It is given that:

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

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 $2 \times 3$ matrix.

Therefore, $X$ has to be a $2 \times 2$ matrix.

Now, let $X=\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]$

Therefore, we have :

$\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$

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

Equating the corresponding elements of two matrices, we have

$a+4 c=-7$,      $2 a+5 c=-8$,       $3 a+6 c=-9$

$\mathrm{ab}+4 \mathrm{d}=2$,    $2 \mathrm{b}+5 \mathrm{d}=4$, $3 \mathrm{b}+6 \mathrm{d}=6$

Now, $a +4 c=-7$   $ \Rightarrow a=-7-4 c$          ……….. $(1)$

$2 a+5 c=-8 \Rightarrow-14-8 c+5 c=-8$                $[$From $(1)$ $]$

$\Rightarrow-3 c=6$

$\Rightarrow c=-2$

$\therefore  $ $a=-7-4(-2)=-7+8=1$

Now, $\mathrm{b}+4 \mathrm{d}=2 \Rightarrow \mathrm{b}=2-4 d$

$\therefore $  $2 b+5 d=4 $   $\Rightarrow $  $4-8 d+5 d=4$

$\Rightarrow $  $-3 d=0$

$\Rightarrow  $  $d=0$

$\therefore $ $b=2-4(0)=2$

Thus, $a=1$ , $b=2$,  $c=-2$,  $d=0$

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