Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 1

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Question 1. Find the adjoint of the following matrices:

Verify that (adj A)A = |A|I = A(adj A) for the above matrices:

(i) $\begin{bmatrix}-3&5\\2&4\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}-3&5\\2&4\end{bmatrix}$
Cofactors of A are:
C₁₁= 4 C₁₂= -2
C₂₁= -5 C₂₂= -3
adj A = $\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T$
(adj A) = $\begin{bmatrix}4&-2\\-5&-3\end{bmatrix}^T$
= $\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
(adj A)A = $\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}\begin{bmatrix}-3&5\\2&4\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}$
|A|I = $\begin{bmatrix}-3&5\\2&4\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix} =(-22)\begin{bmatrix}1&0\\0&1\end{bmatrix}$ = $\begin{bmatrix}-22&0\\0&-22\end{bmatrix}$
A(adj A) = $\begin{bmatrix}-3&5\\2&4\end{bmatrix}\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}=\begin{bmatrix}-22&0\\0&-22\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

(ii) $\begin{bmatrix}a&b\\c&d\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}a&b\\c&d\end{bmatrix}$
Cofactors of A are:
C₁₁= d C₁₂= -c
C₂₁= -b C₂₂= a
(adj A) = $\begin{bmatrix}d&-c\\-b&-a\end{bmatrix}^T$
= $\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
(adj A)A = $\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ad-bc&bd-bd\\-ac+ac&-bc+ad\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}$
|A|I = $\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix} =(ad-bc)\begin{bmatrix}1&0\\0&1\end{bmatrix} =\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}$
A(adj A) = $\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

(iii) $\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}$
Cofactors of A are:
C₁₁= cos α C₁₂= -sin α
C₂₁= -sin α C₂₂= cos α
(adj A) = $\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}^T$
= $\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
(adj A)A = $\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}$
= $\begin{bmatrix}cos^2α-sin^2α&cosαsinα-sinαcosα\\-cosαsinα+sinαcosα&-sin^2α+cos^2α\end{bmatrix}$
= $\begin{bmatrix}cos 2α&0\\0&cos2α\end{bmatrix}$
|A|I = $\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
= $(cos^2α-sin^2α)\begin{bmatrix}1&0\\0&1\end{bmatrix}$
= $\begin{bmatrix}cos^2α-sin^2α&0\\0&cos^2α-sin^2α\end{bmatrix}$
= $\begin{bmatrix}cos2α&0\\0&cos2α\end{bmatrix}$
A(adj A) = $\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}\begin{bmatrix}cos α&-sin α\\-sin α&cosα\end{bmatrix}$
= $\begin{bmatrix}cos^2α-sin^2α&-cosαsinα+sinαcosα\\sinαcosα-cosαsinα&-sin^2α+cos^2α\end{bmatrix}$
= $\begin{bmatrix}cos 2α&0\\0&cos2α\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

(iv) $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}$
Cofactors of A are:
C₁₁= 1 C₁₂= -(-tan α/2) = tan α/2
C₂₁= -tan α/2 C₂₂= 1
adj A = $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}^T$
= $\begin{bmatrix}1&-tan α/2\\tan α/2&1\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
|A| = $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}$
= 1 + tan² α/2
= sec²α/2
(adj)A = $\begin{bmatrix}1&-tan α/2\\tan α/2&1\end{bmatrix}\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}$
= $\begin{bmatrix}1+tan^2 α/2&tan α/2-tan α/2\\tan α/2-tan α/2&tan^2 α/2+1\end{bmatrix}$
= $\begin{bmatrix}sec^2 α/2&0\\0&sec^2 α/2\end{bmatrix}$
|A|I = (sec²α/2) $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
= $\begin{bmatrix}sec^2 α/2&0\\0&sec^2 α/2\end{bmatrix}$
A(adj A) = $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}\begin{bmatrix}1&-tan α/2\\tan α/2&1\end{bmatrix}$
= $\begin{bmatrix}1+tan^2 α/2&-tan α/2+tan α/2\\-tan α/2+tan α/2&tan^2 α/2+1\end{bmatrix}$
= $\begin{bmatrix}sec^2 α/2&0\\0&sec^2 α/2\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

Question 2. Compute the adjoint of each of the following matrices:

Verify that (adj A)A = |A|I = A(adj A) for the above matrices:

