RD Sharma Class 12 Ex 7.1 Solutions Chapter 7 Adjoint and Inverse of a Matrix

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter7
Exercise7.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 7.1 Solutions Chapter 7 Adjoint and Inverse of a Matrix

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 = 3AT.

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}

AT= \begin{bmatrix}-1&2&2\\-2&1&-2\\-2&-2&1\end{bmatrix}

Now, 3AT = 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

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}                  

Question 10. For the following parts of matrices verify that (AB)-1 = B-1A-1.

(i) A = \begin{bmatrix}3&2\\7&5\end{bmatrix} and B = \begin{bmatrix}4&6\\3&2\end{bmatrix}

Solution:

To prove (AB)-1= B-1A-1

We take LHS

AB = \begin{bmatrix}3&2\\7&5\end{bmatrix}\begin{bmatrix}4&6\\3&2\end{bmatrix}   

\begin{bmatrix}18&22\\43&52\end{bmatrix}

|AB| = 18 × 52 – 22 × 43

= 936 – 946 = -10

adj(AB) = \begin{bmatrix}52&-22\\-43&18\end{bmatrix}

AB-1= adj(AB)/|AB| = \frac{1}{(-10)}\begin{bmatrix}52&-22\\-43&18\end{bmatrix}

\frac{1}{10}\begin{bmatrix}-52&22\\43&-18\end{bmatrix}

Now,

A = \begin{bmatrix}3&2\\7&5\end{bmatrix}                

|A| = 15 – 14 = 1

adj A = \begin{bmatrix}5&-2\\-7&3\end{bmatrix}

Therefore, A-1 = adj A/|A| = \frac{1}{1}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}

B = \begin{bmatrix}4&6\\3&2\end{bmatrix}                

|B| = 8 – 18 = -10

adj B = \begin{bmatrix}2&-6\\-3&4\end{bmatrix}

Therefore, B-1= adj B/|B| = \frac{1}{-10}\begin{bmatrix}2&-6\\-3&4\end{bmatrix}

Now, we take RHS

B-1A-1 = \frac{-1}{10}\begin{bmatrix}2&-6\\-3&4\end{bmatrix}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}

\frac{-1}{10}\begin{bmatrix}52&-22\\-43&18\end{bmatrix}

\frac{1}{10}\begin{bmatrix}-52&22\\43&-18\end{bmatrix}

LHS = RHS 

Hence, Proved 

(ii) A = \begin{bmatrix}2&1\\5&3\end{bmatrix}  and B = \begin{bmatrix}4&5\\3&4\end{bmatrix}

Solution:

To prove (AB)-1 = B-1A-1

We take LHS

AB = \begin{bmatrix}2&1\\5&3\end{bmatrix}\begin{bmatrix}4&5\\3&4\end{bmatrix}         

\begin{bmatrix}11&14\\29&27\end{bmatrix}

|AB| = 11 × 27 – 29 × 14

= 407 – 406 = 1

adj(AB) = \begin{bmatrix}37&-14\\-29&11\end{bmatrix}          

AB-1= adj(AB)/|AB| = \frac{1}{1}\begin{bmatrix}37&-14\\-29&11\end{bmatrix}          

=\begin{bmatrix}37&-14\\-29&11\end{bmatrix}

Now,

A = \begin{bmatrix}2&1\\5&3\end{bmatrix}                

|A| = 6 – 5 = 1

adj A= \begin{bmatrix}3&-1\\-5&2\end{bmatrix}          

Therefore, A-1 = adj A/|A| = \frac{1}{1}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}          

B = \begin{bmatrix}4&5\\3&4\end{bmatrix}                

|B| = 16 – 15 = 1

adj B = \begin{bmatrix}4&-5\\-3&4\end{bmatrix}          

Therefore, B-1= adj B/|B| = \frac{1}{1}\begin{bmatrix}4&-5\\-3&4\end{bmatrix}          

Now, we take RHS

B-1A-1 = \begin{bmatrix}4&-5\\-3&4\end{bmatrix}\begin{bmatrix}3&-1\\-5&2\end{bmatrix}          

\begin{bmatrix}37&-14\\-29&11\end{bmatrix}

LHS = RHS 

Hence, Proved 

Question 11.  Let A = \begin{bmatrix}3&2\\7&5\end{bmatrix}  and B = \begin{bmatrix}6&7\\8&9\end{bmatrix}  . Find (AB)-1.

