc}
3 & -1 \\
0 & 2
\end{array}\right|\) = (3 × 2) – (1 × 0) = 6;
M₁₃ = \(\left|\begin{array}{cc}
3 & 5 \\
0 & 1
\end{array}\right|\) = (3 × 1) – (5 × 0) = 3
M₂₁ = \(\left|\begin{array}{cc}
0 & 4 \\
1 & 2
\end{array}\right|\) (0 × 2) – (4 × 1) = 4; M₂₂ = \(\left|\begin{array}{cc}
1 & 4 \\
0 & 2
\end{array}\right|\) = (1 × 2) – (4 × 0) = 2
M₂₃ = \(\left|\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right|\) = (1 × 1) – (0 × 0) = 1;
M₃₁ = \(\left|\begin{array}{cc}
0 & 4 \\
5 & -1
\end{array}\right|\) = (0 × -1)- (4 × 5) = -20. M₃₂ = \(\left|\begin{array}{cc}
1 & 4 \\
3 & -1
\end{array}\right|\) =(1 × -1) – (4 × 3) = -13
M₃₃ = \(\left|\begin{array}{cc}
1 & 0 \\
3 & 5
\end{array}\right|\) = (1 × 5) – (0 × 3) = 5
Now, cofactor of a_ij is A_ij = (-1)sup>i + j M_ij
A₁₁ =(-1)^{1 + 1}(11) = 11; A₁₂ = (-1)^{1 + 2} (6) = -6; A₁₃ = (-1)^{1 + 3} (3) = 3
A₂₁ = (-1)^{2 + 1} (-4) = 4; A₂₂ = (-1)^{2 + 2} (2) = 2; A₂₃ = (-1)^{2 + 3} (1) = -1
A₃₁ =(-1)^{3 + 1} (-20) = -20; A₃₂ = (-1)^{3 + 2} (13)=13; A₃₃ = (-1)^{3 + 3} (5) = 5

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4b

I.

Question 1.
Find area of the triangle with vertices (1, 0), (6, 0), (4, 3)
Solution:
Area of the triangle with vertices A(x₁, y₁) = (1, 0), B(x₂, y₂) = (6,0), C(x₃, y₃) = (4, 3) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
1 & 0 & 1 \\
6 & 0 & 1 \\
4 & 3 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[1(0 – 3) – 0(6 – 4) + 1(18 – 0)]|
= \(\frac{1}{2}\)|[-3 + 18]|
= \(\frac{1}{2}\)[15] = \(\frac{15}{2}\) Sq.units

Question 2.
Find area of the triangle with vertices (2, 7), (1, 1), (10,8)
Solution:
Area of the triangle with vertices A(x₁, y₁) (2, 7), B(x₂, y₂) (1, 1), C(x₃, y₃)= (10, 8) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[2(1 – 8) – 7(1 – 10) + 1(8 – 10)]|
= \(\frac{1}{2}\) |[2(-7) – 7(-9) + 1(-2)]|
= \(\frac{1}{2}\)|-14 + 63 – 2| = \(\frac{1}{2}\)[47]
= \(\frac{47}{2}\) Sq.units

Question 3.
Find area of the triangle with vertices (-2, -3), (3, 2), (-1, -8)
Solution:
Area of the triangle with vertices A(x₁, y₁) (-2, -3), B(x₂, y₂) (3, 2), C(x₃, y₃)= (-1, -8) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)]|
= \(\frac{1}{2}\) |[-2(10) + 3(4) + 1(-22)]|
= \(\frac{1}{2}\)|[-20 + 12 – 22]| = \(\frac{1}{2}\)|-30|
= 15 Sq.units
Hence, area of the triangle is 15 Sq. units

Question 4.
Show that points A (a, h + c), B (b, c + a), C (c, a + b) are collinear.
Answer:
Area of the triangle with vertices A(x₁, y₁) = (a, b + c), B(x₂, y₂) (b,c + a), C (x₃, y₃) (c, a + b) is given by (We apply row operations to simplify easily)
∆ = \(=\frac{1}{2}\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
a & b+c & 1 \\
b-a & a-b & 0 \\
c-a & a-c & 0
\end{array}\right| \)
R₂ → R₂ – R₁
R₃ → R₃ – R₁
= \(\frac{1}{2}\)(a – b) (c – a) \(\left|\begin{array}{ccc}
a & b+c & 1 \\
-1 & 1 & 0 \\
1 & -1 & 0
\end{array}\right|\)
= \(\frac{1}{2}\)(a – b)(c – a)| (-1)(-1) – (1)(1)| = \(\frac{1}{2}\)(a – b)(c – a)(0) = 0
Thus, the area of the triangle formed by the given points is zero.
Hence, the given 3 points are collinear.

Question 5.
Find the values of k if area of triangle is 4 sq. units and vertices are (k, 0). (4, 0), (0, 2).
Solution:
Area of ∆ ABC with vertices A(x₁, y₁) (k, 0), B(x₂, y₂) = (4, 0), C(x₃, y₃) = (0, 2) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
\mathrm{x}_1 & \mathrm{y}_1 & 1 \\
\mathrm{x}_2 & \mathrm{y}_2 & 1 \\
\mathrm{x}_3 & \mathrm{y}_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
\mathrm{k} & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) |[k(0 – 2) – 0(4 – 0) + 1(8 – 0)]
= \(\frac{1}{2}\)|[-2k + 8]| = |-k + 4|
∴ |-k + 4| = 4 ⇒ -k + 4 = ± 4
-k + 4 = 4 ⇒ k = 4 + 4 = 8
when -k + 4 = 4 ⇒ k = 0
∴ k = 0, 8

Question 6.
Find values of k if area of triangle is 4 sq. units and vertices are (-2, 0), (0, 4), (0, k)
Solution:
Area of ∆ABC with vertices A(x₁, y₁) = (-2, 0), B(x₂, y₂) = (0, 4), C(x₃, y₃) = (0, k) is
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
\mathrm{x}_1 & \mathrm{y}_1 & 1 \\
\mathrm{x}_2 & \mathrm{y}_2 & 1 \\
\mathrm{x}_3 & \mathrm{y}_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
-2 & 0 & 1 \\
0 & 4 & 1 \\
0 & \mathrm{k} & 1
\end{array}\right|\)
= \(\frac{1}{2}\)|[-2(4 – k)]| = |k – 4|
∴ |k – 4| = 4 ⇒ k + 4 = ±4
When k – 4 = 4 ⇒ k = 4 + 4 = 8
When k – 4 = -4 ⇒ k = 0
∴ k = 0, 8

Question 7.
Kind equation of line joining (1, 2) and (3, 6) using determinants.
Solution:
Let P(x, y) be a point on the line joining points and A (x₁, y₁) = (1, 2) and B(x₂, y₂) = (3, 6).
Then, the points A,B and P are collinear.
Hence, the area of triangle ABP is zero.
∴ ∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 6 & 1 \\
x & y & 1
\end{array}\right|=\) = 0
⇒ \(\frac{1}{2}\)[1(6 – y) – 2(3 – x) + 1(3y – 6x)] = 0
⇒ 6 – y – 6 + 2x + 3y – 6x = 0
⇒ 2y – 4x = 0 ⇒ y = 2x
∴ The equation of the line joining the given points is y = 2x.