(i) $\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
Cofactors of A are
C₁₁= $\begin{bmatrix}1&2\\2&1\end{bmatrix}$ = -3
C₂₁= $\begin{bmatrix}2&2\\2&1\end{bmatrix}$ = 2
C₃₁= $\begin{bmatrix}2&2\\1&2\end{bmatrix}$ = 2
C₁₂= $\begin{bmatrix}2&2\\2&1\end{bmatrix}$ = 2
C₂₂= $\begin{bmatrix}1&2\\2&1\end{bmatrix}$ =-3
C₃₂= $\begin{bmatrix}1&2\\2&2\end{bmatrix}$ = 2
C₁₃= $\begin{bmatrix}2&1\\2&2\end{bmatrix}$ = 2
C₂₃= $\begin{bmatrix}1&2\\2&2\end{bmatrix}$ = 2
C₃₃= $\begin{bmatrix}1&2\\2&1\end{bmatrix}$ = -3
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}^T$
= $\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
|A| = -3 + 4 + 4 = 5
(adj A)A = $\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$
|A|I= (5) $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ = $\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$
A(adj A) = $\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

(ii) $\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}$
Cofactors of A are
C₁₁= $\begin{bmatrix}3&1\\1&1\end{bmatrix}$ = 2
C₁₂= $\begin{bmatrix}2&1\\-1&1\end{bmatrix}$ = -3
C₁₃= $\begin{bmatrix}2&3\\-1&1\end{bmatrix}$ = 5
C₂₁= $\begin{bmatrix}2&5\\1&1\end{bmatrix}$ = 3
C₂₂= $\begin{bmatrix}1&5\\-1&1\end{bmatrix}$ = 6
C₂₃= $\begin{bmatrix}1&2\\-1&1\end{bmatrix}$ = -3
C₃₁= $\begin{bmatrix}2&5\\3&1\end{bmatrix}$ = -13
C₃₂= $\begin{bmatrix}1&5\\2&1\end{bmatrix}$ = 9
C₃₃= $\begin{bmatrix}1&2\\2&3\end{bmatrix}$ = -1
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}2&-3&5\\3&6&-3\\-13&9&-1\end{bmatrix}^T$
= $\begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
|A| = 1(3 – 1) – 2(2 + 1) + 5(2 + 3)
= 2 – 6 + 25 = 21
(adj A)A = $\begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}$
|A|I = (21) $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}$
A(adj A) = $\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}\begin{bmatrix}2&3&-13\\-3&6&9\\5&-3&-1\end{bmatrix}=\begin{bmatrix}21&0&0\\0&21&0\\0&0&21\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

(iii) $\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}$
Cofactors of A are
C₁₁= $\begin{bmatrix}2&5\\4&-1\end{bmatrix}$ = -22
C₁₂= – $\begin{bmatrix}4&5\\0&-1\end{bmatrix}$ = 4
C₁₃= $\begin{bmatrix}4&2\\0&4\end{bmatrix}$ = 16
C₂₁= – $\begin{bmatrix}-1&3\\4&-1\end{bmatrix}$ = 11
C₂₂= $\begin{bmatrix}2&3\\0&-1\end{bmatrix}$ = -2
C₂₃= – $\begin{bmatrix}2&-1\\0&4\end{bmatrix}$ = -8
C₃₁= $\begin{bmatrix}-1&3\\2&5\end{bmatrix}$ = -11
C₃₂= – $\begin{bmatrix}2&3\\4&5\end{bmatrix}$ = 2
C₃₃= $\begin{bmatrix}2&-1\\4&2\end{bmatrix}$ = 8
adj A = $\begin{bmatrix}-22&4&16\\11&-2&-8\\-11&2&8\end{bmatrix}^T$
= $\begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
|A| = 2(-2 – 20) + 1(-4 – 0) + 3(16 – 0)
= -44 – 4 + 48 = 0
(adj A)A = $\begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
|A|I = $(0)\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
A(adj A) = $\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}\begin{bmatrix}-22&11&-11\\4&-2&2\\16&-8&8\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

