Let M and N be two matrices over mathbb{R} of | vedclass

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(D) Let $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $N = \begin{bmatrix} p & q \ r & s \end{bmatrix}$.Then,$MN = \begin{bmatrix} ap+br & a

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None of these options are true in general

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(D) Let $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $N = \begin{bmatrix} p & q \ r & s \end{bmatrix}$.Then,$MN = \begin{bmatrix} ap+br & aq+bs \ cp+dr & cq+ds \end{bmatrix}$ and $NM = \begin{bmatrix} ap+qc & pb+qd \ ar+cs & br+ds \end{bmatrix}$.For $MN = NM$,we must have $br = qc$,$aq+bs = pb+qd$,$cp+dr = ar+cs$,and $cq+ds = br+ds$.These conditions do not hold for all matrices $M$ and $N$. For example,if $M = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$ and $N = \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}$,then $MN = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$ while $NM = \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$,so $MN 
eq NM$.Since none of the specific conditions $(A)$,$(B)$,or $(C)$ are necessary or sufficient for $MN = NM$ to hold for any arbitrary matrices $M$ and $N$,the correct answer is $(D)$.

Let $M$ and $N$ be two matrices over $\mathbb{R}$ of order $2$. Then,$MN = NM$ if .......

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