Question 8.
Kind equation of line joining (3, 1) and (9, 3) using determinants.
Solution:
Let P(x, y) be a point on the line joining points and A (x₁, y₁) = (3, 1) and B(x₂, y₂) =(9, 3).
Then, the points A,B and P are collinear.
Hence, the area of triangle ABP will be zero.
∴ ∆ = \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{lll}
3 & 1 & 1 \\
9 & 3 & 1 \\
x & y & 1
\end{array}\right|\) = 0
⇒ \(\frac{1}{2}\) |[3(3 – y) – 1(9 – x) + 1(9y – 3x)] = 0
⇒ 9 – 3y – 9 + x + 9y – 3x = 0
⇒ 6y – 2x = 0
⇒ x – 3y = 0
∴ The equation of the line joining the given points is x – 3y = 0

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4a

Question 1.
Evaluate \(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
Solution:
|A| = \(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\) = 2(-1) – 4(-5) = -2 + 20 = 18 [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]

Question 2.
Evaluate \(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\) = (cos θ)(cos θ) – (-sin θ)(sin θ) = cos²θ + sin²θ = 1

Question 3.
Find the determinant of \(\left[\begin{array}{cc}
2 & 1 \\
1 & -5
\end{array}\right]\)
Solution:
det A = ad – bc = 2(-5) – 1(1) = -10 – 1 = -11

Question 4.
Find the determinant of \(\left[\begin{array}{cc}
4 & 5 \\
-6 & 2
\end{array}\right]\)
Solution:
det A = ad – bc = 4(2) – 5(-6) = 8 + 30 = 38

Question 5.
Find the determinant of \(\left[\begin{array}{cc}
i & 0 \\
0 & -i
\end{array}\right]\)
Solution:
det A = i(-i) – 0 = -i² = -(-1) = 1

Question 6.
Evaluate \(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\)
Solution:
\(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|\) = (x² – x + 1)(x + 1) – (x – 1)(x + 1) [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]
= x³ + x² – x² + x – x + 1 – (x² – 1)
= x³ + 1 – x² + 1
= x³ – x² + 2

Question 7.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\), then show that |2A| = 4|A|
Solution:
The given matrix is A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
∴ 2A = \(2\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\) [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]
L.H.S = |2A| = \(\left|\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right|\) = 2 × 4 – 4 × 8 = 8 – 32 = -24
Now, |A| = \(\left|\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right|\) = 1 × 2 – 2 × 4 = 2 – 8 = -6
∴ RHS = 4|A| = 4(-6) = -24
∴ |2A| = 4|A|

Question 8.
Evaluate \(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 0 & -1 \\
3 & -5 & 0
\end{array}\right|\)
Solution:
Let A = \(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 0 & -1 \\
3 & -5 & 0
\end{array}\right|\)
On expanding along the second row R₂, we get
|A| = \(-0\left|\begin{array}{cc}
-1 & -2 \\
-5 & 0
\end{array}\right|+0\left|\begin{array}{cc}
3 & -2 \\
3 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
3 & -1 \\
3 & -5
\end{array}\right|\)
= (-15 + 3) = -12

Question 9.
Evaluate \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
Solution:
Let A = \(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
|A| = \(3\left|\begin{array}{cc}
1 & -2 \\
3 & 1
\end{array}\right|+4\left|\begin{array}{cc}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{cc}
1 & 1 \\
2 & 3
\end{array}\right|\)
= 3(1 + 6) + 4(1 + 4) + 5(3 – 2)
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5 = 46

Question 10.
Evaluate \(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
Solution:
Let A = \(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
∴ |A| = \(0\left|\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right|-1\left|\begin{array}{cc}
-1 & -3 \\
-2 & 0
\end{array}\right|+2\left|\begin{array}{cc}
-1 & 0 \\
-2 & 3
\end{array}\right|\)
= 0 – 1(0 – 6) + 2(-3 – 0) = -1(-6) + 2(-3)
= 6 – 6 = 0

Question 11.
Evaluate \(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right| .\)
Solution:
Let A = \(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\)
∴ |A| = \(2\left|\begin{array}{cc}
2 & -1 \\
-5 & 0
\end{array}\right|-0\left|\begin{array}{cc}
-1 & -2 \\
-5 & 0
\end{array}\right|+3\left|\begin{array}{cc}
-1 & -2 \\
2 & -1
\end{array}\right|\)
= 2(0 – 5) – 0 + 3(1 + 4)
= -10 + 15 = 5

Question 12.
If A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\), find |A|
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)
∴ |A| = \(1\left|\begin{array}{ll}
1 & -3 \\
4 & -9
\end{array}\right|-1\left|\begin{array}{ll}
2 & -3 \\
5 & -9
\end{array}\right|-2\left|\begin{array}{ll}
2 & 1 \\
5 & 4
\end{array}\right|\)
= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)
= 1(3) – 1(-3) – 2(3)
= 3 + 3 – 6 = 0

Question 13.
Find the values of x, if \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
Solution:
Given that \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\) [∵ \(\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|\) = ad – bc]
⇒ 2 × 1 – 5 × 4 = 2x × x – 6 × 4
⇒ 2 – 20 = 2x² – 24
⇒ 2x² = 6
⇒ x² = 3
⇒ x = ±\(\sqrt{3}\)

Question 14.
Find the values of x, if \(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
Solution:
Given that \(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
⇒ 2 × 5 – 3 × 4 = x × 5 – 3 × 2x
⇒ 10 – 12 = 5x – 6x
⇒ -2 = -x
⇒ x = 2

II.

Question 1.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\), then show that |3A| = 27|A|
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7a

I. Find an anti derivative (or integral) of the following functions (1 to 5) by the method of inspection

Question 1.
sin 2x
Solution:
Method of Inspection:
We know that \(\frac{d}{d x}\)(cos2x) = -2sin2x dx
⇒ \(\frac{-1}{2} \frac{\mathrm{~d}}{\mathrm{dx}}\)(cos2x) = sin2x ⇒ \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)\) = sin 2x
By definition of integral, anti-derivative of sin2x is \(\frac{-1}{2}\)cos2x.

Question 2.
cos 3x
Solution:
Method of Inspection:
We know that \(\frac{d}{d x}\)(sin3x) = 3 cos3x dx
⇒ \(\frac{1}{3} \frac{\mathrm{~d}}{\mathrm{dx}}\)(sin3x)= cos3x ⇒ \(\frac{d}{d x}\left(\frac{1}{3} \sin 3 x\right)\) = cos 3x
By definition of integral, anti-derivative of cos3x is \(\frac{1}{3}\)sin3x.

Question 3.
e^2x
Solution:
We know that \(\frac{d}{d x} e^{2 x}=e^{2 x} \frac{d}{d x}(2 x)=2 e^{2 x} \Rightarrow \frac{1}{2} \frac{d}{d x} e^{2 x}=e^{2 x} \Rightarrow \frac{d}{d x}\left(\frac{1}{2} e^{2 x}\right)\) = 2e^2x
∴ An antiderivative of e^2x is \(\frac{1}{2}\) e^2x.

Question 4.
(ax + b)²
Solution:
We know that \(\frac{d}{d x}\)(ax + b)³ = 3(ax + b)²\(\frac{d}{d x}\)(ax + b) = 3(ax + b)²a
⇒ \(\frac{1}{3 a} \frac{d}{d x}\)(ax + b)³ = (ax + b)² ⇒ \(\frac{d}{d x}\left[\frac{1}{3 a}(a x+b)^3\right]\) = (ax + b)²
∴ An antiderivative of (ax + b)² is \(\frac{1}{3a}\) (ax+b)³.

