Srinivas

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

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

AP Inter 2nd Year Maths Inverse Trigonometric Functions Solutions Exercise 2c

I.

Question 1.
Find the value of cos^-1 \(\left(\cos \frac{13 \pi}{6}\right)\)
Solution:
\(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left(\cos \frac{12 \pi+\pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \frac{\pi}{6}\right]=\frac{\pi}{6}\)

Question 2.
Find the value of tan^-1 \(\left(\tan \frac{7 \pi}{6}\right)\)
Solution:
\(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left(\tan \frac{6 \pi+\pi}{6}\right)=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{6}\right)\right]=\tan ^{-1}\left(\tan \frac{\pi}{6}\right)=\frac{\pi}{6}\)

Question 3.
Find the value of sin^-1 \(\left(\sin \frac{4 \pi}{3}\right)\)
Solution:
\(\sin ^{-1}\left(\sin \frac{4 \pi}{3}\right)=\sin ^{-1}\left[\sin \left(\pi+\frac{\pi}{3}\right)\right]=\sin ^{-1}\left[-\sin \frac{\pi}{3}\right]=-\sin ^{-1}\left[\sin \frac{\pi}{3}\right]=-\frac{\pi}{3}\)

Question 4.
Find the value of cos^-1 \(\left(\cos \frac{4 \pi}{3}\right)\)
Solution:
\(\cos ^{-1}\left(\cos \frac{4 \pi}{3}\right)=\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{3}\right)\right]=\cos ^{-1}\left[-\cos \frac{\pi}{3}\right]=\pi-\cos ^{-1}\left[\cos \frac{\pi}{3}\right]=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}\)

Question 5.
Find the value of tan^-1 \(\left(\tan \frac{4 \pi}{3}\right)\)
Solution:
\(\tan ^{-1}\left(\tan \frac{4 \pi}{3}\right)=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{3}\right)\right)=\tan ^{-1}\left(\tan \frac{\pi}{3}\right)=\frac{\pi}{3}\)

II.

Question 1.
Prove that 2sin^-1\(\left(\frac{3}{5}\right)\) = tan^-1\(\left(\frac{24}{7}\right)\)
Solution:
Put sin^-1 \(\left(\frac{3}{5}\right)\) = α

Now to show 2α = tan^-1 \(\frac{24}{7}\) we know that tan 2α = tan^-1\(\frac{24}{7}\)
Now sin α = \(\frac{3}{5}\). Hence tan α = \(\frac{3}{4}\)
∴ tan 2α = \(\frac{2 \tan \alpha}{1-\tan ^2 \alpha}=\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^2}\right)=\frac{3}{2} \times \frac{16}{7}=\frac{24}{7}\)
Hence proved.

Question 2.
Prove that sin^-1\(\frac{8}{17}\) + sin^-1\(\frac{3}{5}\) = tan^-1\(\frac{77}{36}\)
Solution:
Put sin<sup-1\(\frac{8}{17}\) = α and sin^-1\(\frac{3}{5}\) = β
Now to show α + β = tan^-1\(\frac{77}{36}\) we show that tan(α + β) = \(\frac{77}{36}\)

Now sin^-1 \(\frac{8}{17}\) = α ⇒ sin α = \(\frac{8}{17}\) ⇒ tan α = \(\frac{8}{15}\) …………. (1)
Now sin^-1 \(\frac{3}{5}\) = β ⇒ sin β = \(\frac{3}{5}\) ⇒ tan β = \(\frac{3}{4}\) …………. (2)
From (1) & (2), tan(α + β) = \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \cdot \frac{3}{4}}=\left[\frac{\frac{32+45}{60}}{\frac{60-24}{60}}\right]=\frac{77}{36}\)
Hence proved.

Question 3.
Prove that + cos^-1\(\frac{4}{5}\) + cos^-1\(\frac{12}{13}\) = cos^-1\(\frac{33}{65}\)
Solution:
Put cos^-1\(\left(\frac{4}{5}\right)\) = α, cos^-1\(\left(\frac{12}{13}\right)\) = β
Now to show α + β = cos^-1\(\left(\frac{33}{65}\right)\) we show that cos(α + β) = \(\frac{33}{65}\)

Now cos^-1\(\left(\frac{4}{5}\right)\) = α ⇒ cos α = \(\frac{4}{5}\) ⇒ sin α = \(\frac{3}{5}\)
Now cos^-1\(\left(\frac{12}{13}\right)\) = β ⇒ cos β = \(\frac{12}{13}\) ⇒ sin β = \(\frac{5}{13}\)
Now cos (α + β) = cosαcosβ – sinαsinβ = \(\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)-\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)=\frac{48-15}{65}=\frac{33}{65}\)
Hence, proved.

