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If A=[aij]3×3=[23−11230−11]A=[a_{ij}]_{3\times3}=\begin{bmatrix}2 & 3 & -1\\1 & 2 & 3\\0 & -1 & 1\end{bmatrix} and B=[bij]3×2=[3−21412]B=[b_{ij}]_{3\times2}=\begin{bmatrix}3 & -2\\1 & 4\\1 & 2\end{bmatrix} then the value of a12b21+a32b31a_{12}b_{21}+a_{32}b_{31} is

Solution

✅ Correct Option: 4

Reading the entries, a12=3a_{12}=3, b21=1b_{21}=1, a32=−1a_{32}=-1 and b31=1b_{31}=1.

Hence a12b21+a32b31=3(1)+(−1)(1)=2a_{12}b_{21}+a_{32}b_{31}=3(1)+(-1)(1)=2.

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