Find matrix $C$ if: $$ A = \begin{bmatrix}1&3&2\\-1&2&1\\0&1&0 \end{bmatrix} \space B = \begin{bmatrix}3&2&1\\1&1&1\\1&2&3 \end{bmatrix}, \space BC = A $$
I started off this question by attempting to find the inverse of $B$, but then I discovered that $B$ is not invertible. Is it still possible to isolate for $C$ and solve or is this answer no solution?
On
Yes let $c$ be the first column of $C$, and $a$ the first column of $A$, then solve the system
$$Bc=a.$$ do the same doe the other columns. Simplify the work by writing $B,A$ and do row reduction on $B$.
On
It depends.
Find a row vector $\vec v$ for which $\vec vB=\vec0$.
Then $\vec vBC=\vec0C=\vec0$.
If $\vec vA\ne\vec0$ then there is no solution.
On
The answer is no, twice. If $B$ is singular, it is a zero divisor. Your equation does not have a unique solution.
However, this doesn't matter since your equation has no solutions at all. $B$ is singular, so $BC$ must be as well. This tells us $A$ is singular. However, $A$ is definitely invertible. This is a contradiction, so no solutions exist.
Just as the equation $B\mathbf x=\mathbf b$ has a solution iff $\mathbf b$ is an element of the column space of $B$, the equation $BC=A$ can be solved iff every column of $A$ is in the column space of $B$. So augment $B$ with $A$ and row-reduce: $$\left[\begin{array}{ccc|ccc} 3&2&1 & 1&2&3 \\ 1&1&1 & -1&2&1 \\ 1&2&3 & 0&1&0 \end{array}\right] \to \left[\begin{array}{ccc|ccc} 1&0&-1 & -2&3&2 \\ 0&1&2 & 1&-1&-1 \\ 0&0&0 & 5&-5&1 \end{array}\right].$$ No column of $A$ is in $B$’s column space, so there’s no solution. Perhaps there’s an error in the problem or you’ve misread it.