Question 4.
Prove that + cos^-1\(\frac{12}{13}\) + sin^-1\(\frac{3}{5}\) = sin^-1\(\frac{56}{65}\)
Solution:
Put cos^-1\(\left(\frac{12}{13}\right)\) = α, sin^-1\(\left(\frac{3}{5}\right)\) = β
Now to show α + β = sin^-1\(\left(\frac{56}{65}\right)\) we show that sin(α + β) = \(\frac{56}{65}\)

Now cos^-1\(\left(\frac{12}{13}\right)\) = α ⇒ cos α = \(\left(\frac{12}{13}\right)\) ⇒ sin α = \(\frac{5}{13}\)
Now sin^-1\(\left(\frac{3}{5}\right)\) = β ⇒ sin β = \(\left(\frac{3}{5}\right)\) ⇒ cos β = \(\frac{4}{5}\)
Now sin (α + β) = sinαcosβ – cosαsinβ = \(\left(\frac{5}{13}\right)\left(\frac{4}{5}\right)+\left(\frac{12}{13}\right)\left(\frac{3}{5}\right)=\frac{20+36}{65}=\frac{56}{65}\)
Hence, proved.

Question 5.
Prove that + tan^-1\(\frac{63}{16}\) = sin^-1\(\frac{5}{13}\) = cos^-1\(\frac{3}{5}\)
Solution:
Put sin^-1\(\left(\frac{5}{13}\right)\) = α, cos^-1\(\left(\frac{3}{5}\right)\) = β
Now to show α + β = tan^-1\(\left(\frac{63}{16}\right)\) we show that tan(α + β) = \(\frac{63}{16}\)
Now sin^-1\(\left(\frac{5}{13}\right)\) = α ⇒ sin α = \(\left(\frac{5}{13}\right)\) ⇒ tan α = \(\frac{5}{12}\)
Also cos^-1\(\left(\frac{3}{5}\right)\) = β ⇒ cos β = \(\left(\frac{3}{5}\right)\) ⇒ tan β = \(\frac{4}{3}\)
Now tan (α + β) = \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\left(\frac{5}{12}\right)\left(\frac{4}{3}\right)}=\left[\frac{\frac{15+48}{36}}{\frac{36-20}{36}}\right]=\frac{63}{16}\)
∴ tan^-1\(\frac{63}{16}\) = sin^-1\(\frac{5}{13}\) + cos^-1\(\frac{3}{5}\)
Hence, proved.

Question 6.
Prove that sin^-1\(\frac{4}{5}\) + sin^-1\(\frac{7}{25}\) = sin^-1\(\frac{117}{125}\)
Solution:
Put sin^-1\(\frac{4}{5}\) = α and sin^-1\(\frac{7}{25}\) = β
Required To Prove (RTP): α + β = sin^-1\(\frac{117}{125}\) ⇒ sin(α + β) = \(\frac{117}{125}\)

sin^-1\(\frac{4}{5}\) = α ⇒ sinα = \(\frac{4}{5}\) ⇒ cosα = \(\frac{3}{5}\)
sin^-1\(\frac{7}{25}\) = β ⇒ sinβ = \(\frac{7}{25}\) ⇒ cosβ = \(\frac{24}{25}\)
∴ sin(α + β) = sinαcosβ + cosαsinβ = \(\frac{4}{5} \times \frac{24}{25}+\frac{3}{5} \times \frac{7}{25}=\frac{96+21}{125}=\frac{117}{125}\)
Hence proved

Question 7.
Prove that cot^-19 + cosec^-1\(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:
Put cot^-19 = α and cosec^-1\(\frac{\sqrt{41}}{4}\) = β
Required To Prove (RTP): α + β = \(\frac{\pi}{4}\) (or) tan(α + β) = 1

Now cot^-19 = α ⇒ cot α = 9 ⇒ tan α = \(\frac{1}{9}\)
Also, cosec^-1\(\frac{\sqrt{41}}{4}\) = β ⇒ cosecβ = \(\frac{\sqrt{41}}{4}\) ⇒ tan β = \(\frac{4}{5}\)
Now, tan(α + β) = \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9} \cdot \frac{4}{5}}=\frac{\frac{41}{45}}{\frac{41}{45}}=1=\tan \frac{\pi}{4}\)
Hence proved

Question 8.
Prove that sin^-1\(\left(\frac{3}{5}\right)\) + sin^-1\(\left(\frac{8}{17}\right)\) = cos^-1\(\left(\frac{36}{85}\right)\)
Solution:
Put sin^-1\(\frac{3}{5}\) = α and sin^-1\(\frac{8}{17}\) = β
Required To Prove (RTP): α + β = cos^-1\(\frac{36}{85}\) ⇒ cos(α + β) = \(\frac{36}{85}\)

sin^-1\(\frac{3}{5}\) = α ⇒ sinα = \(\frac{3}{5}\) ⇒ cosα = \(\frac{4}{5}\)
sin^-1\(\frac{8}{17}\) = β ⇒ sinβ = \(\frac{8}{17}\) ⇒ cosβ = \(\frac{15}{17}\)
∴ cos(α + β) = cosαcosβ – sinαsinβ = \(\frac{4}{5} \times \frac{15}{17}-\frac{3}{5} \times \frac{8}{17}=\frac{60-24}{85}=\frac{36}{85}\)
Hence proved

