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Matrices — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsMatrices3 Marks
Question
Given the matrices $A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right], B=\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right]$ and $C=\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]$ Find $1) \ce{ABC}, 2) \ce{ACB}$ State whether $\ce{ABC = ACB}.$
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Answer

$A B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right]=\left[\begin{array}{cc}6-1 & 8-2 \\ 12-2 & 16-4\end{array}\right]=\left[\begin{array}{cc}5 & 6 \\ 10 & 12\end{array}\right]$
$A B C=\left[\begin{array}{cc}5 & 6 \\ 10 & 12\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-15+0 & 5-12 \\ -30+0 & 10-24\end{array}\right]=\left[\begin{array}{cc}-15 & -7 \\ -30 & -14\end{array}\right]$
$2 AC =\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]=\left[\begin{array}{cc}-6+0 & 2-2 \\ -12+0 & 4-4\end{array}\right]=\left[\begin{array}{cc}-6 & 0 \\ -12 & 0\end{array}\right]  $
$ A C B=\left[\begin{array}{cc}-6 & 0 \\ -12 & 0\end{array}\right]\left[\begin{array}{cc}3 & 4 \\ -1 & -20\end{array}\right]=\left[\begin{array}{ll}-18-0 & -24-0 \\ -36-0 & -48-0\end{array}\right]=\left[\begin{array}{cc}-18 & -24 \\ -36 & -48\end{array}\right]$
hence $\ce{ABC ? ACB}$

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