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Class 12 RD Sharma Solutions – Chapter 7 Adjoint and Inverse of a Matrix – Exercise 7.1 | Set 2

  • Last Updated : 28 Mar, 2021

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 = (a2 + 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 = (a2 + 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 = (a2 + 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 A2 – 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}

A2 – 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, A2 – 4A + I = O

 A2 – 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 A2 + 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}



A2 + 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, A2 + 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 A2 – 5A + 7I = O. Hence find A-1.

Solution:

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

A2 \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, A2 – 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, A2 – 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 A2 – 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, A2 – 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

A2 – 9A + 14I = 0

⇒ 9A = A2 + 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 A2 = 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 A2 = λA – 2I 

λA = A2 + 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, A2 = 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 x2 – 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, A2 – 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, A2 – 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 x2 – 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, A2 – 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, A2 – 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 A3 – 6A2 + 5A + 11I3 = O.  Hence, find A-1. 

Solution:

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

A2 \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}

A3 – 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, A3 – 6A2 + 5A + 11I = O.



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

⇒ A2 – 6A + 5I = -11A-1

⇒ -11A-1 = (A2 – 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}

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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