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Matrices — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsMatrices4 Marks
Question
If $A=\left[\begin{array}{cc}2 & -1 \\ -4 & 5\end{array}\right]$ and $B=\left[\begin{array}{c}-3 \\ 2\end{array}\right]$ find the matrix $C$ such that $A C=B$
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Answer

Given
$\begin{aligned}& A=\left[\begin{array}{cc}2 & -1 \\-4 & 5\end{array}\right] \end{aligned}$
$ B=\left[\begin{array}{c}-3 \\2\end{array}\right]$
Let martix $C=\left[\begin{array}{l}x \\ y\end{array}\right]$
$\therefore AC =\left[\begin{array}{cc}2 & -1 \\-4 & 5\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{c}2 x y \\-4 x+5 y\end{array}\right]$
But $AC = B$
$\therefore\left[\begin{array}{c}2 x-y \\-4 x+5 y\end{array}\right]=\left[\begin{array}{c}-3 \\2\end{array}\right]$
Comparing the corresponding elements
$\begin{array}{r}2 x-y=-3 \\-4 x+5 y=2\end{array}$
Multiplying (i) by 5 and (ii) by 1
$10 x-5 y=-15$
$ -4 x+5 y=2$
$ \text { Adding, we get }$
$ 6 x=-13$
$ \Rightarrow x=\frac{-13}{6}$
Adding, we get
$6 x =-13$
$ \Rightarrow x =\frac{-13}{6}$
Substituting the value of $x$ in $(i)$
$2\left(\frac{-13}{6}\right)-y=-3$
$ \Rightarrow \frac{-13}{3}-y=-3$
$ -y=-3+\frac{13}{3}$
$ =\frac{-9+13}{3}$
$ =\frac{4}{3}$
$\therefore y=-\frac{4}{3}$
$\therefore$ Matrix $C=\left[\begin{array}{c}\frac{-13}{6} \\ -\frac{4}{3}\end{array}\right]$.

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