Question

If $2X-3Y=\left[\begin{array}{cc}3& 2\\ 1& 0\end{array}\right]$ and $X+2Y=\left[\begin{array}{cc}0& 1\\ 2& 3\end{array}\right]$, and $kX=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right],$ then the value of $k$ is

Answer 1

Given: $2X-3Y=\left[\begin{array}{cc}3& 2\\ 1& 0\end{array}\right]\cdots \left(i\right)$ and $X+2Y=\left[\begin{array}{cc}0& 1\\ 2& 3\end{array}\right]\cdots \left(ii\right)$
On multiplying $\left(ii\right)$ with $2$, we have:
$\Rightarrow 2X-3Y=\left[\begin{array}{cc}3& 2\\ 1& 0\end{array}\right]\cdots \left(iii\right)$ and $2X+4Y=\left[\begin{array}{cc}0& 2\\ 4& 6\end{array}\right]\cdots \left(iv\right)$
Now, on solving, $\left(iv\right)-\left(iii\right)$, we have: $7Y=\left[\begin{array}{cc}-3& 0\\ 3& 6\end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{cc}-\frac{3}{7}& 0\\ \frac{3}{7}& \frac{6}{7}\end{array}\right]$
Now, substituting the $Y$ matrix in $\left(ii\right)$ we have:
$X=\left[\begin{array}{cc}0& 1\\ 2& 3\end{array}\right]-2Y$
$\Rightarrow X=\left[\begin{array}{cc}0& 1\\ 2& 3\end{array}\right]-2\left[\begin{array}{cc}-\frac{3}{7}& 0\\ \frac{3}{7}& \frac{6}{7}\end{array}\right]$
$\Rightarrow X=\left[\begin{array}{cc}\frac{6}{7}& 1\\ \frac{8}{7}& \frac{9}{7}\end{array}\right]$
Hence we have: $7X=\left[\begin{array}{cc}6& 7\\ 8& 9\end{array}\right]$