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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 $B$.

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    $\begingroup$ 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. $\endgroup$ Oct 8 '18 at 12:04
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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 ($1+(-7)\neq3\times(-3)$).

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  • $\begingroup$ Oh I see. Thank you, that clears things up. $\endgroup$
    – Ryan
    Oct 8 '18 at 13:12

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