Question
If $A=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]$ and $C=\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]$ then show that $(B-A) C$ $=B C-A C)$
✓
Answer
$B-A=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]-\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 2 \\ 4 & 1\end{array}\right]$
$(B-A) C=\left[\begin{array}{ll}0 & 2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]$
$=\left[\begin{array}{cc}0 & 0+4 \\ 4+0 & 16+2\end{array}\right]=\left[\begin{array}{cc}0 & 4 \\ 4 & 18\end{array}\right]$
$\begin{array}{l}B C=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+6 \\ 4+0 & 16+2\end{array}\right]=\left[\begin{array}{ll}2 & 14 \\ 4 & 18\end{array}\right] \end{array}$
$ A C=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+2 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right] $
$ B C-A C=\left[\begin{array}{ll}2 & 14 \\ 4 & 18\end{array}\right]-\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 4 \\ 4 & 18\end{array}\right]$
Hence $(B - A)C = BC - AC$
$(B-A) C=\left[\begin{array}{ll}0 & 2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]$
$=\left[\begin{array}{cc}0 & 0+4 \\ 4+0 & 16+2\end{array}\right]=\left[\begin{array}{cc}0 & 4 \\ 4 & 18\end{array}\right]$
$\begin{array}{l}B C=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+6 \\ 4+0 & 16+2\end{array}\right]=\left[\begin{array}{ll}2 & 14 \\ 4 & 18\end{array}\right] \end{array}$
$ A C=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2+0 & 8+2 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right] $
$ B C-A C=\left[\begin{array}{ll}2 & 14 \\ 4 & 18\end{array}\right]-\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 4 \\ 4 & 18\end{array}\right]$
Hence $(B - A)C = BC - AC$
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