Solution:

AB = \begin{bmatrix}3&2\\7&5\end{bmatrix}\begin{bmatrix}4&6\\3&2\end{bmatrix}         

=\begin{bmatrix}34&39\\82&94\end{bmatrix}          

|AB| = 34 × 94 – 39 × 82 = -2

adj(AB) = \begin{bmatrix}94&-39\\-82&34\end{bmatrix}

AB-1 = adj(AB)/|AB| = \frac{-1}{2}\begin{bmatrix}94&-39\\-82&34\end{bmatrix}

\begin{bmatrix}-47&39/2\\41&-17\end{bmatrix}

Question 12. Given A = \begin{bmatrix}2&-3\\-4&7\end{bmatrix}, Compute A-1 and show that 2A-1 = 9I – A. 

Solution:

A = \begin{bmatrix}2&-3\\-4&7\end{bmatrix}                      

|A| = 14 – 12 = 2

adj A = \begin{bmatrix}7&3\\4&2\end{bmatrix}

Therefore, A-1 = adj A/|A| = \frac{1}{2}\begin{bmatrix}7&3\\4&2\end{bmatrix}

 To show 2A-1 = 9I – A.                   

LHS = 2 × (1/2) \begin{bmatrix}7&3\\4&2\end{bmatrix}

\begin{bmatrix}7&3\\4&2\end{bmatrix}

Now we take RHS 

= 9I – A

\begin{bmatrix}9&0\\0&9\end{bmatrix}   – \begin{bmatrix}2&-3\\-4&7\end{bmatrix}

=\begin{bmatrix}7&3\\4&2\end{bmatrix}

LHS = RHS

Hence Proved

Question 13. If A = \begin{bmatrix}4&5\\2&1\end{bmatrix}, then show that A – 3I = 2(I + 3A-1).

Solution:

Here, A = \begin{bmatrix}4&5\\2&1\end{bmatrix}

|A| = 4 – 10 = -6

adj A = \begin{bmatrix}1&-5\\-2&4\end{bmatrix}

Therefore, A-1 = adj A/|A| = \frac{1}{(-6)}\begin{bmatrix}1&-2\\-5&4\end{bmatrix}          

To show, A – 3I = 2(I + 3A-1)

Now we take LHS 

= A – 3I

=\begin{bmatrix}4&5\\2&1\end{bmatrix}   – 3\begin{bmatrix}1&0\\0&1\end{bmatrix}

=\begin{bmatrix}1&5\\2&-2\end{bmatrix}

Now we take RHS 

= 2I + 6A-1

= 2\begin{bmatrix}1&0\\0&1\end{bmatrix}   + 6 × (1/6)\begin{bmatrix}-1&5\\2&-4\end{bmatrix}

\begin{bmatrix}1&5\\2&-2\end{bmatrix}

LHS = RHS 

Hence Proved        

Question 14. Find the inverse of the matrix A = \begin{bmatrix}a&b\\c&(1+bc)/a\end{bmatrix}  and show that aA-1 = (a+ bc + 1)I – aA.

Solution:

Here, A = \begin{bmatrix}a&b\\c&(1+bc)/a\end{bmatrix}

|A| = (a + abc)/a – bc = 1

Therefore, 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}(1+bc)/a &-c\\-b&a\end{bmatrix}^T

\begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}

A-1 = 1/|A|. adj A

= 1/1 \begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}

\begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}

To show that 

aA-1 = (a+ bc + 1)I – aA.

LHS = aA-1

= a\begin{bmatrix}(1+bc)/a &-b\\-c&a\end{bmatrix}

\begin{bmatrix}1+bc &-ab\\-ac&a^2\end{bmatrix}

RHS = (a+ bc + 1)I – aA

\begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1\end{bmatrix}   – a\begin{bmatrix}a&b\\c&(1+bc)/a\end{bmatrix}

\begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1\end{bmatrix}   – \begin{bmatrix}a^2&ab\\ac&1+bc\end{bmatrix}

\begin{bmatrix}1+bc&-ab\\-ac&a^2\end{bmatrix}

LHS = RHS 

Hence Proved         

Question 15. Given A = \begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix}, B-1 \begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}  , Compute (AB)-1.

Solution:

We know (AB)-1 = B-1A-1

Here, A = \begin{bmatrix}5&0&4\\2&3&2\\1&2&1\end{bmatrix}

|A| = 5(3 – 4) + 4(4 – 3) = -5 + 4 = -1 

Co-factors of A are:

C11 = -1      C12 = 0      C13 = 1

C21 = 8       C22 = 1      C23 = -10

C31 = -12    C32 = -2    C33 = 15

adj A = \begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}

A-1 = 1/|A|. adj A

Hence, A-1 = \frac{1}{(-1)}\begin{bmatrix}-1&8&-12\\0&1&-2\\1&-10&15\end{bmatrix}

\begin{bmatrix}1&-8&12\\0&-1&2\\-1&10&-15\end{bmatrix}

 (AB)-1 = B-1A-1

\begin{bmatrix}1&3&3\\1&4&3\\1&3&4\end{bmatrix}\begin{bmatrix}1&-8&12\\0&-1&2\\-1&10&-15\end{bmatrix}

\begin{bmatrix}-2&19&-27\\-2&18&-25\\-3&29&-42\end{bmatrix}       

Question 16. Let F(α) = \begin{bmatrix}cosα&-sinα&0\\sinα&cosα&0\\0&0&1\end{bmatrix} and G(β) = \begin{bmatrix}cosβ&0&sinβ\\0&1&0\\-sinβ&0&cosβ\end{bmatrix}  , Show that

(i) [F(α)]-1 = F(-α) 

Solution:

We have F(α) = \begin{bmatrix}cosα&-sinα&0\\sinα&cosα&0\\0&0&1\end{bmatrix}

|F(α)| = cos2α + sin2α = 1

Therefore, inverse of F(α) exists

Cofactors of F(α) are:  

C11 = cosα    C12 = -sinα     C13 = 0

C21 = sinα     C22 = cosα     C23 = 0

C31 = 0          C32 = 0           C33 = 1

Adj F(α) = \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}cosα&-sinα&0\\sinα&cosα&0\\0&0&1\end{bmatrix}^T

=\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}

[F(α)]-1 = 1/|F(α)|. adj F(α)

Hence, [F(α)]-1 = 1/1\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}

=\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}            

Now, F(-α) = \begin{bmatrix}cos(-α)&-sin(-α)&0\\sin(-α)&cos(-α)&0\\0&0&1\end{bmatrix}

=\begin{bmatrix}cosα&sinα&0\\-sinα&cosα&0\\0&0&1\end{bmatrix}

So, [F(α)]-1 = F(-α)

Hence, Proved

(ii) [G(β)]-1 = G(-β) 

Solution:

We have G(β) = \begin{bmatrix}cosβ&0&sinβ\\0&1&0\\-sinβ&0&cosβ\end{bmatrix}

|G(β)| = cos2β + sin2β = 1

Therefore, inverse of G(β) exists

Cofactors of G(β) are:  

C11 = cosβ   C12 = 0      C13 = sinβ

C21 = 0         C22 = 1      C23 = 0

C31 = -sinβ   C32 = 0     C33 = sinβ

Adj G(β) = \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}cosβ&0&sinβ\\0&1&0\\-sinβ&0&cosβ\end{bmatrix}^T       

=\begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}

[G(β)]-1 = 1/|G(β)|. adj G(β)

Hence, [G(β)]-1 = 1/1\begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}

\begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}

Now, G(-β) =\begin{bmatrix}cos(-β)&0&sin(-β)\\0&1&0\\-sin(-β)&0&cos(-β)\end{bmatrix}

\begin{bmatrix}cosβ&0&-sinβ\\0&1&0\\sinβ&0&cosβ\end{bmatrix}

So, [G(β)]-1 = G(-β)

Hence, Proved

(iii) [F(α)G(β)]-1 = F(-α)G(-β)

Solution:

We already know that S[G(β)]-1 = G(-β) 

 [F(α)]-1 = F(-α) 

Taking LHS = [F(α)G(β)]-1

= [F(α)]-1[G(β)]-1

= F(-α)G(-β) = RHS 

Hence, Proved

Question 17. If A = \begin{bmatrix}2&3\\1&2\end{bmatrix}  , Verify that A– 4A + I = O, where I = \begin{bmatrix}1&0\\0&1\end{bmatrix}   and O = \begin{bmatrix}0&0\\0&0\end{bmatrix}  , Hence, find A-1. 

Solution: 

Here, A = \begin{bmatrix}2&3\\1&2\end{bmatrix}

A2 = \begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}

\begin{bmatrix}7&12\\4&7\end{bmatrix}

4A = 4\begin{bmatrix}2&3\\1&2\end{bmatrix}

\begin{bmatrix}8&12\\4&8\end{bmatrix}

A– 4A + I = O

\begin{bmatrix}7&12\\4&7\end{bmatrix}   –\begin{bmatrix}8&12\\4&8\end{bmatrix}   + \begin{bmatrix}1&0\\0&1\end{bmatrix}

\begin{bmatrix}7-8+1&12-2+0\\4-4+0&7-8+1\end{bmatrix}

Hence, = \begin{bmatrix}0&0\\0&0\end{bmatrix}

Now, A– 4A + I = O

 A– 4A = -I

Multiplying both side by A-1 both sides we get 

A.A(A-1) – 4AA-1 = -IA-1

AI – 4I = -A-1

A-1 = 4I – AI    

\begin{bmatrix}4&0\\0&4\end{bmatrix}   – \begin{bmatrix}2&3\\1&2\end{bmatrix}      

\begin{bmatrix}2&-3\\-1&2\end{bmatrix}

Question 18.  Show that A = \begin{bmatrix}-8&5\\2&4\end{bmatrix}   satisfies the equation A+ 4A – 42I = O. Hence, Find A-1