(iv) $\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}$
Cofactors of the A are
C₁₁= $\begin{bmatrix}1&0\\1&3\end{bmatrix}$ = 3
C₁₂= – $\begin{bmatrix}5&0\\1&3\end{bmatrix}$ = -15
C₁₃= $\begin{bmatrix}5&1\\1&1\end{bmatrix}$ = 4
C₂₁= $\begin{bmatrix}0&-1\\1&3\end{bmatrix}$ = -1
C₂₂= $\begin{bmatrix}2&-1\\1&3\end{bmatrix}$ = 7
C₂₃= $\begin{bmatrix}2&0\\1&1\end{bmatrix}$ = -2
C₃₁= $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ = 1
C₃₂= $\begin{bmatrix}2&-1\\5&0\end{bmatrix}$ = -5
C₃₃= $\begin{bmatrix}2&0\\5&1\end{bmatrix}$ = 2
adj A = $\begin{bmatrix}3&-15&4\\-1&7&-2\\1&-5&2\end{bmatrix}^T$
= $\begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}$
To Prove, (adj A)A = |A|I = A(adj A)
|A| = 2(3 – 0) – 0(15 – 0) – 1(5 – 1)
= 6 – 4 = 2
(adj A)A = $\begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
|A|I = (2) $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
A (adj A) = $\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&7&-5\\4&-2&2\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
Therefore, (adj A)A = |A|I = A(adj A)
Hence Proved

Question 3. For the matrix A = $\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}$ , show that A(adj A) = O.

Solution:

Cofactor of A are,
C₁₁= 30 C₁₂ = -20 C₁₃= -50
C₂₁= 12 C₂₂ = -8 C₂₃= -20
C₃₁= -3 C₃₂= 2 C₃₃= 5
adj A = $\begin{bmatrix}30&-20&-50\\12&-8&-20\\-3&2&5\end{bmatrix}^T$
= $\begin{bmatrix}30&12&-3\\-20&-8&2\\-50&-20&5\end{bmatrix}$
A(adj A) = $\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix} \begin{bmatrix}30&12&-3\\-20&-8&2\\-50&-20&5\end{bmatrix}$
= $\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}(0)$
= 0
Hence Proved

Question 4. If A = $\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}$ , show that adj A = A.

Solution:

Here, A = $\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}$
Cofactor of A are,
C₁₁= -4 C₁₂ = 1 C₁₃= 4
C₂₁= -3 C₂₂ = 0 C₂₃= 4
C₃₁= 4 C₃₂= 4 C₃₃= 3
adj A = $\begin{bmatrix}-4&1&4\\-3&0&4\\-3&1&3\end{bmatrix}^T$
= $\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}$
Therefore, adj A = A

Question 5. If A = $\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}$ , show that adj A = 3A^T.

Solution:

Here, A = $\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}$
Cofactor of A are,
C₁₁= -3 C₁₂ = -6 C₁₃= -6
C₂₁= 6 C₂₂ = 3 C₂₃= -6
C₃₁= 6 C₃₂= -6 C₃₃= 3
adj A = $\begin{bmatrix}-3&-6&-6\\6&3&-6\\6&-6&3\end{bmatrix}^T$
= $\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}$
A^T= $\begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix}$
Now, 3A^T = 3 $\begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix}$ = $\begin{bmatrix}-3&6&6\\-6&3&-6\\-6&-6&3\end{bmatrix}$
adj A = 3.A^T
Hence Proved

Question 6. Find A(adj A) for the matrix A = $\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}$ .

Solution:

Here, A = $\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}$
Cofactor of A are,
C₁₁= 9 C₁₂ = 4 C₁₃= 8
C₂₁= 19 C₂₂ = 14 C₂₃= 3
C₃₁= -4 C₃₂= 1 C₃₃= 2
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}9&4&8\\19&14&3\\-4&1&2\end{bmatrix}^T$
= $\begin{bmatrix}9&19&-4\\4&14&1\\8&3&2\end{bmatrix}$
A(adj A) = $\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix} \begin{bmatrix}9&19&-4\\4&14&1\\8&3&2\end{bmatrix}$
= $\begin{bmatrix}25&0&0\\0&25&0\\0&0&25\end{bmatrix}$
= 25 $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
= 25I₃

Question 7. Find the inverse of each of the following matrices:

(i) $\begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}$
|A| = cos²θ + sin²θ = 1
Hence, inverse of A exist
Cofactors of A are,
Cofactor of A are,
C₁₁ = cos θ C₁₂ = sin θ
C₂₁= -sin θ C₂₂ = cos θ
adj A = $\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T$
= $\begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}^T$
= $\begin{bmatrix}cos θ&-sinθ\\sin θ&cos θ\end{bmatrix}$
A^-1= 1/|A|. adj A
=1/1. $\begin{bmatrix}cos θ&-sinθ\\sin θ&cos θ\end{bmatrix}$

