Solving matrix equations involving non-square matrices

Question

If $$ \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \\ \end{bmatrix} A = \begin{bmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \\ \end{bmatrix} $$

How do we find $A$?

my work:

From this post talking about inverses, I figured out that I could do this by finding right inverse of A but I'm not sure how to do that. That led me to this post which speaks about solving non-square matrices,I do not really understand the procedure which the accepted post is suggesting. It'd be helpful if someone could frame those answers in simpler words.

As a further question, how do I know if a non-square matrix has an inverse or not? we really don't have a determinant here to check..

could you illustrate/ suggest how exactly that I would use this? — Clemens Bartholdy, Aug 22 '20 at 19:28
If you have linear matrix equation $A X = B$, vectorization gives you $(I \otimes A) \mbox{vec} (X) = \mbox{vec} (B)$, where operator $\mbox{vec}$ merely stacks the columns of a matrix into a big column vector. — Rodrigo de Azevedo, Aug 23 '20 at 07:27

score 3 · Answer 1 · answered Aug 22 '20 at 11:49

Well, if you're interested only in finding $A$, for matrices of low order (here $2\times2$), you may proceed with a matrix with variables like

$$A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}$$

So,
$\begin{bmatrix}2 & -1 \\ 1 & 0 \\-3 &4 \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}=\begin{bmatrix}-1 & -8 \\ 1 & -2 \\9&22\end{bmatrix} \Rightarrow \begin{bmatrix}2a-c & 2b-d \\ a & b\\-3a+4c &-3b+4d \end{bmatrix}=\begin{bmatrix}-1 & -8 \\ 1 & -2 \\9&22\end{bmatrix}$

Clearly, $\color{blue}{a=1, b = -2}$ and from the remaining equations, $\color{blue}{c=3, d = 4}$.

Thus, $A = \begin{bmatrix}1 & -2\\ 3 & 4 \end{bmatrix}$.
This may be tedious for higher order matrices, but in this case it works.

I considered this but didn't follow through because it was too brute forcey — Clemens Bartholdy, Aug 22 '20 at 19:27

Solving matrix equations involving non-square matrices

1 Answers1