Question 9.
Prove that sin^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{12}{13}\right)\) = cos^-1\(\left(\frac{33}{65}\right)\)
Solution:
Put sin^-1\(\frac{3}{5}\) = α and cos^-1\(\frac{12}{13}\) = β
Required To Prove (RTP): α + β = cos^-1\(\frac{33}{65}\) ⇒ cos(α + β) = \(\frac{33}{65}\)

Then sinα = \(\frac{3}{5}\) and cosβ = \(\frac{12}{13}\)
∴ cosα = \(\frac{4}{5}\), sinβ = \(\frac{5}{13}\)
Now cos(α + β) = cosαcosβ – sinαsinβ = \(\frac{4}{5} \times \frac{12}{13}-\frac{3}{5} \times \frac{5}{13}=\frac{48-15}{13}=\frac{33}{65}\)
Hence proved

Question 10.
Prove that tan^-1 \(\sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\), x ∈ [0, 1]
Solution:
Put \(\sqrt{x}\) = tan θ
then x = tan² θ. Also \(\sqrt{x}\) = tan θ ⇒ θ = tan^-1\(\sqrt{x}\)
∴ \(\left(\frac{1-x}{1+x}\right)=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\) = cos 2θ
Now R.H.S = \(\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan ^{-1} \sqrt{x}\) = L.H.S
Hence Proved

Question 11.
Prove that cot^-1 \(\left(\frac{\sqrt{1+\sin \mathrm{x}}+\sqrt{1-\sin \mathrm{x}}}{\sqrt{1+\sin \mathrm{x}}-\sqrt{1-\sin \mathrm{x}}}\right)=\frac{\mathrm{x}}{2}, \mathrm{x} \in\left(0, \frac{\pi}{4}\right)\)
Solution:

Question 12.
Prove that tan^-1\(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1\)
Solution:

Question 13.
Solve 2tan^-1 (cos x) = tan^-1 (2 cosec x)
Solution:
Given that 2tan^-1 (cos x) = tan^-1 (2 cosec x). We know, 2 tan^-1 = tan^-1\(\left(\frac{2 x}{1-x^2}\right)\)
∴ 2 tan^-1 (cos x) = tan^-1\(\left(\frac{2 \cos x}{1-\cos ^2 x}\right)\)
∴ tan^-1\(\left(\frac{2 \cos x}{1-\cos ^2 x}\right)\) = tan^-1(2cosec x) ⇒ \(\left(\frac{2 \cos x}{1-\cos ^2 x}\right)\) = 2 cosec x
⇒ \(\frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x}\) ⇒ cot x = 1 ⇒ tan x = 1 ⇒ tan x = tan \(\frac{\pi}{4}\)
∴ x = nπ + \(\frac{\pi}{4}\), where n ∈ Z [∵ tan θ = tan α ⇒ θ = nπ + α, n ∈ Z]

Question 14.
Solve tan^-1\(\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\) tan^-1x. (x > 0)
Solution:
We know that tan^-1x – tan^-1y = tan^-1\(\left(\frac{x-y}{1+x y}\right)\)
⇒ tan^-1\(\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\) tan^-1x ⇒ tan^-11 – tan^-1x = \(\frac{1}{2}\)tan^-1x
⇒ tan^-11 = tan^-1x + \(\frac{1}{2}\)tan^-1x + \(\frac{3}{2}\)tan^-1x
⇒ \(\frac{\pi}{4}\) = \(\frac{3}{2}\)tan^-1x ⇒ tan^-1x = \(\frac{\pi}{6}\) ⇒ x = tan\(\frac{\pi}{6}\)
⇒ x = \(\frac{1}{\sqrt{3}}\)

Question 15.
Solve arc sin\(\left(\frac{5}{x}\right)\) + arc sin\(\frac{12}{x}=\frac{\pi}{2}\) (x > 0)
Solution:
Given that arc sin\(\left(\frac{5}{x}\right)\) + arc sin\(\frac{12}{x}=\frac{\pi}{2}\) ⇒ sin^-1\(\frac{5}{x}\) + sin^-1\(\frac{12}{x}=\frac{\pi}{2}\)
Put sin^-1\(\frac{5}{x}\) = α and sin^-1\(\frac{12}{x}\) = β ⇒ sin α = \(\frac{5}{x}\) and sin β = \(\frac{12}{x}\)
Now, α + β = \(\frac{\pi}{2}\) α = \(\frac{\pi}{2}\) – β ⇒ sin α = sin (\(\frac{\pi}{2}\) – β) = cos β = \(\sqrt{1-\sin ^2 \beta}\)
On Squaring, we get sin² α = 1 = sin²β
⇒ \(\left(\frac{5}{x}\right)^2=1-\left(\frac{12}{x}\right)^2 \Rightarrow\left(\frac{5}{x}\right)^2=\frac{x^2-144}{x^2} \Rightarrow \frac{25}{x^2}=\frac{x^2-144}{x^2}\)
⇒ x² = 169
⇒ x = ± 13
But x = -13 does not satisfy the given equation
∴ x = 13 is only solution of the given equation

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

AP Inter 2nd Year Maths Inverse Trigonometric Functions Solutions Exercise 2b

I.

