Question

The matrix $X$ such that $X\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}$ is

• A. $\begin{bmatrix}1 & 2 \\ 2 & 0\end{bmatrix}$
• B. $\begin{bmatrix}1 & -2 \\ 2 & 0\end{bmatrix}$
• C. $\begin{bmatrix}1 & 2 \\ -2 & 0\end{bmatrix}$
• D. $\begin{bmatrix}1 & 2 \\ 0 & -2\end{bmatrix}$

Answer 1

Answer: (B)

Here, $X \begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}$
The matrix given on the RHS of the equation is a $2 \times 3$ matrix and the one given on the LHS of the equation is a $2 \times 3$ matrix. Therefore, $X$ has to be a $2 \times 2$ matrix. Now, let $X=\begin{bmatrix}a & c \\ b & d\end{bmatrix}$
Therefore, we have
$\begin{bmatrix}a & c \\ b & d\end{bmatrix}\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}$
$\rightarrow\begin{bmatrix}a+4 c & 2 a+5 c & 3 a+6 c \\ b+4 d & 2 b+5 d & 3 b+6 d\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9 \\ 2 & 4 & 6\end{bmatrix}$
On equating the corresponding elements of the two matrices, we have
$ 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 $
$ \text { Now, } a+4 c=-7$
$ \rightarrow a=-7-4 c $
$ 2 a+5 c=-8 $
$ \rightarrow -14-8 c+5 c=-8$
$ \rightarrow -3 c=6 $
$ \rightarrow C =-2$
$ \therefore a--7-4(-2)--7+8-1 $
$ \text { Now, } b+4 d=2 $
$ \rightarrow b=2-4 d $
$ \text { and } 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$
$ \text { Thus, } a=1, b=2, c=-2, d=0 $
$ \text { Hence, the required matrix } X \text { is }\begin{bmatrix}1 & -2 \\2 & 0\end{bmatrix}$