(ii) $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}0&1\\1&0\end{bmatrix}$
|A| = -1
Hence, inverse of A exist
Cofactor of A are,
C₁₁= 0 C₁₂ = -1
C₂₁= -1 C₂₂ = 0
adj A = $\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T$
= $\begin{bmatrix}0&-1\\-1&0\end{bmatrix}^T$
= $\begin{bmatrix}0&-1\\-1&0\end{bmatrix}$
A^-1= 1/|A|. adj A
= $\frac{1}{-1}\begin{bmatrix}0&-1\\-1&0\end{bmatrix}$
= $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

(iii) $\begin{bmatrix}a&b\\c&\frac{(a+bc)}{a}\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}a&b\\c&\frac{(a+bc)}{a}\end{bmatrix}$
|A| = a(1 + bc)/a – bc = 1 + bc – bc = 1
Hence, inverse of A exists.
Cofactor of A are,
C₁₁= (1 + bc)/a C₁₂ = -c
C₂₁= -b C₂₂ = a
adj A = $\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T$
= $\begin{bmatrix}\frac{(a+bc)}{a} &-c\\-b&a\end{bmatrix}^T$
= $\begin{bmatrix}\frac{(a+bc)}{a}&-b\\-c&a\end{bmatrix}$
A^-1= 1/|A|. adj A
= 1/1 $\begin{bmatrix}\frac{(a+bc)}{a}&-b\\-c&a\end{bmatrix}$
= $\begin{bmatrix}\frac{(a+bc)}{a}&-b\\-c&a\end{bmatrix}$

(iv) $\begin{bmatrix}2&5\\-3&1\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}2&5\\-3&1\end{bmatrix}$
|A| = 2 + 15 = 17
Hence, inverse of A exists.
Cofactor of A are,
C₁₁= 1 C₁₂ = 3
C₂₁= -5 C₂₂ = 2
adj A = $\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}^T$
= $\begin{bmatrix}1&3\\-5&2\end{bmatrix}^T$
= $\begin{bmatrix}1&-5\\3&2\end{bmatrix}$
A^-1= 1/|A|. adj A
= $\frac{1}{17}\begin{bmatrix}1&-5\\3&2\end{bmatrix}$
= $\begin{bmatrix}\frac{1}{17}&\frac{-5}{17}\\\frac{3}{17}&\frac{2}{17}\end{bmatrix}$

Question 8. Find the inverse of each of the following matrices.

(i) $\begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}$
|A| = 1(6 – 1) – 2(4 – 3) + 3(2 – 9)
= 5 – 2 – 21 = -18
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 5 C₁₂ = -1 C₁₃= -7
C₂₁= -1 C₂₂ = -7 C₂₃= 5
C₃₁= -7 C₃₂= 5 C₃₃= -1
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}^T$
= $\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{-18}\begin{bmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{bmatrix}$
= $\begin{bmatrix}-5/18&1/18&7/18\\1/18&7/18&-5/18\\7/18&-5/18&1/18\end{bmatrix}$

(ii) $\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}$
|A| = 1(1 + 3) – 2(-1 + 2) + 5(3 + 2)
= 4 – 2 – 25 = 27
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 4 C₁₂ = -1 C₁₃= 5
C₂₁= -17 C₂₂ = -11 C₂₃= 1
C₃₁= 3 C₃₂= 6 C₃₃= -3
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}4&-1&5\\17&-11&1\\3&6&-3\end{bmatrix}^T$
= $\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{27}\begin{bmatrix}4&17&3\\-1&-11&6\\5&1&-3\end{bmatrix}$
= $\begin{bmatrix}4/27&17/27&3/27\\-1/27&-11/27&6/27\\5/27&1/27&-3/27\end{bmatrix}$
= $\begin{bmatrix}4/27&17/27&1/9\\-1/27&-11/27&2/9\\5/27&1/27&-1/9\end{bmatrix}$

(iii) $\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
|A| = 2(4 – 1) – (-1)(-2 + 1) + 1(1 – 2)
= 6 – 1 – 1 = 4
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 3 C₁₂ = 1 C₁₃= -1
C₂₁= 1 C₂₂ = 3 C₂₃= 1
C₃₁= -1 C₃₂= 1 C₃₃= 3
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}^T$
= $\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$
= $\begin{bmatrix}3/4&1/4&-1/4\\1/4&3/4&1/4\\-1/4&1/4&3/4\end{bmatrix}$

(iv) $\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
|A| = 2(3 – 0) – 0 + 1(5)
= 6 – 5 = 1
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 3 C₁₂ = -15 C₁₃= 5
C₂₁= -1 C₂₂ = 6 C₂₃= -2
C₃₁= 1 C₃₂= -5 C₃₃= 2
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}3&-15&5\\-1&6&-2\\1&-5&2\end{bmatrix}^T$
= $\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{1}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
= $\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$