Question 1.
Prove that 3sin^-1x = sin^-1(3x – 4x³), x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
Solution:
Put x = sinθ ⇒ θ = sin^-1x
Now RHS = sin^-1(3x – 4x³) = sin^-1(3sinθ – 4sin³θ) = sin^-1(sin 3θ)
= 3θ = 3 sin^-1x = LHS

Question 2.
Prove that 3cos^-1x = cos^-1 (4x³ – 3x), x ∈ [\(\frac{1}{2}\), 1]
Solution:
Put x = cosθ ⇒ θ = cos^-1x
Now RHS = cos^-1 (4x³ = 3x) = cos^-1(4cos³θ – 3cosθ)
= cos^-1(cos 3θ) = 3θ = 3 cos^-1x = LHS [∵ cos^-1(cosθ) = θ]

Question 3.
Find the value of sin^-1 \(\left(\sin \frac{2 \pi}{3}\right)\).
Solution:
Here 2π/3 = 120° is not in the P.V range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) of sin^-1x
∴ \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3} \text { and } \frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Question 4.
Find the value of tan^-1 \(\left(\tan \frac{3 \pi}{4}\right)\).
Solution:
Here 3π/4 = 135° is not in the P.V range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) of sin^-1x
∴ \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]=\tan ^{-1}\left[-\tan \left(\frac{\pi}{4}\right)\right]\)
= \(-\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \text { and }-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

II.

Question 1.
Find the value of tan(sin^-1\(\frac{3}{5}\) + cot^-1\(\frac{3}{2}\))
Solution:
Put sin^-1\(\frac{3}{5}\) = θ ⇒ sin θ = \(\frac{3}{5}\). Hence tan θ = \(\frac{3}{4}\).
∴ sin^-1\(\frac{3}{5}\) = tan^-1\(\frac{3}{4}\) …………… (1)
cot^-1\(\frac{3}{2}\) = tan^-1\(\frac{2}{3}\) …………… (2)
∴ tan (sin^-1\(\frac{3}{5}\) + cot^-1\(\frac{3}{2}\)) = tan(tan^-1\(\frac{3}{4}\) + tan^-1\(\frac{2}{3}\))
= \(\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right]=\tan \left(\tan ^{-1} \frac{17}{6}\right)=\frac{17}{6}\) [∵ tan^-1x + tan^-1y = \(\tan ^{-1}\left(\frac{x+y}{1-x y}\right) .\)]

Question 2.
Write the function tan^-1\(\), x ≠ 0 in the simplest form.
Solution:
Put x = tanθ ⇒ θ = tan^-1x
\(\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= \(\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x\) [∵ 1 – cos 2θ = 2 sin² θ; sin 2θ = 2 cos θ sin θ]

Question 3.
Write the function tan^-1\(\left(\sqrt{\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}}\right)\), 0 < x < π in the simplest form
Solution:
We know that 1 – cosx = 2 sin²\(\frac{x}{2}\) and 1 + cosx = 2 cos²\(\frac{x}{2}\). Also tan^-1(tan θ) = θ
∴ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}}\right)=\tan ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}\)

Question 4.
Write the function tan^-1\(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\), \(\frac{-\pi}{4}\) < x < \(\frac{3\pi}{4}\) in the simplest form
Solution:
G.E = \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\tan ^{-1}\left(\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{}}\right)=\tan ^{-1}\left(\frac{\frac{\cos x}{\sin x}}{\cos x}\right)\)
= \(\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)\) = tan^-1 (1) – tan^-1 (tan x) = \(\frac{\pi}{4}\) – x [∵ tan^-1\(\left(\frac{1-x}{1+x}\right)\) = tan^-1 1 – tan^-1 2]

Question 5.
Write the function tan^-1\(\frac{x}{\sqrt{a^2-x^2}}\), |x| < a in the simplest form.
Solution:
Put x = a sin θ ⇒ sin θ = \(\frac{\mathrm{x}}{\mathrm{a}}\) ⇒ θ = sin^-1 \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\)
∴ \(\tan ^{-1} \frac{x}{\sqrt{a^2-x^2}}=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right)=\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^2 \theta}}\right)\)
= \(\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)=\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1}\left(\frac{x}{a}\right)\)

