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. 

Since the second column is not a multiple of the first column, the matrix $\pmatrix{1 & 8\cr 3 & 5\cr 2 & 2\cr}$ has rank $2$.  That implies that a solution exists.  One way to get a solution is to take any two rows of $A$ (check that they are linearly independent) and use the inverse of that $2 \times 2$ matrix for the corresponding columns of $B$, and $0$'s in the other column.  Thus if you take the first two rows of $A$, the inverse of $\pmatrix{1 & 8\cr 3 & 5\cr}$ is $\pmatrix{-5/19 & 8/19\cr 3/19 & -1/19\cr}$, corresponding to 
$B = \pmatrix{-5/19 & 8/19 & 0\cr 3/19 & -1/19 & 0\cr}$. 
For the general solution, write the system using block matrices as 
$$ [ B_1 \ b ] \left[\matrix{A_1\cr a^T\cr} \right] = B_1 A_1 + b a^T = I$$
(where $B_1$ and $A_1$ are $2 \times 2$, $b$ is $2 \times 1$ and $a^T$ is $1 \times 2$).  We can solve for $B_1$ in terms of $b$: $B_1 = (I - b a^T) A_1^{-1}$.  We already have $A_1^{-1} = \pmatrix{-5/19 & 8/19 \cr 3/19 & -1/19 \cr}$, so with $b = \pmatrix{b_1\cr b_2}$ and $a^T = (2, 2)$ we get
$$ B_1 = \pmatrix{1 - 2 b_1 & - 2 b_1\cr -2 b_2 & 1 - 2 b_2\cr}  \pmatrix{-5/19 & 8/19 \cr 3/19 & -1/19 \cr} = \pmatrix{ (-5 + 4 b_1)/19 & (8 - 14 b_1)/19\cr
(3 + 4 b_2)/19 & (-1 - 14 b_2)/19\cr}$$
i.e. 
$$ B =  \pmatrix{ (-5 + 4 b_1)/19 & (8 - 14 b_1)/19 & b_1\cr
(3 + 4 b_2)/19 & (-1 - 14 b_2)/19 & b_2\cr}$$
where $b_1$ and $b_2$ are arbitrary.
