AP Inter 2nd Year Maths Exercise 4d Solutions

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)