Question 6.
Write the function tan^-1\(\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)\), a > 0; \(\frac{-a}{\sqrt{3}}\) < x < \(\frac{a}{\sqrt{3}}\) in the simplest form.
Solution:
Put x = a tan θ ⇒ tan θ = \(\frac{\mathrm{x}}{\mathrm{a}}\) ⇒ θ = tan^-1 \(\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\)
∴ \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=\tan ^{-1}\left(\frac{3 a^2 \cdot(a \tan \theta)-a^3 \tan ^3 \theta}{a^3-3 a \cdot\left(a^2 \tan ^2 \theta\right)}\right)\)
= \(\tan ^{-1}\left(\frac{3 a^3 \tan \theta-a^3 \tan ^3 \theta}{a^3-3 a^3 \tan ^2 \theta}\right)=\tan ^{-1}(\tan 3 \theta)=3 \theta=3 \tan ^{-1}\left(\frac{x}{a}\right)\)

Question 7.
Find the value of tan^-1\(\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)
Solution:
\(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]=\tan ^{-1}\left[2 \cos \left(2 \cdot \frac{\pi}{6}\right)\right]=\tan ^{-1}\left[2\left(\frac{1}{2}\right)\right]=\tan ^{-1} 1=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}\)

Question 8.
Find the value of tan \(\frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]\), |x| < 1, y > 0 and xy < 1
Solution:
Put x = tan θ ⇒ θ = tan^-1 x
Now sin^-1\(\frac{2 x}{+x^2}\) = sin^-1\(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = sin^-1 (sin 2θ) = 2θ = 2 tan^-1 x
Put y = tan Φ ⇒ Φ = tan^-1 y
Now \(\cos ^{-1} \frac{1-y^2}{1+y^2}=\cos ^{-1}\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)=\cos ^{-1}(\cos 2 \phi)=2 \phi=2 \tan ^{-1} y\)
∴ \(\tan \frac{1}{2}\left(\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right)=\tan \frac{1}{2}\left(2 \tan ^{-1} x+2 \tan ^{-1} y\right)\)
= \(\tan \left(\tan ^{-1} x+\tan ^{-1} y\right)=\tan \left[\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]=\frac{x+y}{1-x y}\)

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

AP Inter 2nd Year Maths Inverse Trigonometric Functions Solutions Exercise 2a

I.

Question 1.
Find the principal value of sin^-1\(\left(-\frac{1}{2}\right)\)
Solution:
We know that sin^-1(-x) = -sin^-1x
∴ \(\sin ^{-1}\left(\frac{-1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)=-\sin ^{-1}\left[\sin \left(\frac{\pi}{6}\right)\right]=-\frac{\pi}{6}\) Here \(-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Question 2.
Find the principal value of cos^-1\(\left(\frac{\sqrt{3}}{2}\right)\)
Solution:
\(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\) Here \(\frac{\pi}{6}\) ∈ [0, π]

Question 3.
Find the principal value of cosec^-1 (2)
Solution:
cosec^-1 (2) = cosec^-1[cosec\(\left(\frac{\pi}{6}\right)\)] = \(\frac{\pi}{6}\) Here \(\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

Question 4.
Find the principal value of tan^-1 (-\(\sqrt{3}\))
Solution:
We know that tan^-1(-x)= -tan^-1 x
tan^-1 (-\(\sqrt{3}\)) = -tan^-1 (\(\sqrt{3}\)) = \(-\tan ^{-1}\left(\tan \frac{\pi}{3}\right)=-\frac{\pi}{3}\) Here \(-\frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Question 5.
Find the principal value of cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:
We know that cos^-1(-x) = π = cos^-1x
\(\cos ^{-1}\left(\frac{-1}{2}\right)=\pi-\cos ^{-1}\left(\frac{1}{2}\right)=\pi-\cos ^{-1}\left[\cos \left(\frac{\pi}{3}\right)\right]=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}\) Here \(-\frac{\pi}{3} \in[0, \pi]\)

Question 6.
Find the principal value of tan^-1 (-x)
Solution:
We know that tan^-1 (-x) = -tan^-1 x
tan^-1 (-1) = -tan^-1 1 = \(-\tan ^{-1}\left[\tan \left(\frac{\pi}{4}\right)\right]=-\frac{\pi}{4}\) Here \(-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Question 7.
Find the principal value of sec^-1 \(\left(\frac{2}{\sqrt{3}}\right)\)
Solution:
\(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\sec ^{-1}\left(\sec \frac{\pi}{6}\right)=\frac{\pi}{6}\) Here \(\frac{\pi}{6} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}\)

Question 8.
Find the principal value of cot^-1 (\(\sqrt{3}\))
Solution:
cot^-1 (\(\sqrt{3}\)) = \(\cot ^{-1}\left[\cot \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\) Here \(\frac{\pi}{6}\) ∈ (0, π)

