Question
If $A=\left[\begin{array}{cc}3 & a \\ -4 & 8\end{array}\right], B=\left[\begin{array}{cc}c & 4 \\ -3 & 0\end{array}\right], C=\left[\begin{array}{cc}-1 & 4 \\ 3 & b\end{array}\right]$ and $3 A -2 C =6 B$, find the values of $a, b, c$.

Answer

$3A - 2C = 6B$
$\begin{array}{l}3\left[\begin{array}{cc}3 & a \\ -4 & 8\end{array}\right]-2\left[\begin{array}{cc}-1 & 4 \\ 3 & b\end{array}\right]=6\left[\begin{array}{cc}c & 4 \\ -3 & 0\end{array}\right] \end{array} $
$ {\left[\begin{array}{cc}9 & 3 a \\ -12 & 24\end{array}\right]-\left[\begin{array}{cc}-2 & 8 \\ 6 & 2 b\end{array}\right]=\left[\begin{array}{cc}6 c & 24 \\ -18 & 0\end{array}\right]} $
$ {\left[\begin{array}{cc}11 & 3 a-8 \\ -18 & 24-2 b\end{array}\right]=\left[\begin{array}{cc}6 c & 24 \\ -18 & 0\end{array}\right]}$
Comparing the corresponding elements we get
$3 a-8=24 $
$\Rightarrow 3 a=32 $
$\Rightarrow a=\frac{32}{3}=10 \frac{2}{3}$
$24-2 b=0 $
$\Rightarrow 2 b=24$
$\Rightarrow b=12$
$11=6 c $
$\Rightarrow c=\frac{11}{6}=1 \frac{5}{6}$
 

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