Find matrix $X$ so that: $X \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{pmatrix}$

$\begin{pmatrix} 1 & -2 \\ 2 & 0 \end{pmatrix}$

The correct answer is Option (2) → $\begin{pmatrix} 1 & -2 \\ 2 & 0 \end{pmatrix}$ ##

Let $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

Then, $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{pmatrix}$

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

Equating and solving:

$a+4b = -7$ and $2a+5b = -8 ⇒a=1, b=-2$

$c+4d = 2$ and $2c+5d = 4 ⇒c=2, d=0$

$X = \begin{pmatrix} 1 & -2 \\ 2 & 0 \end{pmatrix}$