Question 5.
sin 2x – 4e^3x
Solution:
We know that \(\frac{d}{d x}\)(cos2x) = -2sin2x dx
⇒ \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)\) = sin 2x ……(i)
Again \(\frac{d}{d x}\)e^3x = 3e^3x
∴ \(\frac{d}{d x}\left(\frac{1}{3} e^{3 x}\right)=e^{3 x} \Rightarrow \frac{d}{d x}\left(\frac{-4}{3} e^{3 x}\right)\) = -4e^3x ………(ii)
Adding eqns. (i) and (ii) \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)+\frac{d}{d x}\left(\frac{-4}{3} e^{3 x}\right)\) = sin 2x – 4e^3x
⇒ \(\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\right)\) = sin 2x – 4e^3x
∴ An antiderivative of sin 2x – 4e^3x is \(\frac{-1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\)

Question 6.
Find ∫(4e^3x + 1) dx
Solution:
∫(4e^3x + 1) dx = ∫4e^3x dx + ∫1 dx
= 4∫e^3x dx + x = 4\(\left(\frac{e^{3 x}}{3}\right)\) + x + c. [∵ ∫e^ax dx \(\frac{e^{a x}}{a}\) and ∫ 1 dx = x]

Question 7.
Find ∫ x²(1 – \(\frac{1}{x^2}\)) dx
Solution:
∫ x²(1 – \(\frac{1}{x^2}\)) dx = ∫(x² – \(\frac{x^2}{x^2}\)) dx = ∫ (x² – 1) dx
= ∫ x² dx – ∫ 1 dx = \(\frac{x^3}{3}\) – x + c. [∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) if n ≠ -1]

Question 8.
Find ∫ (ax² + bx + c) dx
Solution:
∫(ax² + bx + c) dx = ∫ ax² dx + ∫bx dx + ∫ x dx
= a ∫ x² dx + b ∫x¹ dx + c∫ 1 dx = a\(\frac{x^3}{3}\) + b\(\frac{x^2}{2}\) + cx + c₁
where c₁ is the constant of integration.

Question 9.
Find ∫ (2x² + e^x) dx
Solution:
∫(2x² + e^x)dx = ∫2x² dx + ∫ e^x dx
= 2∫x² dx + ∫e^x dx = 2\(\frac{x^{2+1}}{2+1}\) + e^x + c = \(\frac{2}{3}\)x³ + e^x + c.

Question 10.
Find \(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\) dx
Solution:
\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\) dx
\(\int\left((\sqrt{x})^2+\left(\frac{1}{\sqrt{x}}\right)^2-2 \sqrt{x} \frac{1}{\sqrt{x}}\right)\) dx [∵ (a – b)² = a² – b² – 2ab]
= ∫(x + \(\frac{1}{x}\) – 2) dx = ∫x dx + ∫\(\frac{1}{x}\) dx – ∫2dx = \(\frac{x^2}{2}\) + log|x| – 2x + c. [∵∫2dx = 2∫1 dx = 2x]

Question 11.
Find \(\int \frac{x^3+5 x^2-4}{x^2} d x\)
Solution:
\(\int \frac{x^3+5 x^2-4}{x^2} d x=\int\left(\frac{x^3}{x^2}+\frac{5 x^2}{x^2}-\frac{4}{x^2}\right)\) dx
= ∫(x + 5 – 4x^-2) dx = ∫x¹ dx + ∫5 dx – ∫4x^-2 dx = \(\frac{x^2}{2}\) + 5 ∫1 dx – 4∫x^-2 dx
= \(\frac{x^2}{2}\) + 5x – 4\(\frac{x^{-2+1}}{-2+1}\) + c = \(\frac{x^2}{2}\) + 5x + \(\frac{4}{x}\) + c

Question 12.
Find \(\int \frac{x^3+3 x+4}{\sqrt{x}}\) dx
Solution:
\(\int \frac{x^3+3 x+4}{\sqrt{x}}\) dx = \(\int\left(\frac{x^3}{x^{1 / 2}}+\frac{3 x}{x^{1 / 2}}+\frac{4}{x^{1 / 2}}\right)\) dx
= ∫(x^3-1/2 + 3x^1-1/2 + 4x^-1/2) dx = ∫(x^5/2 + 3x^1/2 + 4x^-1/2) dx
= ∫x^5/2 dx + 3∫x^1/2 dx + 4∫x^{-1/2 dx}
= \(\frac{x^{5 / 2+1}}{\frac{5}{2}+1}+3 \frac{x^{1 / 2+1}}{\frac{1}{2}+1}+4 \frac{x^{-1 / 2+1}}{\frac{-1}{2}+1}+c=\frac{x^{7 / 2}}{\frac{7}{2}}+3 \frac{x^{3 / 2}}{\frac{3}{2}}+4 \frac{x^{1 / 2}}{\frac{1}{2}}+c\)
= \(\frac{2}{7}\) x^7/2 + 2x^3/2 + 8x^1/2 + c.

Question 13.
Find \(\int \frac{x^3-x^2+x-1}{x-1}\) dx
Solution:
\(\int \frac{x^3-x^2+x-1}{x-1} d x=\int \frac{x^2(x-1)+(x-1)}{x-1} d x=\int \frac{(x-1)\left(x^2+1\right)}{(x-1)} d x=\int\left(x^2+1\right) d x\)
= \(\int x^2 d x+\int 1 d x=\frac{x^{2+1}}{2+1}+x+c=\frac{x^3}{3}+x+c\)

Question 14.
Find ∫(1 – x)\(\sqrt{\mathbf{x}}\) dx
Solution:
∫(1 – x)\(\sqrt{\mathbf{x}}\) dx = \(\int(\sqrt{\mathrm{x}}-\mathrm{x} \sqrt{\mathrm{x}}) \mathrm{dx}=\int\left(\mathrm{x}^{1 / 2}-\mathrm{x}^1 \mathrm{x}^{1 / 2}\right) \mathrm{dx}=\int\left(\mathrm{x}^{1 / 2}-\mathrm{x}^{1+1 / 2}\right) \mathrm{dx}\)
= \(\int\left(x^{1 / 2}-x^{3 / 2}\right) d x=\frac{x^{1 / 2+1}}{\frac{1}{2}+1}-\frac{x^{3 / 2+1}}{\frac{3}{2}+1}+c=\frac{x^{3 / 2}}{\frac{3}{2}}-\frac{x^{5 / 2}}{\frac{5}{2}}+c=\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}+c \ldots\)

Question 15.
Find ∫\(\sqrt{x}\)(3x² + 2x + 3) dx
Solution:
∫\(\sqrt{x}\)(3x² + 2x + 3) dx = ∫x^1/2(3x² + 2x + 3) dx
= ∫(3x²x^1/2 + 2xx^1/2 + 3x^1/2) dx = ∫(3x^5/2 + 2x^3/2 + 3x^1/2) dx
= 3∫x^5/2 dx + 2∫x^3/2 dx + 3∫x^1/2 dx (∵\(2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}, 1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}\))
= \(3 \frac{x^{5 / 2+1}}{\frac{5}{2}+1}+2 \frac{x^{3 / 2+1}}{\frac{3}{2}+1}+3 \frac{x^{1 / 2+1}}{\frac{1}{2}+1}+c=3 \frac{x^{7 / 2}}{\frac{7}{2}}+2 \frac{x^{5 / 2}}{\frac{5}{2}}+3 \frac{x^{3 / 2}}{\frac{3}{2}}+c\)
= \(\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}+2 x^{3 / 2}+c\)

Question 16.
Find ∫(2x – 3cosx + e^x) dx
Solution:
∫(2x – 3cosx + e^x) dx = ∫2x dx – ∫3cos x dx + ∫e^x dx
= 2 ∫x¹ dx – 3∫cosx dx + e^x dx = 2\(\frac{x^2}{2}\) – 3sin x + e^x + c = x² – 3sin x + e^x + c

Question 17.
Find ∫(2x² – 3sin x + 5\(\sqrt{x}\)) dx
Solution:
∫(2x² – 3sin x + 5\(\sqrt{x}\)) dx = 2∫x² dx – 3∫sinx dx + 5∫x^1/2 dx
= \(2 \frac{x^{2+1}}{2+1}-3(-\cos x)+5 \frac{x^{1 / 2+1}}{\frac{1}{2}+1}+c=2 \frac{x^3}{3}+3 \cos x+5 \frac{x^{3 / 2}}{\frac{3}{2}}+c\)
= \(\frac{2}{3}\)x³ + 3cos x + \(\frac{10}{3}\)x^3/2 + c.

