# What is the set of all matrices A such that AB=C?

I am trying to find all matrices $$A$$ such that $$AB = C$$, where $$B=\begin{pmatrix} 1 & 3 & 8 \\ -3 & -4 & -9 \\ 2 & 4 & 10 \end{pmatrix}$$ and $$C= \begin{pmatrix} 2 & 1 & 1 \\ 1 & -3 & -7 \end{pmatrix}$$ I am explicitly told not to do any "excessive work" such as row reductions or computing inverses. However, I am given that $$\begin{array}{ccc|ccc} 1 & 3 & 8 & 1 & 0 & 0 \\ -3 & -4 & -9 & 0 & 1 & 0 \\ 2 & 4 & 10 & 0 & 0 & 1 \\ \end{array}$$ row reduces to $$\begin{array}{ccc|ccc} 1 & 0 & -1 & 12 & -7 & -16\\ 0 & 1 & 3 & -5 & 3 & 7 \\ 0 & 0 & 0 & -4 & 2 & 5 \\ \end{array}$$ I can immediately see that $$B$$ is not invertible, which means there are either an infinite number of $$A$$'s or no $$A$$'s that satisfy the equation. At this point I'm not sure what to do. If I was given the row reduction of $$B^T$$ then it would be easy to solve for $$AB = C \Longrightarrow B^tA^t = C^t$$ for each column vector of $$A^t$$. As this is not the case, I don't know how I would solve this using just the row reduction of $$B$$.

• Just write $A$ as a matrix with "unknowns" $x_i$ and do explicit matrix multiplication $AB$. This is far from being "excessive".Then compare both sides of $AB=C$, giving easy equations in the $x_i$. Solve the equations. Oct 8 '18 at 12:04

The problem has no solution. Note that if you add the first and the third columns of $$B$$, you get $$3$$ times the second one. So, for any matrix $$A$$, $$AB$$ has that property too. But $$C$$ doesn't ($$1+(-7)\neq3\times(-3)$$).