Solution:

Here, A = \begin{bmatrix}-8&5\\2&4\end{bmatrix}

A2 = \begin{bmatrix}-8&5\\2&4\end{bmatrix}\begin{bmatrix}-8&5\\2&4\end{bmatrix}

=\begin{bmatrix}64+10&-40+20\\-16+8&10+16\end{bmatrix}          

=\begin{bmatrix}74&-20\\-8&26\end{bmatrix}

4A = 4\begin{bmatrix}-8&5\\2&4\end{bmatrix}      

\begin{bmatrix}-32&20\\8&16\end{bmatrix}

A+ 4A – 42I = \begin{bmatrix}74&-20\\-8&26\end{bmatrix}   + \begin{bmatrix}-32&20\\8&16\end{bmatrix}   – \begin{bmatrix}42&0\\0&42\end{bmatrix}

=\begin{bmatrix}74-74&-20+20\\-8+8&42-42\end{bmatrix}

Hence, \begin{bmatrix}0&0\\0&0\end{bmatrix}

Now, A+ 4A – 42I = 0

⇒ A-1A.A + 4A-1.A – 42A-1I = 0

⇒ IA + 4I – 42A-1 = 0

⇒ A-1 = 1/42 [A + 4I]

⇒ A-1 = \frac{1}{42}\begin{bmatrix}-4&5\\2&8\end{bmatrix}

Question 19. If A = \begin{bmatrix}3&1\\-1&2\end{bmatrix}  , show that A– 5A + 7I = O. Hence find A-1.

Solution:

Here, A = \begin{bmatrix}3&1\\-1&2\end{bmatrix}

A\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}

=\begin{bmatrix}8&5\\-5&3\end{bmatrix}

Now, A– 5A + 7I = \begin{bmatrix}8&5\\-5&3\end{bmatrix}   + 5\begin{bmatrix}3&1\\-1&2\end{bmatrix}   + 7\begin{bmatrix}1&0\\0&1\end{bmatrix}

=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}

=\begin{bmatrix}0&0\\0&0\end{bmatrix}

Now, A– 5A + 7I = O

Multiplying by A-1 both sides

⇒ A-1AA + 5AA – 1 + 7IA-1 = 0

⇒ A-1 = 1/7[5I – A]

⇒ A-1 \frac{1}{7}\begin{bmatrix}5&0\\0&5\end{bmatrix}-\begin{bmatrix}3&1\\-1&2\end{bmatrix}

⇒ A-1 \frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}

Question 20. If A = \begin{bmatrix}4&3\\2&5\end{bmatrix}  , find x and y such that A– xA + yI = O. Hence, evaluate A-1.

Solution:

Here, A = \begin{bmatrix}4&3\\2&5\end{bmatrix}

A2 = \begin{bmatrix}4&3\\2&5\end{bmatrix}\begin{bmatrix}4&3\\2&5\end{bmatrix}  

\begin{bmatrix}22&27\\18&31\end{bmatrix}

Now, A– xA + yI = O

⇒ \begin{bmatrix}22&27\\18&31\end{bmatrix}   –  \begin{bmatrix}4x&3x\\2x&5x\end{bmatrix}    + \begin{bmatrix}y&0\\0&y\end{bmatrix}   

\begin{bmatrix}0&0\\0&0\end{bmatrix}

⇒ 22 – 4x + y = 0

⇒ 4x – y = 22  ………(i)

or 

18 – 2x = 0

⇒ x = 9

Putting x = 9 in eq (i)

⇒ y = 14

A– 9A + 14I = 0

⇒ 9A = A+ 14I

⇒ 9A-1A = A-1AA + 14A-1

⇒ 9I = IA + 14A-1

⇒ A-1 = 1/14[9I – A] = 1/14(\begin{bmatrix}9&0\\0&9\end{bmatrix}-\begin{bmatrix}4&3\\2&4\end{bmatrix}  )

⇒ A-1\frac{1}{14}\begin{bmatrix}5&-3\\-2&4\end{bmatrix}        

Question 21. If A = \begin{bmatrix}3&-2\\4&-2\end{bmatrix} , find the value of λ so that A= λA – 2I. Hence, find A-1

Solution:

Here, A = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}

A2 = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}

\begin{bmatrix}1&-2\\4&-4\end{bmatrix}

If A= λA – 2I 

λA = A+ 2I

⇒ λ \begin{bmatrix}3&-2\\4&-2\end{bmatrix}  = \begin{bmatrix}1&-2\\4&-4\end{bmatrix} + \begin{bmatrix}2&0\\0&2\end{bmatrix}