Question 18.
Find ∫secx(secx + tanx) dx
Solution:
∫secx(secx + tanx) dx = ∫(sec² x + sec x tan x) dx
= ∫sec² x dx + ∫secx tanx dx = tan x + sec x + c.

Question 19.
Find \(\int \frac{\sec ^2 x}{{cosec}^2 x}\) dx
Solution:
\(\int \frac{\sec ^2 x}{{cosec}^2 x} d x=\int \frac{\left(\frac{1}{\cos ^2 x}\right)}{\left(\frac{1}{\sin ^2 x}\right)} d x=\int \frac{\sin ^2 x}{\cos ^2 x} d x\)
= ∫tan²x dx = ∫(sec² x – 1)dx = tan x – x + c (∵ sec²x – tan² x = 1 ⇒ sec²x – 1 = tan²x)

Question 20.
Find \(\int \frac{2-3 \sin x}{\cos ^2 x} d x\)
Solution:
\(\int \frac{2-3 \sin x}{\cos ^2 x} d x=\int\left(\frac{2}{\cos ^2 x}-\frac{3 \sin x}{\cos ^2 x}\right) d x\)
= \(\int\left(2 \sec ^2 x-\frac{3 \sin x}{\cos x \cos x}\right) d x\) = ∫(2 sec² x – 3tan x sec x) dx
= 2∫sec² xdx – 3∫secx tanxdx = 2 tanx – 3 secx + c

Question 21.
Find the integral of \(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\)
Solution:
\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}=\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}\)
= \(\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}=\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b}\)
⇒ \(\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x\)
= \(\frac{1}{(a-b)}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+C\)

Question 22.
Find the integral of \(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\)
Solution:
Given integral is \(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}=\frac{e^{4 \log x}\left(e^{\log x}-1\right)}{e^{2 \log x}\left(e^{\log x}-1\right)}\) = e^{2log x} = e^{log x2} = x²
∴ \(\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int x^2 d x=\frac{x^3}{3}+C\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Matrices Solutions Exercise 3d

I.

Question 1.
For what values of x : \(\left[\begin{array}{lll}
1 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 0 \\
2 & 0 & 1 \\
1 & 0 & 2
\end{array}\right]\left[\begin{array}{l}
0 \\
2 \\
x
\end{array}\right]\) = 0
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), show that A² – 5A + 7I = 0
Solution:

Question 3.
Find x, if \(\left[\begin{array}{lll}
x & -5 & -1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
4 \\
1
\end{array}\right]\) = 0
Solution:

II.

Question 1.
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Solution:
Given that A and B are symmetric matrices. Then A’ =A and B’ =B
Now (AB – BA)’ = (AB)’ – (BA) [∵ (A – B)’ = A’ – B’]
= B’A’ – A’B’ [∵ (AB) = B’A’] = BA – AB [∵ B’ = B and A’= A] = -(AB – BA)
∴ (AB – BA) = -(AB – BA).
Thus, AB – BA is a skew symmetric matrix.

Question 2.
Find the values of x, y, z if the matrix A = \(\left[\begin{array}{ccc}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]\) satisfy the equation A’A = I
Solution:

Question 3.
Find the matrix X so that X\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
Solution:
Given that X\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
The order of the matrix on R.H.S. is 2×3 and that of L.H.S is 2×3 .
So, X has to be a 2×2 matrix. Let X = \(\left[\begin{array}{ll}
\mathrm{a} & \mathrm{c} \\
\mathrm{~b} & \mathrm{~d}
\end{array}\right]\)
∴ \(\left[\begin{array}{ll}
a & c \\
b & d
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\) ⇒ \(\left[\begin{array}{ccc}
\mathrm{a}+4 \mathrm{c} & 2 \mathrm{a}+5 \mathrm{c} & 3 \mathrm{a}+6 \mathrm{c} \\
\mathrm{~b}+4 \mathrm{~d} & 2 \mathrm{~b}+5 \mathrm{~d} & 3 \mathrm{~b}+6 \mathrm{~d}
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
Equating the corresponding elements of the two matrices, we have:
a + 4c = -7…..(1)
2a + 5c = – 8 …………. (2)
3a + 6c = -9 …………. (3)
b + 4d = 2 ….(4)
2b + 5d = 4 …. (5)
3b + 6d = 6 …………. (6)
Solving (1) and (2) we get a,c
(1) ⇒ a + 4c = – 7 ⇒ a = – 7 – 4c
(2) ⇒2a + 5c = – 8
⇒ 2(- 7 – 4c) + 5c = -8 ⇒ -14 – 8c + 5c = -8 ⇒ -3c = 6 ⇒ c = -2
∴ a = -7-4(-2) = -7 + 8 = 1 ⇒ a = 1
Solving (4) and (5) we get b, d
(4) ⇒ b + 4d = 2 ⇒ b = 2 – 4d
(5) ⇒ 2b + 5d = 4 ⇒ 2(2 – 4d) + 5d = 4
⇒ 4 – 8d + 5d = 4 ⇒ -3d = 0 ⇒ d = 0
∴ b = 2 – 4d = 2 – 4(0) = 2 ⇒ b = 2
Thus, a = 1, b = 2, c = -2 and d = 0
Hence, the required matrix X = \(\left[\begin{array}{cc}
1 & -2 \\
2 & 0
\end{array}\right]\)

III.

Question 1.
Show that the matrix B’AB k symmetric or skew symmetric according as A is symmetric or skew symmetric.
Solution:
We suppose that A is a symmetric matrix, then A’ = A (1)
Consider ( B’AB)’ = [B'(AB)]’ = (AB)'(B’)’ [: (AB)’ B’A’]
= B’A'(B)[∵ (B’)’ = B]
= B'(A’B)= B'(AB)[ using(1)]
∴ (B’AB)’ = B’AB
Thus, if A is symmetric matrix, then B’AB is a symmetric matrix.
Now, we suppose that A is a skew symmetric matrix, then A’ = -A ……….. (2)
Consider, (B’AB)’ = [B'(AB)]’ = (AB)'( B’)’ = (B’A’)B = B'(-A )B [Using (2)] = -B’AB
∴ (B’AB)’ = -B’AB
Thus, if A is a skew symmetric matrix, then B’AB is a skew symmetric matrix.
Hence, if A is symmetric or skew symmetric matrix, then B’AB is symmetric or skew symmetric accordingly.

