Q.Find the matrix $A$ such that $\begin{bmatrix}2 & -1\\ 1 & 0\\ -3 & 4\end{bmatrix} A = \begin{bmatrix}-1 & -8 & -10\\ 1 & -2 & -5\\ 9 & 22 & 15\end{bmatrix}$.

Idea: $BA=C$ with $B$ of size $3\times2$ and $C$ of size $3\times3$ forces $A$ to be $2\times3$. Each column of $A$ satisfies $B(\text{column})=(\text{matching column of }C)$, so solve three small consistent systems exactly — no inverse needed.

Let $A=\begin{bmatrix}a&b&c\\d&e&f\end{bmatrix}$, $B=\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}$. The middle row of $B$ is $[1,\;0]$, which reads the top entry off directly.

Col 1 $=\begin{bmatrix}-1\\1\\9\end{bmatrix}$: $a=1$ (row 2); $2a-d=-1\Rightarrow d=3$; check $-3a+4d=-3+12=9$. ✓ …

Fix the size of $A$

We are given $BA=C$ with

For the product to be defined and give a $3\times3$ result, $A$ must be $2\times3$. Write

Split into columns

The $k$-th column of $BA$ equals $B$ times the $k$-th column of $A$. So matching columns of $C$ turns the problem into three independent systems in two unknowns each. $B$ is only $3\times2$, so an ordinary square inverse does not apply — but the systems are consistent and solve directly. Notice the middle row of $B$ is $[1,\ 0]$, which immediately gives the top entry of each column of $A$.

Column 1: $B\begin{bmatrix}a\\d\end{bmatrix}=\begin{bmatrix}-1\\1\\9\end{bmatrix}$

• Row 2: $a=1$.
• Row 1: $2a-d=-1\Rightarrow 2-d=-1\Rightarrow d=3$.
• Row 3 (check): $-3(1)+4(3)=9$. ✓

Column 2: $B\begin{bmatrix}b\\e\end{bmatrix}=\begin{bmatrix}-8\\-2\\22\end{bmatrix}$

• Row 2: $b=-2$.
• Row 1: $2(-2)-e=-8\Rightarrow -4-e=-8\Rightarrow e=4$.
• Row 3 (check): $-3(-2)+4(4)=6+16=22$. ✓ …

Method: Solving $BA=C$ for an unknown matrix, column by column

When $B$ is not square you cannot simply write $A=B^{-1}C$. Instead, determine the order of the unknown matrix from compatibility, introduce unknown entries, and solve the small linear systems that each column of the equation produces.

Steps

Step 1: Fix the order of the unknown from dimension rules.

If $B$ is $3\times2$ and $C$ is $3\times3$, then in $BA=C$ the matrix $A$ must be $2\times3$ (columns of $B$ = rows of $A$; result columns = columns of $C$).

Step 2: Write $A$ with unknown entries (e.g. $A=\begin{bmatrix}a&b&c\\ d&e&f\end{bmatrix}$).

Step 3: Split into columns. …

Common Mistakes

Mistake 1: Assuming $A=B^{-1}C$ when $B$ is not square.

Why it's wrong: a $3\times2$ matrix has no inverse, so this formula does not apply. Correct approach: solve column-by-column linear systems instead.

Mistake 2: Getting the order of $A$ wrong.

Why it's wrong: for $BA=C$ with $B$ being $3\times2$ and $C$ being $3\times3$, $A$ must be $2\times3$; any other shape makes the product undefined. Correct approach: match inner dimensions ($B$'s columns to $A$'s rows) and outer dimensions to $C$. …