Question

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

Answer 1

$X_{2\times 2}{\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]}_{2\times 3}={\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]}_{2\times 3}$

$X$ should be of order $2\times 2$ to make multiplication possible

Let $X$ be $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\therefore \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}a+4b& 2a+5b& 3a+6b\\ c+4d& 2c+5d& 3c+6d\end{array}\right]$

Comparing we get,

$a+4b=-7⟶\left(1\right)\times 2c+4d=2⟶\left(3\right)\times 22a+5b=-8⟶\left(2\right)\times 12c+5d=4⟶\left(4\right)\times 1\therefore 2a+8b=-142c+8d=42a+5b=-82c+5d=4\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_3b=-63d=0b=-2d=0\therefore a=1\therefore c=2\therefore X=\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$