Question 2.
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Solution:
(a) The unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00 respectively.
Consequently, the total revenue in market I can be represented in the form of a matrix as:
\(\left[\begin{array}{lll}
10000 & 2000 & 18000
\end{array}\right]\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right]\) = 10000 × 2.50 + 2000 × 1.50 + 18000 × 1.00
= 25000 + 3000 + 18000 = 46000
The total revenue in market II can be represented in the form of a matrix as:
\(\left[\begin{array}{lll}
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{l}
2.50 \\
1.50 \\
1.00
\end{array}\right]\) = 6000 × 2.50 + 20000 × 1.50 + 8000 × 1.00
= 15000 + 30000 + 8000 = 53000
Thus, the total revenue in market I is ₹ 46000 and the total revenue in market.II is ₹ 53000.

(b) The unit costs of x, y and z are ₹ 2.00, ₹ 1.00 and 50 paise respectively.
Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:
\(\left[\begin{array}{lll}
10000 & 2000 & 18000
\end{array}\right]\left[\begin{array}{l}
2.00 \\
1.00 \\
0.50
\end{array}\right]\) = 10000 × 2.00 + 2000 × 1.00 + 18000 × 0.50
= 20000 + 2000 + 9000 = 31000
Since the total revenue in market I is ₹ 46000,
the gross profit in this market in ₹ is 46000 – 31000=15000
The total cost prices of all the products in market II can be represented in the form of a matrix as:
\(\) = 6000 × 2.00 + 20000 × 1.00 + 8000 × 0.50
= 12000 + 20000 + 4000 = 36000
Since the total revenue in market I is ₹ 53000 , the gross profit in this market in ₹ is 53000 – 36000 = 17000
Thus, the gross profit in market I is ₹ 15000 and in market II is ₹ 17000

Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Matrices Solutions Exercise 3c

I.

Question 1.
Find the transpose of matrix \(\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]\) ⇒ A’ = \(\left[\begin{array}{lll}
5 & \frac{1}{2} & -1
\end{array}\right]\)

Question 2.
Find the transpose of matrix \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\) ⇒ A’ = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right]\)

Question 3.
Find the transpose of matrix \(\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]\) ⇒ A’ = \(\left[\begin{array}{ccc}
-1 & \sqrt{3} & 2 \\
5 & 5 & 3 \\
-6 & 6 & -1
\end{array}\right]\)

Question 4.
If A’ = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right]\), then find (A + 2B)’
Solution:
Let A’ = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]\) ⇒ A = (A’)’ = \(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]\)
Now A + 2B = \(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+2\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{cc}
-2 & 0 \\
2 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
-4 & 1 \\
5 & 6
\end{array}\right]\)
⇒ (A + 2B)’ = \(\left[\begin{array}{cc}
-4 & 5 \\
1 & 6
\end{array}\right]\)

Question 5.
Show that the matrix A = \(\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) is a symmetric matrix.
Solution:
Given that A = \(\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\)
Now A’ = \(\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) = A
Hence, A is a symmetric matrix

Question 6.
Show that the matrix A = \(\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\) is a skew symmetric matrix.
Solution:
A = \(\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\)
Now A’ = \(\left[\begin{array}{ccc}
0 & -1 & 1 \\
1 & 0 & -1 \\
-1 & 1 & 0
\end{array}\right]\) = –\(\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\) = -A
Hence, A is a skew symmetric matrix.

Question 7.
For the matrix A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), verify that (A + A’) is a symmetric matrix.
Solution:
Given that A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\) A’ = \(\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\)
A + A’ = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]\)
Also [(A + A’)]’ = \(\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]\) = (A + A’)
∴ (A + A’) is a symmetric matrix

Question 8.
For the matrix A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), verify that (A – A’) is a skew symmetric matrix
Solution:
A – A’ = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\) = \(\left[\begin{array}{ll}
0 & -1 \\
1 & 0
\end{array}\right]\)
∴ (A – A’)’ = \(\left[\begin{array}{ll}
0 & 1 \\
-1 & 0
\end{array}\right]\) = –\(\left[\begin{array}{ll}
0 & -1 \\
1 & 0
\end{array}\right]\) = -(A – A’)
∴ (A – A’) is a skew symmetric matrix

Question 9.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) then verify that A’A = I
Solution:

Question 10.
If A = \(\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\), then verify that A’A = I
Solution:

II.

Question 1.
Find \(\frac{1}{2}\)(A + A’) and \(\frac{1}{2}\)(A – A’) when A = \(\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\)
Solution:

Question 2.
Express the matrix \(\left[\begin{array}{cc}
3 & 5 \\
1 & -1
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix.
Solution:

Hence Q’ = \(\left[\begin{array}{cc}
0 & -2 \\
2 & 0
\end{array}\right]\) = -Q
Thus, Q = \(\frac{1}{2}\)(A – A’) is a skew symmetric matrix.
Expressing A as the sum of P and Q:
P + Q = \(\left[\begin{array}{cc}
3 & 3 \\
3 & -1
\end{array}\right]+\left[\begin{array}{cc}
0 & 2 \\
-2 & 0
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & 5 \\
1 & -1
\end{array}\right]\) = A

Question 3.
Express the matrix \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix.
Solution:

Expressing A as the sum of P and Q
P + Q = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) = A

Question 4.
Express the matrix \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\) as the sum of a symmetric and a skew symmetricmatrix
Solution:

Question 5.
Express the matrix \(\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix.
Solution:

Question 6.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]\), then verify that (A + B)’ = A’ + B’
Solution:

Now, A’ + B’ = \(\left[\begin{array}{ccc}
-1 & 5 & -2 \\
2 & 7 & 1 \\
3 & 9 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-4 & 1 & 1 \\
1 & 2 & 3 \\
-5 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-5 & 6 & -1 \\
3 & 9 & 4 \\
-2 & 9 & 2
\end{array}\right] .\) …………… (2)
∴ From (1) & (2), (A + B)’ = A’ + B’

Question 7.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]\) then verify that (A – B)’ = A’ – B’
Solution:

∴ From (1) & (2), (A – B)’ = A’ – B’

Question 8.
If A’ = \(\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\), then verify that (A + B)’ = A’ + B’
Solution:

∴ From (1) & (2), (A + B)’ = A’ + B’

Question 9.
If A = \(\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right] .\) then verify that (A – B)’ = A’ – B’
Solution:

∴ From (1) & (2), (A – B)’ = A’ – B’

Question 10.
For the matrices A and B, verify that (AB)’ = B’A’, where A = \(\left[\begin{array}{c}
1 \\
-4 \\
3
\end{array}\right]\), B = \(\left[\begin{array}{lll}
-1 & 2 & 1
\end{array}\right]\)
Solution:

∴ From (1) & (2), (AB)’ = B’A’

Question 11.
For the matrices A and B, verify that (AB)’ = B’A’, where A = \(\left[\begin{array}{c}
0 \\
1 \\
2
\end{array}\right]\), B = \(\left[\begin{array}{lll}
1 & 5 & 7
\end{array}\right]\)
Solution:

∴ From (1) & (2), (AB)’ = B’A’

Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Matrices Solutions Exercise 3b

I.

