Find the matrix X so that

Solution:

It is given that:

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

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

Therefore, we have:

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

Equating the corresponding elements of the two matrices, we have:

$\begin{array}{lll}a+4 c=-7, & 2 a+5 c=-8, & 3 a+6 c=-9 \\ b+4 d=2, & 2 b+5 d=4, & 3 b+6 d=6\end{array}$

Now, $a+4 c=-7 \Rightarrow a=-7-4 c$

$\begin{aligned} \therefore 2 a+5 c=-8 & \Rightarrow-14-8 c+5 c=-8 \\ & \Rightarrow-3 c=6 \\ & \Rightarrow c=-2 \end{aligned}$

$\therefore a=-7-4(-2)=-7+8=1$

$\Rightarrow-3 c=6$

$\Rightarrow c=-2$

$\therefore a=-7-4(-2)=-7+8=1$

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

$\begin{aligned} \therefore 2 b+5 d=4 & \Rightarrow 4-8 d+5 d=4 \\ & \Rightarrow-3 d=0 \\ & \Rightarrow d=0 \end{aligned}$

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

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

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