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I am new to Linear Algebra and this question is giving me problems because I end up with more unknowns than equations when trying to determine the unknown matrix. This is what I am given:

Matrix $A$: \begin{bmatrix} 1 & 3 & 2 \\ 0 & -1 & 1 \end{bmatrix}

Matrix $AB$: \begin{bmatrix} 3 & 4 \\ -1 & 0 \end{bmatrix}

Matrix $A$ is $2\times3$ and matrix $AB$ is $2\times2$. So I know that Matrix $B$ needs to be $3\times2$. So I let $B$ be an arbitrary $3\times2$ matrix \begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix} and then multiply it out. I end up with these four equations:

$a + 3c + 2e = 3;\quad b + 3d + 2f = 4;\quad -c + e = 0;\quad -d + f = 0$

After this I don't know how to continue as I have more unknowns than equations. I tried googling and I found a different matrix question on this site. This Matrix problem doesn't have the same more unknowns than equations problem as this one.

I also attempted finding the inverse of $A$ and that was instantly shut down as a rectangular matrix does not have an inverse. I am wondering if it is possible to find the inverse of $AB$ and somehow work it out like that, but I don't know how that would work.

Any help will be appreciated.

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    $\begingroup$ Why do you assume that there is a unique solution? In particular, it is possible that for matrices $A,B,C$, we have $AB= AC$. Yes, if $A$ is square and invertible then the solution is unique, but otherwise $A(B-C) = 0$ can happen without $B= C$. In the above case the solution is therefore not unique, and is obtainable by solving the equations using one or more of the variables as a parameter. $\endgroup$ Commented Mar 4, 2018 at 9:28

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You seem to think that there should be a single answer. Why? There are infinitely many solutions. In fact, for each $e$ and each $f$,$$B=\begin{pmatrix}-5e&4-5f\\1+e&f\\e&f\end{pmatrix}$$is a solution. How did I get it? Simply solving the system of linear equations that you mentioned.

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Consider the augmented matrix:

$$\left[ \begin{array}{ccc|cc} 1 & 3 & 2 & 3 & 4\\ 0 & -1 & 1& -1 & 0\end{array} \right]$$

Perform $R_1+3R_2$:

$$\left[ \begin{array}{ccc|cc} 1 & 0 & 5 & 0 & 4\\ 0 & -1 & 1& -1 & 0\end{array} \right]$$

Perform $-R_2$:

$$\left[ \begin{array}{ccc|cc} 1 & 0 & 5 & 0 & 4\\ 0 & 1 & -1& 1 & 0\end{array} \right]$$

We have more variables than equations, we can let $e$ and $f$ be our free variables.

We have $$a + 5e=0, b+5f=4 $$ $$c-e=1, d-f=0$$

Express everythign in terms of $e$ and $f$,

$$B = \begin{bmatrix} -5e & 4-5f \\ e+1 & f \\ e & f\end{bmatrix}$$

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