Question 1.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find A + B
Solution:
A + B
= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) + \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\) = \(\left[\begin{array}{ll}
2+1 & 4+3 \\
3-2 & 2+5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 & 7 \\
1 & 7
\end{array}\right]\)

Question 2.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find A – B
Solution:
A – B
= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) – \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\) = \(\left[\begin{array}{ll}
2-1 & 4-3 \\
3+2 & 2-5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 1\\
5 & -3
\end{array}\right]\)

Question 3.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), C = \(\left[\begin{array}{ll}
-2 & 5 \\
3 & 4
\end{array}\right]\). Find 3A – C
Solution:
3A – C
= 3\(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) – \(\left[\begin{array}{ll}
-2 & 5 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 \times 2 & 3 \times 4 \\
3 \times 3 & 3 \times 2
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
6+2 & 12-5 \\
9-3 & 6-4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
8 & 7 \\
6 & 2
\end{array}\right]\)

Question 4.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find AB
Solution:
AB
= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2(1)+4(-2) & 2(3)+4(5) \\
3(1)+2(-2) & 3(3)+2(5)
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2-8 & 6+20 \\
3-4 & 9+10
\end{array}\right]\) = \(\left[\begin{array}{ll}
-6 & 26 \\
-1 & 19
\end{array}\right]\)

Question 5.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find BA
Solution:
BA
= \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 4 \\
3 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1(2)+3(3) & 1(4)+3(2) \\
-2(2)+5(3) & -2(4)+5(2)
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2+9 & 4+6 \\
-4+15 & -8+10
\end{array}\right]\) = \(\left[\begin{array}{cc}
11 & 10 \\
11 & 2
\end{array}\right]\)

Question 6.
Compute \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\) + \(\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\) + \(\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\) = \(\left[\begin{array}{cc}
\mathrm{a}+\mathrm{a} & \mathrm{~b}+\mathrm{b} \\
-\mathrm{b}+\mathrm{b} & \mathrm{a}+\mathrm{a}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2 \mathrm{a} & 2 \mathrm{~b} \\
0 & 2 \mathrm{a}
\end{array}\right]\)

Question 7.
Compute \(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]\) + \(\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]\) + \(\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a^2+b^2+2 a b & b^2+c^2+2 b c \\
a^2+c^2-2 a c & a^2+b^2-2 a b
\end{array}\right]\)
= \(\left[\begin{array}{ll}
(a+b)^2 & (b+c)^2 \\
(a-c)^2 & (a-b)^2
\end{array}\right]\)

Question 8.
Compute \(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]\) + \(\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]\) + \(\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]\)

Question 9.
Compute \(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin ^2 x+\cos ^2 x & \cos ^2 x+\sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)

Question 10.
Compute \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]\) = \(\left[\begin{array}{cc}
\mathrm{a}(\mathrm{a})+\mathrm{b}(\mathrm{~b}) & \mathrm{a}(-\mathrm{b})+\mathrm{b}(\mathrm{a}) \\
-\mathrm{b}(\mathrm{a})+\mathrm{a}(\mathrm{~b}) & -\mathrm{b}(-\mathrm{b})+\mathrm{a}(\mathrm{a})
\end{array}\right]\)
= \(\left[\begin{array}{cc}
a^2+b^2 & -a b+a b \\
-a b+a b & b^2+a^2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
a^2+b^2 & 0 \\
0 & a^2+b^2
\end{array}\right]\)

Question 11.
Compute \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]|233|\)
Solution:
\(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]|233|\) = \(\left[\begin{array}{lll}
1(2) & 1(3) & 1(4) \\
2(2) & 2(3) & 2(4) \\
3(2) & 3(3) & 3(4)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
2 & 3 & 4 \\
4 & 6 & 8 \\
6 & 9 & 12
\end{array}\right]\)

Question 12.
Compute \(\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\
2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-3 & -4 & 1 \\
8 & 13 & 9
\end{array}\right]\)

Question 13.
Compute \(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\
3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\
4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
14 & 0 & 42 \\
18 & -1 & 56 \\
22 & -2 & 70
\end{array}\right]\)

Question 14.
Compute \(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
1 & 2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\
3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\
-1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & 4 & 5 \\
-2 & 2 & 0
\end{array}\right]\)

Question 15.
Compute \(\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & 3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & 3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\
-1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1)
\end{array}\right]\)
= \(\left[\begin{array}{cc}
14 & -6 \\
4 & 5
\end{array}\right]\)

Question 16.
If A = \(\left[\begin{array}{lll}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{array}\right]\), then compute 3A – 5B
Solution:
3A – 5B = 3\(\left[\begin{array}{lll}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]\) – 5\(\left[\begin{array}{ccc}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]\) – \(\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 17.
Simplify cosθ \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sinθ \(\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:
cosθ \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sinθ \(\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta & \cos \theta \sin \theta \\
-\sin \theta \cos \theta & \cos ^2 \theta
\end{array}\right]\) + \(\left[\begin{array}{cc}
\sin ^2 \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin ^2 \theta
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta+\sin ^2 \theta & \sin \theta \cos \theta-\sin \theta \cos \theta \\
-\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^2 \theta+\sin ^2 \theta
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

Question 18.
Find X and Y, if X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
Solution:
Given X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) …………… (1)
X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) …………… (2)
(1) + (2) ⇒ 2X = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) + \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) = \(\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right]\)
⇒ X = \(\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
⇒ X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\)
⇒ Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) – \(\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
⇒ Y = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]\)

Question 19.
Find X and Y, if 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) and 3X + 2Y = \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\)
Solution:
Given 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) ………….. (1)
3X + 2Y = \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\) ……….. (2)
Multiplying equation(1) by 2, we have 2(2X +3Y) = 2\(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) ⇒ 4X + 6Y = \(\left[\begin{array}{ll}
4 & 6 \\
8 & 0
\end{array}\right]\) …..(3)
MultIplying equation (2) by 3, we have 3(3X + 2Y) = 3 \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\) ⇒ 9X + 6Y = \(\left[\begin{array}{cc}
6 & -6 \\
-3 & 15
\end{array}\right]\) …………. (4)
From (3) and(4), we have (4X + 6Y)(9X + 6Y) = \(\left[\begin{array}{ll}
4 & 6 \\
8 & 0
\end{array}\right]\) – \(\left[\begin{array}{cc}
6 & -6 \\
-3 & 15
\end{array}\right]\)

Question 20.
Find X, if Y = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) and 2X + Y = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\)
Solution:
2X + Y = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\) ⇒ 2X + \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\)
⇒ 2X = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\) – \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right]\)
⇒ X = \(\frac{1}{2}\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right]\) = \(\left[\begin{array}{ll}
-1 & -1 \\
-2 & -1
\end{array}\right]\)

Question 21.
Find x and y, if 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Solution:
Given 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 & 6 \\
0 & 2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Equating the corresponding elements of these two matrices, 2 + y = 5 ⇒ y = 3
2x + 2 = 8 ⇒ x = 3
∴ x = 3, y = 3

Question 22.
Solve the equation for x, y, z and t, if 2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right] .\) + 3\(\left[\begin{array}{cc}
1 & -1 \\
0 & 2
\end{array}\right]\) = 3\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right] .\) + 3\(\left[\begin{array}{cc}
1 & -1 \\
0 & 2
\end{array}\right]\) = 3\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 \mathrm{x} & 2 \mathrm{z} \\
2 \mathrm{y} & 2 \mathrm{t}
\end{array}\right]\) + \(\left[\begin{array}{cc}
3 & -3 \\
0 & 6
\end{array}\right]\) = \(\left[\begin{array}{cc}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 \mathrm{x}+3 & 2 \mathrm{z}-3 \\
2 \mathrm{y} & 2 \mathrm{t}+6
\end{array}\right]=\) = \(\left[\begin{array}{cc}
9 & 15 \\
12 & 18
\end{array}\right]\)
Equating the corresponding elements of these two matrIces, 2x + 3 = 9 ⇒ 2x = 6 ⇒ x = 3
2y = 12 ⇒ y = 6
2z – 3 = 15 ⇒ 2z = 18 ⇒ z = 9
2t + 6 = 18 ⇒ 2t = 12 ⇒ t = 6;
∴ x = 3, y = 6, z = 9, t = 6

