Question

Find the matrix X so that X $\left[\begin{array}{ccc}1& 2& 3\\ 4& 4& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$.

Answer 1

Here, $X\left[\begin{array}{ccc}1& 2& 3\\ 4& 4& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$
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 as a $2\times 3$ matrix. Therefor, X has to be a $2\times 2$ matrix. Now, let $X=\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]Therefore,wehave\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]\Rightarrow \left[\begin{array}{ccc}a+4c& 2a+5c& 3a+6c\\ b+4d& 2b+5d& 3b+6d\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$
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
Now, $a+4c=-7\Rightarrow a=-7-4c$
$2a+5c=-8\Rightarrow -14-8c+5c=-8\Rightarrow -3c=6\Rightarrow c=-2\therefore a=-7-4(-2)=-7+8=1$
Now, $b+4d=2\Rightarrow b=2-4d$ and $2b+5d=4\Rightarrow 4-8d+5d=4$
$\Rightarrow -3d=0\Rightarrow d=0\therefore b=2-4\left(0\right)=2$
Thus, a=1, b=2, c=-2, d=0
Hence, the required matrix X is $\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$.