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Compute the following sums :\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\)\(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\) |
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Answer» \(\begin{bmatrix}2+1 & 1-2 & 3+3 \\[0.3em]0+2 & 3+6 & 5+1 \\[0.3em]-1+0 & 2-3 & 5+1\end{bmatrix}\) \(=\begin{bmatrix}3 & -1 & 6 \\[0.3em]2 & 9 & 6 \\[0.3em]-1 & -1 & 6\end{bmatrix}\) Hence,\(\begin{bmatrix}2 & 1 & 3 \\[0.3em]0 & 3 & 5 \\[0.3em]-1 & 2 & 5\end{bmatrix}+\) \(\begin{bmatrix}1 & -2 & 3 \\[0.3em]2 & 6 & 1 \\[0.3em]0 & -3 & 1\end{bmatrix}\) =\(\begin{bmatrix}3 & -1 & 6 \\[0.3em]2 & 9 & 6 \\[0.3em]-1 & -1 & 6\end{bmatrix}\) |
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