If A=left[begin{array}{ccc}1 & -2 & 1 2 & 1 | vedclass

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(B) Given,$A=\left[\begin{array}{ccc}1 & -2 & 1 \ 2 & 1 & 3\end{array}ight]$ and $B=\left[\begin{array}{cc}2 & 1 \ 3 & 2 \ 1 & 1\end{array}ight

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$\left[\begin{array}{cc}-3 & 10 \ -2 & 7\end{array}ight]$

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(B) Given,$A=\left[\begin{array}{ccc}1 & -2 & 1 \ 2 & 1 & 3\end{array}ight]$ and $B=\left[\begin{array}{cc}2 & 1 \ 3 & 2 \ 1 & 1\end{array}ight]$.First,calculate the product $AB$:$AB = \left[\begin{array}{ccc}1 & -2 & 1 \ 2 & 1 & 3\end{array}ight] \left[\begin{array}{cc}2 & 1 \ 3 & 2 \ 1 & 1\end{array}ight]$$AB = \left[\begin{array}{cc}(1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1)\end{array}ight]$$AB = \left[\begin{array}{cc}2 - 6 + 1 & 1 - 4 + 1 \ 4 + 3 + 3 & 2 + 2 + 3\end{array}ight]$$AB = \left[\begin{array}{cc}-3 & -2 \ 10 & 7\end{array}ight]$Now,find the transpose $(AB)^{\prime}$ by interchanging rows and columns:$(AB)^{\prime} = \left[\begin{array}{cc}-3 & 10 \ -2 & 7\end{array}ight]$Thus,the correct option is $B$.

If $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ 1 & 1\end{array}\right]$,then $(AB)^{\prime}$ is equal to

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