Question 23.
If \(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\) find the values of x and y.
Solution:
\(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 \mathrm{x} \\
3 \mathrm{x}
\end{array}\right]+\left[\begin{array}{c}
-\mathrm{y} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 \mathrm{x}-\mathrm{y} \\
3 \mathrm{x}+\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
Equating the corresponding elements of these two matrices,
2x – y = 10 ……….. (1)
3x + y = 5 ……….. (2)
By adding these two equations, we get 5x = 15 ⇒ x = 3
Now putting this value in (2)
3x + y = 5 ⇒ y = 5 – 3x
⇒ y = 5 – 3(3) ⇒ y = 5 – 9
⇒ y = -4
∴ x = 3, y = -4

Question 24.
Given, 3\(\left[\begin{array}{cc}
\mathrm{x} & \mathrm{y} \\
\mathrm{z} & \mathrm{w}
\end{array}\right]\) = \(\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]\) + \(\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\) find the values of x, y, z and w.
Solution:
3\(\left[\begin{array}{cc}
\mathrm{x} & \mathrm{y} \\
\mathrm{z} & \mathrm{w}
\end{array}\right]\) = \(\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]\) + \(\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
3 \mathrm{x} & 3 \mathrm{y} \\
3 \mathrm{z} & 3 \mathrm{w}
\end{array}\right]\) = \(\left[\begin{array}{cc}
x+4 & 6+x+y \\
-1+z+w & 2 w+3
\end{array}\right]\)
Equating the corresponding elements of these two matrices,
3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
3y = 6 + x + y ⇒ 2y = 6 + x ⇒ 2y = 6 + 2 ⇒ 2y = 8 ⇒ y = 4
3w = 2w + 3 ⇒ w = 3
3z = -1 + z + w ⇒ 2z = w – 1 ⇒ 2z = 3 – 1 ⇒ 2z = 2 ⇒ z = 1
∴ x = 2, y = 4, z = 1, w = 3 .

Question 25.
Show that \(\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{cc}
2 & 1 \\
3 & 4
\end{array}\right]\) ≠ \(\left[\begin{array}{cc}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\)
Solution:

II.

Question 1.
If F(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\) show that F(x) F(y) = F(x + y)
Solution:

Question 2.
Show that \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\) ≠ \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\)
Solution:

Question 3.
Find A² – 5A + 6I, if A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Solution:

Question 4.
If A = \(\), prove that A³ – 6A² + 7A + 2I = 0
Solution:


Hence, A³ – 6A² + 7A + 2I = 0

Question 5.
If A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find k so that A² = kA – 2I
Solution:

Equating the corresponding elements, we have 3k – 2 = 1
⇒ 3k = 3
⇒ k = 1
∴ k = 1

Question 6.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{u}{2} \\
\tan \frac{u}{2} & 0
\end{array}\right]\) and I is the identify matrix of order 2, show that I + A – (I – A) \(\left[\begin{array}{cc}
0 & -\tan \frac{u}{2} \\
\tan \frac{u}{2} & 0
\end{array}\right]\)
Solution:

Question 7.
A trust fund has ₹ 30,000 that must be invested in two different types of bonds. – The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) ₹ 1800 (b) ₹ 2000
Solution:
(a) Let be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹(3000 – x)
It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year.
Now in order to obtain an annual total interest of ₹ 1800, we have:
\(\left[\begin{array}{ll}
\mathrm{x} & (30000-\mathrm{x})
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]\) = 1800
\(\left[\text { S.I for } 1 \text { year }=\frac{\text { Principal } \text { × } \text { Rate }}{100}\right]\) ⇒ \(\frac{5 x}{100}+\frac{7(30000-x)}{100}\) = 1800
⇒ 5x + 210000 – 7x = 180000
⇒ 210000 – 2x = 180000 .
⇒ -2x = 210000 -180000 ⇒ 2x = 30000
⇒ x = 15000
Thus, in order to obtain an annual total interest of ₹ 1800, the trust fund should invest ₹ 15000 in the first bond and the remaining ₹ 15000 in the second bond.

(b) Let ₹ x be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹ (3000 – x)
Now in order to obtain art annual total interest of ₹ 2000, we have:
\(\left[\begin{array}{ll}
\mathrm{x} & (30000-\mathrm{x})
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]\) = 2000
\(\left[\text { S.I for } 1 \text { year }=\frac{\text { Principal × Rate }}{100}\right]\) ⇒ \(\frac{5 x}{100}+\frac{7(30000-x)}{100}\) = 2000
⇒ 5x + 210000 – 7x = 200000 ⇒ 210000 – 2x = 200000
⇒ 2x = 210000- 200000 ⇒ 2x = 10000
⇒ x = 5000
Thus, in order to obtain an annual total interest of ₹ 2000, the trust fund should invest ₹ 5000 in the first bond and the remaining ₹ 25000 in the second bond.

Question 8.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution:
The total amount of money that will be received from the sale of all these books can be represented in D the matrix form as:
\(12\left[\begin{array}{lll}
10 & 8 & 10
\end{array}\right]\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]\) = 12[10(80) + 8(60) + 10(40)]
= 12(800 + 480 + 400) = 12(1680) = 20160
Thus, the book shop receives ₹ 20160 from the sale of all these books.

Question 9.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]\) then compute (A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Matrices Solutions Exercise 3a

Question 1.
In the matrix A = \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\),write the order of the matrix
Solution:
There are 3 rows and 4 columns in the given matrix.
∴ Order is 3 × 4.

Question 2.
In the matrix A = \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), write the number of elements.
Solution:
Order of the matrix is 3 × 4
∴ Number of elements is 3 × 4 = 12 elements.

Question 3.
In the matrix A \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.
Solution:
a₁₃ = 19, a₂₁ = 35, a₃₃ = -5, a₂₄ = 12, a₂₃ = 5/2

Question 4.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
We know that if a matrix A is of the order m × n, then A has mn elements.
Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the
ordered pairs of natural numbers whose product is 24.
The ordered pairs are: (1, 24),(24, 1),(2, 12),( 12, 2),(3, 8),(8, 3),(4, 6), (6, 4)
Hence, the possible orders of a matrix having 24 elements are: :
(1×24),(24×1),(2×12),(12×2),(3 ×8),(8×3),(4×6) , (6×4).
13 is a prime, so we get only 2 ordered pairs with product 13. They are (1×13) and(13×1)

Question 5.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution:
We know that if a matrix A is of the order m×n , then A has mn elements.
Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1,18),(18,1),(2,9),(9,2),(3,6), (6,3)
Hence, the possible orders of a matrix having 18 elements are:
(1×18),(18×1),(2×9),(9×2),(3×6),(6×3)
5 is a prime, so we get only two ordered pairs with product 5. They are (1 × 5) and (5×1)

Question 6.
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by a_ij = \(\frac{(i+j)^2}{2}\)
Solution:
A 2×2 matrix is given by A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
Given that a_ij = \(\frac{(i+j)^2}{2}\); i, j = 1,2
∴ a₁₁ = \(\frac{(1+1)^2}{2}=\frac{4}{2}\) = 2;
a₁₂ = \(\frac{(1+2)^2}{2}=\frac{9}{2}\)
a₂₁ = \(\frac{(2+1)^2}{2}=\frac{9}{2}\)
a₂₂ = \(\frac{(2+2)^2}{2}=\frac{16}{2}\) = 8
Thus, the required matrix is A = \(\left[\begin{array}{cc}
2 & \frac{9}{2} \\
\frac{9}{2} & 8
\end{array}\right]\)

