Question

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

• A. $\left[\begin{array}{cc}1& 2\\ 2& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 2\\ -2& 0\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 2\\ 0& -2\end{array}\right]$

Answer 1

Answer: (B)

Here, $X\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$
The matrix given on the 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 $X=\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]$
Therefore, we have
$\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$
$\to \left[\begin{array}{ccc}a+4c& 2a+5c& 3a+6c\\ b+4d& 2b+5d& 3b+6d\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$
On equating the corresponding elements of the two matrices, we have
$a+4c=-7,2a+5c=-8,3a+6c=-9$
$b+4d=2,2b+5d=4,3b+6d=6$
$\text{ Now, }a+4c=-7$
$\to a=-7-4c$
$2a+5c=-8$
$\to -14-8c+5c=-8$
$\to -3c=6$
$\to C=-2$
$\therefore a--7-4(-2)--7+8-1$
$\text{ Now, }b+4d=2$
$\to b=2-4d$
$\text{ and }2b+5d=4$
$\to 4-8d+5d=4$
$\to -3d=0\Rightarrow d=0$
$\therefore b-2-4(0)-2$
$\text{ Thus, }a=1,b=2,c=-2,d=0$
$\text{ Hence, the required matrix }X\text{ is }\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$