⇒ λ \begin{bmatrix}3&-2\\4&-2\end{bmatrix}  = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}

⇒ λ = 1

Now, A= λA – 2I

Multiplying both side A-1

⇒ A-1AA = A-1A – 2A-1I

⇒ A = I – 2A-1

⇒ 2A-1 = I – A = \begin{bmatrix}1&0\\0&1\end{bmatrix} - \begin{bmatrix}3&-2\\4&-2\end{bmatrix}

A-1 = \frac{1}{2}\begin{bmatrix}-2&2\\-4&3\end{bmatrix}

Question 22. Show that A = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}  satisfies the equation x– 3x – 7 = 0. Thus, find A-1.

Solution:

Here, A = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}

A2 = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}\begin{bmatrix}5&3\\-1&-2\end{bmatrix} = \begin{bmatrix}22&9\\-3&-1\end{bmatrix}

Now, A– 3A – 7= \begin{bmatrix}22&9\\-3&-1\end{bmatrix}-\begin{bmatrix}15&9\\-3&-6\end{bmatrix}-\begin{bmatrix}7&0\\0&7\end{bmatrix}

=\begin{bmatrix}0&0\\0&0\end{bmatrix}

We have, A– 3A – 7 = 0

⇒ A-1AA – 3A-1A – 7A-1 = 0

⇒ A-3I – 7A-1 = 0

⇒ 7A-1 = A – 3I

⇒ 7A-1 = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}  – \begin{bmatrix}3&0\\0&3\end{bmatrix}

A-1 = \begin{bmatrix}2/7&3/7\\-1/7&-5/7\end{bmatrix}

Question 23. Show that A = \begin{bmatrix}6&5\\7&6\end{bmatrix}  satisfies the equation x– 12x + 1 = 0. Thus, find A-1.

Solution:

Here, A = \begin{bmatrix}6&5\\7&6\end{bmatrix}

A2 = \begin{bmatrix}6&5\\7&6\end{bmatrix}\begin{bmatrix}6&5\\7&6\end{bmatrix}

\begin{bmatrix}71&60\\84&71\end{bmatrix}

Now, A– 12A + I = \begin{bmatrix}71&60\\84&71\end{bmatrix}  – \begin{bmatrix}6&5\\7&6\end{bmatrix} + \begin{bmatrix}1&0\\0&1\end{bmatrix}

= \begin{bmatrix}0&0\\0&0\end{bmatrix}

We have, A– 12A + I = 0

⇒ A – 12I + A-1 = 0 

⇒ A-1 = 12I – A

⇒ A-1 = \begin{bmatrix}12&0\\0&12\end{bmatrix}-\begin{bmatrix}6&5\\7&6\end{bmatrix}

⇒ A-1 = \begin{bmatrix}6&-5\\-7&6\end{bmatrix}

Question 24. For the matrix A =\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}  show that A– 6A2 + 5A + 11I= O.  Hence, find A-1. 

Solution:

Here, A =  \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}

A\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}  \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}

\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}

A3 = \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}  \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}

\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}

A– 6A2 + 5A + 11I

\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}  – 6   \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}+ 5 \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}+11  \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

=\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix} - \begin{bmatrix}24&12&6\\-18&48&-84\\42&-18&84\end{bmatrix}+ \begin{bmatrix}5&5&5\\5&10&-15\\10&-5&15\end{bmatrix}+\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}

\begin{bmatrix}24-24&12-12&6-6\\-18+18&48-48&-84+84\\42-42&-18+18&84-84\end{bmatrix}

\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O

We have, A– 6A2 + 5A + 11I = O.

⇒ A-1(AAA) – 6A-1(AA) + 5A-1A + 11IA-1 = 0

⇒ A– 6A + 5I = -11A-1

⇒ -11A-1 = (A– 6A + 5I)

=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}-6  \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}+5  \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}-\begin{bmatrix}6&6&6\\6&12&-18\\22&-6&18\end{bmatrix}+\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}

=\begin{bmatrix}4-6+5&2-6&1-6\\-3-6&8-12+5&-14+18\\7-12&-3+6&14-18+5\end{bmatrix}

=\begin{bmatrix}3&-4&-5\\-9&1&4\\-5&3&1\end{bmatrix}

Therefore, A-1 = \frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}

Question 25. Show that the matrix A = \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}  satisfies the equation A– A– 3A – I= 0. Hence, find A-1.