Question 9.
Find the principal value of cos^-1 \(\left(-\frac{1}{\sqrt{2}}\right)\)
Solution:
We know that cos^-1(-x) = π – cos^-1x
\(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\pi-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\pi-\cos ^{-1}\left[\cos \left(\frac{\pi}{4}\right)\right]=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}\) Here \(\frac{3 \pi}{4}\) ∈ [0, π]

Question 10.
Find the principal value of cosec^-1 (-\(\sqrt{2}\))
Solution:
We know that cosec^-1(-x) = cosec^-1x
cosec^-1 (-\(\sqrt{2}\)) = -cosec^-1 (\(\sqrt{2}\)) = -cosec^-1[cosec\(\left(\frac{\pi}{4}\right)\)] = \(-\frac{\pi}{4}\) Here \(-\frac{\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

Question 11.
Find the value of \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)
Solution:
\(\cos ^{-1}\left[\cos \frac{\pi}{3}\right]+2 \sin ^{-1}\left[\sin \frac{\pi}{6}\right]=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}\) Here \(\frac{2 \pi}{3}\) ∈ [0, π]

II.

Question 1.
Find the value of tan^-1(1) + cos^-1\(\left(-\frac{1}{2}\right)+\) + sin^-1\(\left(-\frac{1}{2}\right)+\)
Solution:
We know that cos^-1 (-x) = π – cos^-1 x and sin^-1(-x) = -sin^-1 x
∴ \(\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)=\tan ^{-1}(1)+\left[\pi-\cos ^{-1}\left(\frac{1}{2}\right)\right]-\sin ^{-1}\left(\frac{1}{2}\right)\)
= \(\frac{\pi}{4}+\left(\pi-\frac{\pi}{3}\right)-\frac{\pi}{6}\) = 45° + 180° – 60° – 30° = 135° = \(\frac{3 \pi}{4}\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 1 Relations and Functions Exercise 1b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Relations and Functions Solutions Exercise 1b

I.

Question 1.
Check the injectivity and surjectivity of the function f : N → N given by f (x) = x²
Solutions:
f : N → N; f (x) = x²
∴ f is injective, (one-one)
∴ f is not surjective(onto)
(i) f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = x₂
(ii) For 2 ∈ N we can’t find in x in N such that f(x) = x² = 2
Aliter: We know N= {1, 2, 3, ….} and f (x) = x²
(i) f(1) = 1² = 1; f(2) = 2² = 4; f(3) = 3² = 9;
∴ f = {(1, 1), (2, 4), (3, 9), ….} ………….. (1)
Here, distinct elements of the domain N have distinct images.
∴ f is injective.

ii) From(1), Range off = {1, 4, 9, ….}
(set of all second elements in the ordered pairs)
But codomain N= {1, 2, 3, 4, ….}
Thus, Range set ≠ Codomain N.
∴ f is not surjective
Given function f is injective but not surjective.

Question 2.
Check the injectivity and surjectivity of the function f : Z → Z given by f (x) = x²
Solution:
f : Z → Z; f (x) = x²
(i) f(x₁) = f(x₂) ⇒ x₁² = x₁² ⇒ x₁= ±x₂
(ii) For 2 ∈ Z we can’t find in x in Z such that f(x) = x² = 2
Aliter: Now f(1) = 1² = 1; f(-1) = (-1)² = 1;
Thus, distinct elements 1, -1 of the domain Z have the same image 1.
∴ f is not injective,

ii) Range off = {0, 1, 4, 9, , …}
But codomain Z= {…-3, -2, -1, 0, 1, 1.5 2, 3, 4, ….};
Thus, Range set ≠ Codomain Z.
∴ f is not surjective
Given function f is not injective and not surjective.

Question 3.
Check the injectivity and surjectivity of the function f : R → R given by f (x) = x²
i) f: R → R given by f (x) = x²
f(1) = 1² = 1; f(-1) = (-1)² = 1;
Here, distinct elements 1, -1 of the domain R have the same image 1.
∴ f is not injective.

ii) For -2 ∈ R we can’t find in x in R such that f(x) = x² = -2
∴ f is not surjective
Given function f is not injective and not surjective.

Question 4.
Check the injectivity and surjectivity of the function f : N → N given by f (x) = x³
Solution:
f: N → N; f (x) = x³
(i) f (x₁) = f (x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂
∴ f is injective.
(ii) For 2 ∈ N we can’t find in x in N such that f(x) = x³ = 2
∴ f is not surjective
Aliter: We know N= {1, 2, 3, 4, ….} and f(x) = x³

i) f(1) = 1³ = 1; f(2) = 2³ = 8; f(3) = 3³ = 27;
∴ f{(1, 1), (2, 8), (3, 27), …………} ………….. (1)
Here, distinct elements of the domain N have distinct images.
∴ f is injective.
ii) From (1) Range of f contains only the cube values = {1, 8, 27, ….}
But codomain N = {1, 2, 3, 4, ….}
Thus, Range set ≠ Codomain N.
∴ f is not surjective
Given function f is injective but not surjective.

