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Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 3
  • Last Updated : 28 Mar, 2021

Question 25. Show that the matrix A = \begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}  satisfies the equation A3 – A2 – 3A – I3 = 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 A3 – A2 – 3A – I3 \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, A3 – A2 – 3A – I3 = 0

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

⇒ A2 – A – 3I – A-1 = 0

⇒ A-1 = A2 – 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 A3 – 6A2 + 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, A3 – 6A2 + 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, A3 – 6A2 + 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 A2 = 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 A2 – 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, A2 + 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, A2 – 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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