Solution:

Here, A = \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}

A2 = \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}  \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}=  \begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}

A3 = \begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}  \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}=  \begin{bmatrix}-1&-8&-10\\0&7&10\\7&12&7\end{bmatrix}

Now A– A– 3A – I\begin{bmatrix}-1&-8&-10\\0&7&10\\7&12&7\end{bmatrix}-\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix}-3\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

=\begin{bmatrix}-1+5-3-1&-8+8&-10+4+6\\-6+6&7-9+3-1&10-4-6\\7+2-9&12-12&7-3-3-1\end{bmatrix}

\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O

So, A– A– 3A – I= 0

⇒ A-1(AAA) – A-1(AA) – 3A-1A – A-1I = 0

⇒ A– A – 3I – A-1 = 0

⇒ A-1 = A– A – 3I

\begin{bmatrix}-5&-8&-4\\6&9&4\\-2&0&3\end{bmatrix} -  \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}-  \begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}

Therefore, A-1 =\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}

Question 26. If A = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} . Verify that A– 6A+ 9A – 4I = O and hence find A-1.

Solution:

Here, A = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}

A2 = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}    \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}

=\begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix}

\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}

A3 = A2A = \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix} \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}

\begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix}

\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}

Now, A– 6A+ 9A – 4I

\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-6  \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}+9  \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}-4  \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-\begin{bmatrix}36&-30&30\\-30&30&-30\\30&-30&36\end{bmatrix}+\begin{bmatrix}18&-9&9\\-9&18&-9\\9&-9&18\end{bmatrix}-\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}

\begin{bmatrix}40&-30&30\\-30&40&-30\\30&-30&40\end{bmatrix}-\begin{bmatrix}40&-30&30\\-30&40&-30\\30&-30&40\end{bmatrix}

\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O

So, A– 6A+ 9A – 4I = O

Multiplying both side by A-1

⇒ A-1(AAA) – 6A-1(AA) 9A-1A – 4A-1I = O

⇒ AAI – 6AI + 9I = 4A-1

⇒ 4A-1 = A2I – 6AI + 9I

=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-6  \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}+9  \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-\begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix}+\begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}

=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}

⇒ A-1 \frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}

Question 27. If A =\frac{1}{9}\begin{bmatrix}-8&1&4\\4&4&7\\1&-8&4\end{bmatrix} , prove that A-1 = AT.

Solution:

Here, A = \frac{1}{9}\begin{bmatrix}-8&1&4\\4&4&7\\1&-8&4\end{bmatrix}

AT = \frac{1}{9}\begin{bmatrix}-8&4&1\\1&4&-8\\4&7&4\end{bmatrix}

Now, Finding A-1

|A| = 1/9[-8(16 + 56) – 1(16 – 7) + 4(-32 – 4)]

= -81

 Therefore, inverse of A exists

Cofactors of A are:

C11 = 72      C12 = -9       C13 = -36

C21 = -36    C22 = -36    C23 = -63

C31 = -9      C32 = 72      C33 = -36

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}72&-9&-36\\-36&-36&-63\\-9&72&-36\end{bmatrix}^T

=\begin{bmatrix}72&-36&-9\\-9&-36&72\\-36&-63&-36\end{bmatrix}

A-1 = 1/|A|. adj A

Hence, A-1 = \frac{1}{-81}\begin{bmatrix}72&-36&-9\\-9&-36&72\\-36&-63&-36\end{bmatrix}

\frac{1}{9}\begin{bmatrix}-8&4&1\\1&4&-8\\4&7&4\end{bmatrix}     

= AT 

Hence Proved   

Question 28. If A = \begin{bmatrix}3&-3&4\\2&3&4\\0&-1&1\end{bmatrix} , show that A-1 = A3.

Solution:

Here, A = \begin{bmatrix}3&-3&4\\2&3&4\\0&-1&1\end{bmatrix}

|A| = 3(-3 + 4) + 3(2 – 0) + 4(-2 – 0)

= 3 + 6 – 8

= 1 

Therefore, inverse of A exists

Cofactors of A are:

C11 = 1      C12 = -2     C13 = -2

C21 = -1    C22 = 3      C23 = 3

C31 = 0     C32 = -4     C33 = -3

adj A =\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}

A-1= 1/|A|. adj A

\frac{1}{1}\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}

Now, A2 = \begin{bmatrix}3&-3&4\\2&3&4\\0&-1&1\end{bmatrix}\begin{bmatrix}3&-3&4\\2&3&4\\0&-1&1\end{bmatrix}

=\begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}

A3 = \begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}\begin{bmatrix}3&-3&4\\2&3&4\\0&-1&1\end{bmatrix}

=\begin{bmatrix}1&-1&0\\-2&3&-4\\-2&3&-3\end{bmatrix}

= A-1 

Hence Proved

Question 29. If A =\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix} , show that A= A-1.

