NIMCET 2011 Question Paper with solutions

Question : 8 of 120

Marks: +1, -0

A

and

B

are two square matrices such that

B=-A^{-1} B A

, then

(A+B)^{2}

will be equal to

0
$A^{2}+2 A B+B^{2}$
$A^{2}+B^{2}$
$A+B$

Validate

Solution:

Hint: In order to determine the value of

(A+B)^{2}

, multiply the both side of the given equation

B=-A^{-1} B A

with the

A

. Remember that the multiplication of a matrix with its inverse gives the identity
matrix. Now expand the

(A+B)^{2}

(A+B)(A+B)

.simplify it further and use the result obtained earlier in this expansion to get the required result.
Complete step-by-step answer:
We are given two square matrices

A

and

B

having relation between as

B=-A^{-1} B A

Since,

B=-A^{-1} B A

Let's multiply

A

on both sides of the equation, we get

A B=-A A^{-1} B A

As we know when a matrix is multiplied with its inverse matrix, it results into identity matrix. So here

A A^{-1}=I

A B=-I B A

A B=-B A\;\;\;\;\;⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

We have to find the value of

(A+B)^{2}

, lets rewrite this thing

⇒(A+B)^{2}=(A+B)(A+B)

⇒(A+B)^{2}=A ⋅ A+A ⋅ B+B ⋅ A+B ⋅ B

From the equation (1)

A B=-B A

,
putting this into the above expression we get

⇒(A+B)^{2}=A^{2}-B A+B A+B^{2}

⇒(A+B)^{2}=A^{2}+B^{2}

Therefore the

(A+B)^{2}

is equal to

A^{2}+B^{2}

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