If A= [ ll 2 & 1 0 & 0 ], B= [ ll 2 & 3 4 & 1 ] and C= [ ll 1 & 4 0 & 2 ] then show that A(B+ C)=A B+A C

B+C= [ ll 2 & 3 4 & 1 ]+ [ ll 1 & 4 0 & 2 ]= [ ll 3 & 7 4 & 3 ] A(B+C)= [ ll 2 & 1 0 & 0 ] [ ll 3 & 7 4 & 3 ] = [ cc 6+4 & 14+3 0 & 0 ]= [ cc 10 & 1() 7 0 & 0 ] A B= [ ll 2 & 1 0 & 0 ] [ ll 2 & 3 4 & 1 ]= [ cc 4+4 & 6+1 0 & 0 ]= [ ll 8 & 7 0 & 0 ] l A C= [ ll 2 & 1 0 & 0 ] [ ll 1 & 4 0 & 2 ]= [ cc 2+0 & 8+2 0 & 0 ]= [ cc 2 & 10 0 & 0 ] A B+A C= [ ll 8 & 7 0 & 0 ]+ [ cc 2 & 10 0 & 0 ]= [ cc 10 & 17 0 & 0 ] Hence A(B + C) = AB + AC

If A= [ ll 2 & 1 0 & 0 ], B= [ ll 2 & 3 4 & 1 ] and C= [ ll 1 & 4…

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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 $A(B+ C)=A B+A C$

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Answer

$B+C=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}3 & 7 \\ 4 & 3\end{array}\right]$
$A(B+C)=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}3 & 7 \\ 4 & 3\end{array}\right]$
$=\left[\begin{array}{cc}6+4 & 14+3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}10 & 1() 7 \\ 0 & 0\end{array}\right]$
$A B=\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right]=\left[\begin{array}{cc}4+4 & 6+1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}8 & 7 \\ 0 & 0\end{array}\right]$
$\begin{array}{l}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] \end{array} $
$ A B+A C=\left[\begin{array}{ll}8 & 7 \\ 0 & 0\end{array}\right]+\left[\begin{array}{cc}2 & 10 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}10 & 17 \\ 0 & 0\end{array}\right]$
Hence $A(B + C) = AB + AC$