<s>[INST] Find all matrices that satisfy $\mathrm B \mathrm A = \mathrm I_2$ 
Given the matrix $$A=\begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix}$$ find all $2 \times 3$ matrices in $B \in M_{2 \times 3}(\mathbb R)$ with $BA=I_2$.

Here's what I did:
$$\begin{pmatrix}a&b&c\\d&e&f\\ \end{pmatrix} \begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
and then multiplying things out:
$$\begin{pmatrix} {a+3b+2c}&{8a+5b+2c}\\{d+3e+2f}&{8d+5e+2f}\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
So would I just set
$$a+3b+2c=1$$
$$8a+5b+2c=0$$
$$d+3e+2f=0$$
$$8d+5e+2f=1$$
But then this gives $4$ equations in $6$ unknowns, so wouldn't there be infinitely many solutions? Did I do this correctly? Matrices aren't my strong point...
Thanks. 
 [/INST] The variables of the first two equations do not occur in the last two equations and vice versa. You therefore can independently solve two systems of two equations with three unknowns each. This gives you two solutions with one free variable in each of them:
$$
B = \pmatrix{ a & -\frac{1}{2}(7a+1) & \frac{1}{4}(19a+5)\cr
d & \frac{1}{2}(1-7d) & \frac{1}{4}(19d-3)\cr}
$$
</s>