Question
If $\left.A=\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]\right]$ and $C=\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]$ Simplify $A^2+B C$
✓
Answer
$A^2=\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}1+8 & 4+4 \\ 2+2 & 8+1\end{array}\right]=\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right]$
$B C=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-3+0 & 0+4 \\ 4=0 & 0+0\end{array}\right]=\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right]$
$A^2+B C=\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right]+\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right]=\left[\begin{array}{cc}6 & 12 \\ 8 & 9\end{array}\right]$
$B C=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-3+0 & 0+4 \\ 4=0 & 0+0\end{array}\right]=\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right]$
$A^2+B C=\left[\begin{array}{ll}9 & 8 \\ 4 & 9\end{array}\right]+\left[\begin{array}{cc}-3 & 4 \\ 4 & 0\end{array}\right]=\left[\begin{array}{cc}6 & 12 \\ 8 & 9\end{array}\right]$
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