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Engineering
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Let   Then

• Option 1)

there cannot exist any B such that AB = BA

• Option 2)

there exist more than one but finite number B's such that AB = BA

• Option 3)

there exists exactly one B such that AB = BA

• Option 4)

there exist infinitely many B's such that AB = BA

As we learnt in

Multiplication of matrices -

-

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

$B= \begin{bmatrix} a &0\\ 0& b \end{bmatrix}$

$AB= \begin{bmatrix} 1 &2\\ 3& 4 \end{bmatrix}\begin{bmatrix} a &0\\ 0& b \end{bmatrix}= \begin{bmatrix} a &2b\\ 3a& 4b \end{bmatrix}$

$BA= \begin{bmatrix} a &0\\ 0& b \end{bmatrix}\begin{bmatrix} 1 &2\\ 3& 4 \end{bmatrix}= \begin{bmatrix} a &2a\\ 3b& 4b \end{bmatrix}$

$AB= BA$

$\begin{bmatrix} a &2b\\ 3a& 4b \end{bmatrix}= \begin{bmatrix} a &2a\\ 3b& 4b \end{bmatrix}$

$\Rightarrow$    2b = 2a

3a  = 3b

$a=b$

$B=a \begin{bmatrix} 1 &0\\ 0& 1 \end{bmatrix}= aI \; \: \: \: \: \:where A\leftarrow R$

Option 1)

there cannot exist any B such that AB = BA

Incorrect Option

Option 2)

there exist more than one but finite number B's such that AB = BA

Incorrect Option

Option 3)

there exists exactly one B such that AB = BA

Incorrect Option

Option 4)

there exist infinitely many B's such that AB = BA

Correct Option

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Engineering
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If A and B are square matrices of size n x n such that $\dpi{100} A^{2}-B^{2}=(A-B)(A+B),$ then which of the following will be always true?

• Option 1)

A = B

• Option 2)

AB = BA

• Option 3)

either A or B is a zero matrix

• Option 4)

either A or B is an identity matrix

As we learn tin

Multiplication of matrices -

-

$A^{2}-B^{2}= (A-B)(A+B)$

$= AA+ AB- BA -BB$

$\therefore AB=BA$

Option 1)

A = B

Incorrect option

Option 2)

AB = BA

Correct option

Option 3)

either A or B is a zero matrix

Incorrect option

Option 4)

either A or B is an identity matrix

Incorrect option

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