Question 7.
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by a_ij = \(\frac{i}{j}\)
Solution:
A 2×2 matrix is given by A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
Given that a_ij = \(\frac{i}{j}\); i, j = 1, 2
∴ a₁₁ = \(\frac{1}{1}\) = 1;
a₁₂ = \(\frac{1}{2}\)
a₂₁ = \(\frac{2}{1}\) = 2
a₂₂ = \(\frac{2}{2}\) = 1
Thus, the required matrix is A = \(\left[\begin{array}{ll}
1 & \frac{1}{2} \\
2 & 1
\end{array}\right]\)

Question 8.
Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by a_ij = \(\frac{(i+2j)^2}{2}\)
Solution:
A 2×2 matrix is given by A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]\)
Given that a_ij = \(\frac{(i+2j)^2}{2}\); i, j = 1,2
∴ a₁₁ = \(\frac{(1+2)^2}{2}=\frac{9}{2}\)
a₁₂ = \(\frac{(1+4)^2}{2}=\frac{25}{2}\)
a₂₁ = \(\frac{(2+2)^2}{2}\) = 8
a₂₂ = \(\frac{(2+4)^2}{2}\) = 18
Thus, the required matrix is A = \(\left[\begin{array}{cc}
\frac{9}{2} & \frac{25}{2} \\
8 & 18
\end{array}\right]\)

Question 9.
Construct a 3 × 4 matrix, whose elements are given by a_ij = \(\frac{1}{2}\)|-3i + j|
Solution:
A 3 × 4 matrix is given by A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
Given a_ij = \(\frac{1}{2}\)|-3i + j|; i = 1,2,3; j = 1,2,3,4
a₁₁ = \(\frac{1}{2}\)|-3(1) + 1| = \(\frac{1}{2}\)|-3 + 1| = \(\frac{1}{2}\)|-2| = \(\frac{2}{2}\) = 1;
a₂₁ = \(\frac{1}{2}\)|-3(2) + 1| = \(\frac{1}{2}\)|-6 + 1| = \(\frac{1}{2}\)|-5| = \(\frac{5}{2}\)
a₃₁ = \(\frac{1}{2}\)|-3(3) + 1| = \(\frac{1}{2}\)|-9 + 1| = \(\frac{1}{2}\)|-8| = \(\frac{8}{2}\) = 4;
a₁₂ = \(\frac{1}{2}\)|-3(1) + 2| = \(\frac{1}{2}\)|-3 + 2| = \(\frac{1}{2}\)|-1| = \(\frac{1}{2}\) ;
a₂₂ = \(\frac{1}{2}\)|-3(2) + 2| = \(\frac{1}{2}\)|-6 + 2| = \(\frac{1}{2}\)|-4| = \(\frac{4}{2}\) = 2;
a₃₂ = \(\frac{1}{2}\)|-3(3) + 2| = \(\frac{1}{2}\)|-9 + 2| = \(\frac{1}{2}\)|-7| = \(\frac{7}{2}\) ;
a₁₃ = \(\frac{1}{2}\)|-3(1) + 3| = \(\frac{1}{2}\)|-3 + 3| = 0 ;
a₂₃ = \(\frac{1}{2}\)|-3(2) + 3| = \(\frac{1}{2}\)|-6 + 3| = \(\frac{1}{2}\)|-3| = \(\frac{3}{2}\) ;
a₃₃ = \(\frac{1}{2}\)|-3(3) + 3| = \(\frac{1}{2}\)|-9 + 3| = \(\frac{1}{2}\)|-6| = \(\frac{6}{2}\) = 3;
a₁₄ = \(\frac{1}{2}\)|-3(1) + 4| = \(\frac{1}{2}\)|-3 + 4| = \(\frac{1}{2}\)|1| = \(\frac{1}{2}\) ;
a₂₄ = \(\frac{1}{2}\)|-3(2) + 4| = \(\frac{1}{2}\)|-6 + 4| = \(\frac{1}{2}\)|-2| = \(\frac{2}{2}\) = 1;
a₃₄ = \(\frac{1}{2}\)|-3(3) + 4| = \(\frac{1}{2}\)|-9 + 4| = \(\frac{1}{2}\)|-5| = \(\frac{5}{2}\) ;
Thu, the required matrix is A = \(\left[\begin{array}{cccc}
1 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{5}{2} & 2 & \frac{3}{2} & 1 \\
4 & \frac{7}{2} & 3 & \frac{5}{2}
\end{array}\right]\)

Question 10.
Construct a 3 × 4 matrix, whose elements are given by a_ij = 2i – j
Solution:
A 3 × 4 matrix is given by A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
Given a_ij = 2i – j; i = 1,2,3; j = 1,2,3,4
a₁₁ = 2(1) – 1 = 2 – 1 = 1;
a₂₁ = 2(2) – 1 = 4 – 1 = 3
a₃₁ = 2(3) – 1 = 6 – 1 = 5;
a₁₂= 2(1) – 2 = 2 – 2 = 0
a₂₂ = 2(2) – 2= 4 – 2 = 2;
a₃₂ = 2(3) – 2 = 6 – 2 = 4
a₁₃ = 2(1) – 3 = 2 – 3 = -1;
a₂₃ = 2(2) – 3 = 4 – 3 = 1
a₃₃ = 2(3) – 3 = 6 – 3 = 3;
a₁₄ = 2(1) – 4 = 2 – 4 = -2;
a₂₄ = 2(2) – 4 = 4 – 4 = 0;
a₃₄ = 2(3) – 4 = 6 – 4 = 2
Thus, the required matrix is A = \(\left[\begin{array}{cccc}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{array}\right]\)

Question 11.
Find the values of x,y and z from \(\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements, we get x = 1, y = 4 and z = 3

Question 12.
Find the values of x,y and z from \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{cc}
6 & 2 \\
5 & 8
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{cc}
6 & 2 \\
5 & 8
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements,
we get x + y = 6; xy = 8;5 + z = 5 ⇒ z = 0
Now (x – y)² = (x + y)² – 4xy ⇒ (x – y )² = 6² – 4(8) = 36 – 32 = 4
⇒ (x – y)² = 4 ⇒ (x – y) = ±2 ⇒ x – y = 2 or x – y = -2
When x – y = 2 and x + y = 6 we get x = 4, y = 2
When x – y = – 2 and x + y = 6 we get x = 2, y = 4
Thus, x = 4, y = 2, z = 0 or x = 2, y = 4, z = 0

Question 13.
Find the values of x,y and z from \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{c}
9 \\
5 \\
7
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{c}
9 \\
5 \\
7
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements, we get
x + y + z = 9 ………. (1); x + z = 5 …………..(2); y + z = 7 …………… (3)
From (1) and (2), we have y + 5 = 9 ⇒ y = 4
From (3), we have 4 + z = 7 ⇒ z = 3
Now x + z = 5 ⇒ x + 3 = 5 ⇒ x = 2.
Thus, x = 2, y = 4, z = 3

Question 14.
Find the values of a,b, c and d from the equation \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
As the two matrices are equal, equating the corresponding elements, we get
a – b = – 1 …………… (1)
2a – b = 0 …………. (2)
2a + c = 5 ……….. (3)
3c + d = 13 …………… (4)
From (2), b = 2a
Putting this value in (1), ⇒ a – 2a = – 1 ⇒ a = 1 Hence, b = 2
Putting a = 1 in (3) ⇒ 2(1) + c = 5 ⇒ c = 3
Putting c = 3 in(4) ⇒ 3(3) + d = 13 ⇒ d = 4
Thus, a = 1, b = 2, c = 3 and d = 4