(v) $\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$
|A| = 0 – 1(16 – 12) – 1(-12 + 9)
= -4 + 3 = -1
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 0 C₁₂= -4 C₁₃= -3
C₂₁= -1 C₂₂ = 3 C₂₃= 3
C₃₁= 1 C₃₂= -4 C₃₃= -4
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}0&-4&-3\\-1&3&3\\1&-4&-4\end{bmatrix}^T$
= $\begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{-1}\begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}$
= $\begin{bmatrix}0&-1&1\\-4&3&-4\\-3&3&-4\end{bmatrix}$

(vi) $\begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}$
|A| = 0 – 0 – 1(-12 + 8)
= -1(-4) = 4
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= -8 C₁₂ = 11 C₁₃= -4
C₂₁= 4 C₂₂ = -2 C₂₃= 0
C₃₁= 4 C₃₂= -3 C₃₃= 0
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}-8&11&-4\\4&-2&0\\4&-3&0\end{bmatrix}^T$
= $\begin{bmatrix}-8&4&4\\11&-2&-3\\-4&0&-0\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{4}\begin{bmatrix}-8&4&4\\11&-2&-3\\-4&0&-0\end{bmatrix}$
= $\begin{bmatrix}-2&1&1\\11/4&-1/2&-3/4\\-1&0&-0\end{bmatrix}$

(vii) $\begin{bmatrix}1&0&0\\0&cosα &sinα\\0&sinα&-cosα \end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&0&0\\0&cosα &sinα\\0&sinα&-cosα \end{bmatrix}$
|A| = -cos²α – sin²α
= -(cos²α + sin²α) = -1
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= -1 C₁₂ = 0 C₁₃= -0
C₂₁= 0 C₂₂ = -cosα C₂₃= -sinα
C₃₁= 0 C₃₂= -sinα C₃₃= cosα
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}-1&0&0\\0&- cosα &- sinα \\0&-sinα & cosα \end{bmatrix}^T$
= $\begin{bmatrix}-1&0&0\\0&- cosα &- sinα \\0&-sinα & cosα \end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{-1}\begin{bmatrix}-1&0&0\\0&- cosα &- sinα \\0&-sinα & cosα \end{bmatrix}$
= $\begin{bmatrix}1&0&0\\0&cosα &sinα \\0&sinα &-cosα \end{bmatrix}$

Question 9. (i) $\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}$
|A| = 1(16 – 9) – 3(4 – 3) + 3(3 – 4)
= 7 – 3 – 3 = 1
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 7 C₁₂ = -1 C₁₃= -1
C₂₁= -3 C₂₂ = 1 C₂₃= 0
C₃₁= -3 C₃₂= 0 C₃₃= 1
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}^T$
= $\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = 1/1 $\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$
= $\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}$
To verify A^-1A = $\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix} \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}$
= $\begin{bmatrix}7-3-3&21-12-9&21-9-12\\-1+1&-3+4&-3+3\\-1+1&-3+3&-3+4\end{bmatrix}$
= $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

(ii) $\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$

Solution:

Here, A = $\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$
|A| = 2(8 – 7) – 3(6 – 3) + 1(21 – 12)
= 2 – 3(3) + 1(9) = 2
Therefore, inverse of A exists
Cofactors of A are:
C₁₁= 1 C₁₂ = -3 C₁₃= 9
C₂₁= 1 C₂₂ = 1 C₂₃= -5
C₃₁= -1 C₃₂= 1 C₃₃= -1
adj A = $\begin{bmatrix}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{34}\end{bmatrix}^T$
= $\begin{bmatrix}1&-3&9\\1&1&-5\\-1&1&-1\end{bmatrix}^T$
= $\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}$
A^-1= 1/|A|. adj A
Hence, A^-1 = $\frac{1}{2}\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix}$
To verify A^-1A = $\frac{1}{2}\begin{bmatrix}1&1&-1\\-3&1&1\\9&-5&-1\end{bmatrix} \begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$
= $\frac{1}{2}\begin{bmatrix}2+3-3&3+4-7&1+1-2\\-6+3+3&-9+4+7&-3+1+2\\18-15-3&27-20-7&9-5-2\end{bmatrix}$
= $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

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Last Updated : 28 Mar, 2021

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Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 1

Question 1. Find the adjoint of the following matrices:

Verify that (adj A)A = |A|I = A(adj A) for the above matrices:

(i) $\begin{bmatrix}-3&5\\2&4\end{bmatrix}$

(ii) $\begin{bmatrix}a&b\\c&d\end{bmatrix}$

(iii) $\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}$

(iv) $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}$

Question 2. Compute the adjoint of each of the following matrices:

Verify that (adj A)A = |A|I = A(adj A) for the above matrices:

(i) $\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$

(ii) $\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}$

(iii) $\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}$

(iv) $\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}$

Question 3. For the matrix A = $\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}$ , show that A(adj A) = O.