Solution: 

Here, A = \begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}

LHS = A2

\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}\begin{bmatrix}-1&2&0\\-1&1&1\\0&1&0\end{bmatrix}

\begin{bmatrix}-1&0&2\\0&0&1\\-1&1&2\end{bmatrix}

|A| = -1(1 – 0) – 2(-1 – 0) + 0

= 1

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 0     C13 = -1

C21 = 0      C22 = 0     C23 = 1

C31 = 21    C32 = 1      C33 = 1

adj A = \begin{bmatrix}-1&0&-1\\0&0&1\\2&1&1\end{bmatrix}^T

=\begin{bmatrix}-1&0&2\\0&0&1\\-1&1&1\end{bmatrix}

A-1 = 1/|A|. adj A

Hence, A-1 = \frac{1}{1}\begin{bmatrix}-1&0&2\\0&0&1\\-1&1&1\end{bmatrix}

=\begin{bmatrix}-1&0&2\\0&0&1\\-1&1&1\end{bmatrix}

= A

Hence Proved

Question 30. Solve the matrix equation\begin{bmatrix}5&4\\1&1\end{bmatrix}X = \begin{bmatrix}1&-2\\1&3\end{bmatrix} , where X is a 2×2 matrix.

Solution:

We have, \begin{bmatrix}5&4\\1&1\end{bmatrix}X = \begin{bmatrix}1&-2\\1&3\end{bmatrix}

Let A = \begin{bmatrix}5&4\\1&1\end{bmatrix}     and B = \begin{bmatrix}1&-2\\1&3\end{bmatrix}

So. AX = B

⇒ X = A-1B

Now, |A| = 5 – 4 = 1 

Co factors of A are:

C11 = 1         C12 = -1

C21 = -4      C22 = 5

adj A = \begin{bmatrix}1&-1\\-4&5\end{bmatrix}^T

=\begin{bmatrix}1&-4\\1&5\end{bmatrix}

A-1 = \begin{bmatrix}1&-4\\1&5\end{bmatrix}

Therefore, X = \begin{bmatrix}1&-4\\1&5\end{bmatrix}\begin{bmatrix}1&-2\\1&3\end{bmatrix}

X = \begin{bmatrix}-3&-14\\4&17\end{bmatrix}

Question 31. Find the matrix X satisfying the matrix equation: X\begin{bmatrix}5&3\\-1&-2\end{bmatrix}=\begin{bmatrix}14&7\\7&7\end{bmatrix} .

Solution:

We have, \begin{bmatrix}5&3\\-1&-2\end{bmatrix}=\begin{bmatrix}14&7\\7&7\end{bmatrix}

Let A = \begin{bmatrix}5&3\\-1&-2\end{bmatrix}    and B = \begin{bmatrix}14&7\\7&7\end{bmatrix}

So, XA = B

XAA-1 = BA-1

XI = BA-1 ………..(i)  

Now, |A| = -7

Co factors of A are:

C11 = -2       C12 = 1

C21 = -3      C22 = 5

adj A = \begin{bmatrix}-2&1\\-3&5\end{bmatrix}^T

=\begin{bmatrix}-2&-3\\1&5\end{bmatrix}

A-1 = 1/|A|.adj (A)

=\frac{1}{(-7)}\begin{bmatrix}-2&-3\\1&5\end{bmatrix}=\frac{-1}{7} \begin{bmatrix}2&3\\-1&-5\end{bmatrix}

Therefore, X = \begin{bmatrix}14&7\\7&7\end{bmatrix} .1/7.\begin{bmatrix}2&3\\-1&-5\end{bmatrix}

=\frac{7}{7}\begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}2&3\\-1&-5\end{bmatrix}

X = \begin{bmatrix}3&1\\1&-2\end{bmatrix}

Question 32. Find the matrix X for which: \begin{bmatrix}3&2\\7&5\end{bmatrix} X\begin{bmatrix}-1&1\\-1&1\end{bmatrix}=\begin{bmatrix}2&-1\\0&4\end{bmatrix}

Solution:

Let, A = \begin{bmatrix}3&2\\7&5\end{bmatrix}

B = \begin{bmatrix}-1&1\\-2&1\end{bmatrix}

C = \begin{bmatrix}2&-1\\0&4\end{bmatrix}

 Then the given equation becomes

 A × B = C

⇒ X = A-1CB-1

Now |A| = 35 -14 = 21

|B| = -1 + 2 = 1

A-1 = adj (A)/|A| = \frac{1}{21}\begin{bmatrix}5&-2\\-7&3\end{bmatrix}

B-1 = adj (B)/|A| = \begin{bmatrix}1&-1\\2&-1\end{bmatrix}

X = A-1 CB-1 = \frac{1}{21}\begin{bmatrix}5&-2\\-7&3\end{bmatrix} \begin{bmatrix}2&-1\\0&4\end{bmatrix} \begin{bmatrix}1&-1\\2&-1\end{bmatrix}

\begin{bmatrix}-16&3\\24&-5\end{bmatrix}

Question 33. Find the matrix X satisfying the equation:\begin{bmatrix}2&1\\5&3\end{bmatrix}X\begin{bmatrix}5&3\\3&2\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}

