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{\bf Question}

Find $(AB)C$ and $A(BC)$ where

$$A=\left(\begin{array}{cc} -5 &amp; 1\\ 9 &amp; -1 \end{array}\right);\ \
B=\left(\begin{array}{cc} 1 &amp; 2\\ 3 &amp; 2 \end{array}\right);\ \
C=\left(\begin{array}{cc} 1 &amp; 0\\ -1 &amp; 1 \end{array}\right)$$

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{\bf Answer}

$(AB)=\left(\begin{array}{cc} -5 &amp; 1\\9 &amp; -1
\end{array}\right)\left(\begin{array}{cc} 1 &amp; 2\\3 &amp; 2
\end{array}\right)=\left(\begin{array}{cc} -2 &amp; -8\\6 &amp; 16
\end{array}\right)$

$(AB)C=\left(\begin{array}{cc} -2 &amp; -8\\6 &amp; 16
\end{array}\right)\left(\begin{array}{cc} 1 &amp; 0\\-1 &amp; 1
\end{array}\right)=\left(\begin{array}{cc} 6 &amp; -8\\-10 &amp; 16
\end{array}\right)$

NB note order

$A(BC)$ should be the same, but let's check anyway.

$(BC)=\left(\begin{array}{cc} 1 &amp; 2\\3 &amp; 2
\end{array}\right)\left(\begin{array}{cc} 1 &amp; 0\\-1 &amp; 1
\end{array}\right)=\left(\begin{array}{cc} -1 &amp; 2\\1 &amp; 2
\end{array}\right)$

$A(BC)=\left(\begin{array}{cc} -5 &amp; 1\\9 &amp; -1
\end{array}\right)\left(\begin{array}{cc} -1 &amp; 2\\1 &amp; 2
\end{array}\right)=\left(\begin{array}{cc} 6 &amp; -8\\-10 &amp; 16
\end{array}\right)$

So $(AB)C=A(BC)$

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