Question

Find the matrix $X$ so that $X\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

Answer 1

It is given that:
$X\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

The matrix given on the R.H.S. of the equation is a $2\times 3$ matrix and the one given on the L.H.S. of the equation is a $2\times 3$ matrix.

Therefore, $X$ has to be a $2\times 2$ matrix.
Now, let $X=\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]$

Therefore, we have:
$\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}a+4c& 2a+5c& 3a+6c\\ b+4d& 2b+5d& 3b+6d\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

Equating the corresponding elements of the two matrices, we have:
$\begin{array}{ccc}a+4c=-7& 2a+5c=-8& 3a+6c=-9\\ b+4d=2& 2b+5d=4& 3b+6d=6\end{array}$

Now, $a+4c=-7\Rightarrow a=-7-4c$
$\therefore 2a+5c=-8\Rightarrow -14-8c+5c=-8$
$\Rightarrow -3c=6\Rightarrow c=-2$
$\therefore a=-7-4(-2)=-7+8=1$

Now, $b+4d=2\Rightarrow b=2-4d$
$\therefore 2b+5d=4\Rightarrow 4-8d+5d=4$
$\Rightarrow -3d=0$
$\Rightarrow d=0$
$\therefore b=2-4\left(0\right)=2$

Thus, $a=1,b=2,c=-2,d=0$

Hence, the required matrix $X$ is $\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$