Solution:

Let A = \begin{bmatrix}2&1\\5&3\end{bmatrix}   B = \begin{bmatrix}5&3\\3&2\end{bmatrix}

AXB = I

X = A-1B-1

|A| = 6 – 5 = 1

|B| = 10 – 9 = 1

A-1 = adj A /|A| = \begin{bmatrix}3&-1\\-5&2\end{bmatrix}

B-1 = adj B/|B| = \begin{bmatrix}2&-3\\-3&5\end{bmatrix}

X = \begin{bmatrix}3&-1\\-5&2\end{bmatrix}\begin{bmatrix}2&-3\\-3&5\end{bmatrix}

\begin{bmatrix}9&-14\\-16&25\end{bmatrix}   

Question 34. If A = \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix} , Find A-1 and prove that A– 4A – 5I = O.

Solution:

Here, A = \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}

A2 = \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}  \begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}

\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}

Now, A+ 4A – 5I

\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-4\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}

\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}= 0

Now, A– 4A – 5I = O

⇒ A-1AA – 4A-1A – 5A-1I = O 

⇒ 5A-1 = [A – 4I]

\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}-\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}

\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}

 A-1 \frac{1}{5}\begin{bmatrix}-3&2&2\\2&-3&2\\2&2&-3\end{bmatrix}

Question 35. If A is a square matrix of order n, prove that |A adj A| = |A|n.

Solution:

Given, |A adj A| = |A|n

Taking LHS = |A Adj A|

= |A| |Adj A| 

= |A| |A|n-1

= |A|n-1+1

= |A|n = RHS 

Hence Proved

Question 36. If A-1 = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}  and B = \begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix} , find (AB)-1.

Solution:

Here, B = \begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}

|B| = 1(3 – 0) – 2(-1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 1

Therefore, inverse of B exists

Cofactors of B are:

C11 = 3       C12 = 1      C13 = 2

C21 = 2      C22 = 1      C23 = 2

C31 = 6     C32 = 2      C33 = 5

adj A = \begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}^T    

\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}

Therefore,

B-1 \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}

Hence, (AB)-1 = B-1A-1

\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}

\begin{bmatrix}9&-3&5\\-2&1&0\\61&-24&22\end{bmatrix}        

Question 37. If A = \begin{bmatrix}1&-2&3\\0&-1&4\\-2&2&1\end{bmatrix} , find (AT)-1.

Solution:

Assuming B = AT = \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}

|B| = 1(-1 – 8) – 0 – 2(-8 + 3)

= -9 + 10 = 1

Therefore, inverse of B exists

Cofactors of B are:

C11 = -9      C12 = 8      C13 = -5

C21 = -8     C22 = 7      C23 = -4

C31 = -2     C32 = 2      C33 = -1

adj B = \begin{bmatrix}-9&8&-5\\-8&7&-4\\-2&2&-1\end{bmatrix}^T

=\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}

B-1 = \frac{1}{1}\begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}

or (AT)-1 = \begin{bmatrix}-9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}

Question 38. Find the adjoint of the matrix A = \begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}  and hence show that A (adj A) = |A|I3.

Solution:

Here, A = \begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}

|A| = -1(1 – 4) – 2(2 + 4) – 2(-4 – 2)

= 3 + 12 + 12 = 27

Therefore, inverse of A exists

Cofactors 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 (adj A) = \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}

\begin{bmatrix}27&0&0\\0&27&0\\0&0&27\end{bmatrix}

or A (adj A) = 27\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

Hence, A (adj A) = |A|I

Hence Proved

Question 39. If A = \begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix} , A-1 and show that A-1 = 1/2(A2 – 3I).

Solution:

Here, A = \begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}

|A| = 0 – 1(0 – 1) + 1(1 – 0)

= 1 + 1 = 2

Therefore, inverse of A exists

Cofactors of A are:

C11 = -1     C12 = 1      C13 = 1

C21 = 1      C22 = -1    C23 = 1

C31 = 1      C32 = 1      C33 = -1

adj A = \begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}^T

\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}

A-1 = 1/|A|. adj A

Hence, A-1 = \frac{1}{2}\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}    

Now, A2 – 3I = \begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}-3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}               

=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}-\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}

=\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}

Hence, A-1 = 1/2(A2 – 3I) 

Hence Proved

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