Question 5.
Check the injectivity and surjectivity of the function f: Z → Z given by f (x) = x³
Solution:
f : Z → Z; f(x) = x³
(i) f(x₁) = f(x₂) = x₁³ = x₂³ = x₁ = x₂
∴ f is injective.
(ii)For 2 ∈ Z we can’t find in x in Z such that f(x) = x³ = 2
∴ f is not swje1ctive

Question 6.
Prove that the Greatest Integer Function f : R → R, given by f (x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution:
f: R → R is given by f(x) = [x]= Greatest integer ≤ x

Now,f(0) = [0] = 0; f(1) = [1] = 1;
f(1.2) = [1.2] = 1, f(2) = [2] = 2;
f(-o.5) = [-0.5] = -1; f(-1.1) = [-1.1] = -2;
(i) Thus, we have f(1) = 1, f(1.2) = 1.
Thus distinct elements R have the same image.
∴ f is not one-one.

(ii) Thus, we see that the range contains only the integers f={…., -2, -1, 0, 1, 2, 3, ….}= Z
Thus, Range Z ≠ Codomain R.
∴ f is not onto.
∴ The greatest integer function is neither one-one nor onto.

Question 7.
Show that the Modulus Function f : R → R, given by f (x) = | x |, is neither one- one nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative.
Solution:
f: R → R is given as f(x) = |x| =
(i) f(-1) = |-1| = 1 and f(1) = |1| = 1
Thus distinct elements -1, 1 of the domain R have the same image 1.
∴ f is not one-one.

(ii) Range of |x| is [0, ∞]; [ This range donot contain any negative reals)
Thus, range ≠ codomain R
∴ f is not onto.
∴ The modulus function is neither one-one nor onto.

Question 8.
Show that the Signum Function f : R → R, given by

Solution:
(i) From the given function we have f(1) = 1 and f(2) = 1 [∵ f(x) = 1 if x > 1]
Thus distinct elements of R have the same image.
∴ f is not one-one.

(ii) From the given function f we see that the Range of f(x)is {1, 0, -1}.But codomain is R.
Thus, range ≠ codomain R
∴ f is not onto.
∴ The signum function is neither one-one nor onto.

Question 9.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Solution:
Given that A = {1, 2, 3} , B = {4, 5, 6, 7}
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}
∴ f(1) = 4, f(2) = 5, f(3) = 6
Thus, distinct elements of A have distinct images in B under f.
∴ f is one-one.

Question 10.
Define One-one (Injective) function
Solution:
One-one (Injective) function : A function f: A → B is said to be a one one function if x₁, x₂ ∈ A
be such that f(x₁)= f(x₁) ⇒ x₁= x₂

Question 11.
Define onto(Surjective) function
Solution:
Onto(Surjective) function: A function f: A → B is said to be an onto function if its
Range equals to Codomain. [f(A)= B] (or) .
A function f: A → B is an onto function if ∀ y ∈ B ∃ x ∈ A such that f(x) = y

Question 12.
Define bijective function.
Solution:
Bijective function : A function which is both one-one and onto then f is called bijective.

II.

Question 1.
Show that the function f: R → R„ defined by f(x) = \(\frac{1}{x}\) is one-one and onto, where R_* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R_*?
Solution:
a) When f: R_* → R_* is given as f(x) = —. Let x₁, x₂ ∈ R_*
(i) f(x₁) = f(x₂) ⇒ \(\frac{1}{x_1}=\frac{1}{x_2}\) ⇒ x₁ = x₂
∴ f is one-one.

(ii) Given that f(x) = \(\frac{1}{x}\) …………. (1)
Let f(x) = y ⇒ y = \(\frac{1}{x}\) ⇒ x = \(\frac{1}{y}\), exists ∀ y ≠ 0 [ as y ∈ R_*, set of non-zero reals]
∴ f is onto.
Verification: From (1), f (x) = \(\frac{1}{x}\) = \(\frac{1}{\left(\frac{1}{y}\right)}\) = y
∴ Given function f is one-one and onto.

b) We take g: N → R_* as g(x) = \(\frac{1}{x}\). Let x₁, x₂ ∈ N
(i) g(x₁) = g(x₂) ⇒ \(\frac{1}{x_1}=\frac{1}{x_2}\) ⇒ x₁ = x₂
∴ g is one-one.

(ii) For some 1.5 ∈ R_* (codomain) there is no x in N such that
g(x) = \(\frac{1}{1.5}\) {∵ \(\frac{1}{1.5}\) is not a natural}
∴ the other function g is one-one but not onto.

