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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 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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