Question 4. If A = $\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}$ , show that adj A = A.

Question 5. If A = $\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}$ , show that adj A = 3A^T.

Question 6. Find A(adj A) for the matrix A = $\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}$ .

Question 7. Find the inverse of each of the following matrices:

(i) $\begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}$

(ii) $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

(iii) $\begin{bmatrix}a&b\\c&\frac{(a+bc)}{a}\end{bmatrix}$

(iv) $\begin{bmatrix}2&5\\-3&1\end{bmatrix}$

Question 8. Find the inverse of each of the following matrices.

(i) $\begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}$

(ii) $\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}$

(iii) $\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$

(iv) $\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}$

(v) $\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$

(vi) $\begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}$

(vii) $\begin{bmatrix}1&0&0\\0&cosα &sinα\\0&sinα&-cosα \end{bmatrix}$

(ii) $\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 1

Question 1. Find the adjoint of the following matrices:

Verify that (adj A)A = |A|I = A(adj A) for the above matrices:

(i)

(ii)

(iii)

(iv)

Question 2. Compute the adjoint of each of the following matrices:

Verify that (adj A)A = |A|I = A(adj A) for the above matrices:

(i)

(ii)

(iii)

(iv)

Question 3. For the matrix A =, show that A(adj A) = O.

Question 4. If A =, show that adj A = A.

Question 5. If A = , show that adj A = 3AT.

Question 6. Find A(adj A) for the matrix A =.

Question 7. Find the inverse of each of the following matrices:

(i)

(ii)

(iii)

(iv)

Question 8. Find the inverse of each of the following matrices.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(ii)

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

(i) $\begin{bmatrix}-3&5\\2&4\end{bmatrix}$

(ii) $\begin{bmatrix}a&b\\c&d\end{bmatrix}$

(iii) $\begin{bmatrix}cos α&sin α\\sin α&cosα\end{bmatrix}$

(iv) $\begin{bmatrix}1&tan α/2\\-tan α/2&1\end{bmatrix}$

(i) $\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$

(ii) $\begin{bmatrix}1&2&5\\2&3&1\\-1&1&1\end{bmatrix}$

(iii) $\begin{bmatrix}2&-1&3\\4&2&5\\0&4&-1\end{bmatrix}$

(iv) $\begin{bmatrix}2&0&-1\\5&1&0\\1&1&3\end{bmatrix}$

Question 3. For the matrix A = $\begin{bmatrix}1&-1&1\\2&3&0\\18&2&10\end{bmatrix}$ , show that A(adj A) = O.

Question 4. If A = $\begin{bmatrix}-4&-3&-3\\1&0&1\\4&4&3\end{bmatrix}$ , show that adj A = A.

Question 5. If A = $\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}$ , show that adj A = 3A^T.

Question 6. Find A(adj A) for the matrix A = $\begin{bmatrix}1&-2&3\\0&2&-1\\-4&5&2\end{bmatrix}$ .

(i) $\begin{bmatrix}cos θ&sinθ\\-sin θ&cos θ\end{bmatrix}$

(ii) $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

(iii) $\begin{bmatrix}a&b\\c&\frac{(a+bc)}{a}\end{bmatrix}$

(iv) $\begin{bmatrix}2&5\\-3&1\end{bmatrix}$

(i) $\begin{bmatrix}1&2&3\\2&3&1\\3&1&2\end{bmatrix}$

(ii) $\begin{bmatrix}1&2&5\\1&-1&-1\\2&3&-1\end{bmatrix}$

(iii) $\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$

(iv) $\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}$

(v) $\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$

(vi) $\begin{bmatrix}0&0&-1\\3&4&5\\-2&-4&-7\end{bmatrix}$

(vii) $\begin{bmatrix}1&0&0\\0&cosα &sinα\\0&sinα&-cosα \end{bmatrix}$

(ii) $\begin{bmatrix}2&3&1\\3&4&1\\3&7&2\end{bmatrix}$