Question 2.
State whether the function f : R → R defined by f (x) = 3 – 4x is one-one, onto or bijective. Justify your answer.
Solution:
f: R. → R is defined as f(x) = 3 – 4x. Let x₁, x₂ ∈ R
(i) f(x₁) = f(x₂) ⇒ 3 – 4x₁ = 3 – 4x₂ ⇒ -4x₁ = -4x₂ ⇒ x₁ = x₂
∴ f is one-one.

(ii) Given that f (x) = 3 – 4x (1); Put f(x) = y
⇒ y = 3 – 4x ⇒ 4x = 3 – y ⇒ x = \(\frac{3-y}{4}\) exists ∀ y ∈ R( codomain)
∴ f is onto.
Justification: From (1), f(x) = \(\mathrm{f}\left(\frac{3-\mathrm{y}}{4}\right)=3-4\left(\frac{3-\mathrm{y}}{4}\right)\) = 3 – (3 – y) = y
Hence, f is both one-one and onto hence bijective.

Question 3.
State whether the function f : R → R defined by f (x) = 1 + x² is one-one, onto or bijective. Justify your answer.
Solution:
f: R → R is defined as f (x) = 1 + x². Let x₁, x₂ ∈ R
(i) Take two elements 1, -1 in the domain
Now f(1) = 1 + 1² = 2; f(-1) = 1 + (-1) = 1 + 1 = 2;
Two distinct elements in the domain have the same image.
∴ f is not one-one.

(ii) From the given function f (x) = 1 + x² it can be seen that Range of f is always positive.
So negative elements in the codomain R like -2 do not have any preimage in the
domain R such that f(x) = 1 + x²
∴ f is not onto.
Hence, f is neither one-one nor onto.

Question 4.
Let A and B be sets. Show that f: A × B → B × A such that f (a, b) = (b, a) is bijective function.
Solution:
f : A × B → B × A is defined as(a, b) = (b, a).
Let (a₁, b₁), (a₂, b₂) ∈ A × B
(i) f(a₁, b₁)= f(a₂, b₃) (b₁, a₁) = (b₂, a₂)
⇒ (b₂, a₂) = (b₁, a₁) (a₁, b₁)=(a₂, b₂)
∴ f is one—one.

(ii) ∀(b, a) ∈ B × A there exist (a, b) ∈ A × B such that f(a, b) = (b, a)
∴ f is onto.
Here, f is both one-one and onto hence bijective.

Question 5.
Let f : N → N be defined by

for all n ∈ N
State whether the function f is bijective. Justify your answer.
Solution:
Case i: For odd n = 1, 3, 5, …. we have f(n) = \(\frac{\mathrm{n}+1}{2}\)
∴ f(1) = \(\frac{1+1}{2}\) = 1; f(3) = \(\frac{3+1}{2}\) = 2; f(5) = \(\frac{5+1}{2}\) = 3;
⇒ f = {(1, 1), (3, 2), (5, 3), …………} ……. (1)

Case ii: For even n = 2, 4, 6, …. we have f(n) = \(\frac{\mathrm{n}}{2}\)
∴ f(2) = \(\frac{2}{2}\) = 1; f(4) = \(\frac{4}{2}\) = 2; f(6) = \(\frac{6}{2}\) = 3; ⇒ {(2, 1), (4, 2), (6, 3), }….. (2)
From (1) and (2) we get f= {(1, 1), (2, 1), (3, 2), (4, 2), (5, 3), (6, 3), ….} (3)
a) From (3), we have f(1) = 1 and f(2) = 1
Thus distinct elements 1, 2 of the domain N have the same image 1.
∴ f is not one-one

b) From (3) the range set = {1, 1, 2, 2, 3, 3, } = {1, 2, 3, ….}
⇒ Range of = {1, 2, 3, }= codomain N
∴ f is onto.
∴ f is onto but not one-one. Hence, f is not a bijective function.

Question 6.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by f(x) = \(\left(\frac{x-2}{x-3}\right)\) Is f one-one and onto? Justify your answer.
Solution:
Given that f: A-+B is defined as f (x) = \(\left(\frac{x-2}{x-3}\right)\) for A = R -{3}, B = R -{1}
a) Take x₁, x₂ ∈ A such that f(x₁) = f(x₂) ⇒ \(\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}\) ⇒ (x₁ -2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 3x₂ – 2x₁ + 6
⇒ -3x₁ – 2x₂ = -3x₂ – 2x₁
⇒ 3x₁ – 2x₁ = 3x₂ – 2x₂
⇒ x₁ = x₂
∴ f is one-one.

b) Given that f(x) = \(\left(\frac{x-2}{x-3}\right)\) …………. (1). Put f(x) = y
⇒ y = \(\left(\frac{x-2}{x-3}\right)\) ⇒ x – 2 = xy – 3y ⇒ x(1 – y) = -3y + 2 ⇒ x = \(\frac{2-3 y}{1-y}\) exists ∀ y ∈ B = R-{1}
∴ f is onto.