Question

Let $M$ be a matrix such that $M\times \left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}4& -7\end{array}\right]$ Find M

Answer 1

$M\times \left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]=[4-7]$

Order of $\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]$ is $2\times 2$

Order of $[4-7]$ is $1\times 2$

Two matrices can be multiplied if:

The number of columns of the $1^{st}$ matrix must equal the number of rows of the $2^{nd}$ matrix.

And the result will have the same number of rows as the $1^{st}$ matrix, and the same number of columns as the $2^{nd}$ matrix.

$\therefore M$ must be $1\times 2$ matrix.

${\left[M\right]}_{1\times 2}\times {\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]}_{2\times 2}={[4-7]}_{1\times 2}$

${\left[xy\right]}_{1\times 2}\times {\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right]}_{2\times 2}={[4-7]}_{1\times 2}$ for $x,y\in R$

$2x+0=4\Rightarrow x=2$

$x+3y=-7\Rightarrow 3y=-7-x=-7-2=-9$

$\Rightarrow y=\frac{-9}{3}=-3$

$\therefore x=2,y=-3$

$\therefore M=[2-3]$