Question
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right], C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right]$, compute $A(B+C)$
✓
Answer
$\begin{aligned} & A(B+C) \\ & A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \\ & B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right] \\ & C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right] \\ & A(B+C)=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}2+5 & 1+1 \\ 4+7 & 2+4\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}7 & 2 \\ 11 & 6\end{array}\right] \\ & =\left[\begin{array}{cc}7+22 & 2+12 \\ 21+44 & 6+24\end{array}\right] \\ & =\left[\begin{array}{ll}29 & 14 \\ 65 & 30\end{array}\right] .\end{aligned}$
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