Solving matrix equation involving singular matrix Given that $AB=AC$,


*

*$A=\begin{bmatrix} 3 & 6 \\ 1 & 2 \end{bmatrix}$

*$B=\begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}$
Find a matrix $C$ whose elements are all non-zero and is $2\times 2$ matrix.
I attempt to find the inverse of $A$ but it was unsuccessful as it is a singular matrix
I have also attempted to solve the question $AB-AC=0$, $A(B-C)=0$ but still unsuccessful...
Thank you
 A: Hint: Compute $AB$ and $AC$, then solve a system to compute the entries of $C$.
One possible solution is letting $C =\begin{bmatrix} -1 & 1 \\ 1 & 3 \end{bmatrix}$
There are infinite solutions.
Edit: We have
$AB =\begin{bmatrix} 3 & 21 \\ 1 & 7 \end{bmatrix}$. If $C = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $AC=\begin{bmatrix} 3a+6c & 3b+6d \\ a+2c & b+2d \end{bmatrix}$.
So if $AB=AC$, it will follow that


*

*$3a+6c=3$

*$a+2c=1$

*$3b+6d=21$

*$b+2d=7$


Since 1. and 2. are equivalent, we'll work with 2. only. For the same reason we'll take 4. over 3.
Now we have that $a+2c=1$ and $b+2d=3$. Therefore $a=1-2c$ and $b=7-2d$.
This where I think you have trouble: you know $a$ and $b$, but you don't know $c$ and $d$. So what are $c,d$? As it happens $c,d$ can take any value you want! So just choose $c$ and $d$ as you wish, just as long as they satisfy the initial condition that all entries of $C$ are different from $0$. This means that $c$ can't be $0$ or $1\over 2$. Also $d$ can't be $0$ or $7\over 2$. I picked $c=1$ and $d=3$ to get the above mentioned solution.
A: The approach which you began to take will also work. You can rewrite $AB=AC$ as $A(B-C)=0$. This tells you that the columns of $B-C$ must be in the nullspace of $A$.
The nullspace of $A$ is found by solving $Ax=0$, which gives $\mathcal{N}(a)=\operatorname{span}\{x\}$, where $x=[-2,1]^T$.
Denote the $i^{th}$ column of $B$ and $C$ by $B_i$ and $C_i$. Since each column of $B-C$ must be in $\mathcal{N}(a)$, you have $B_i-C_i\in\mathcal{N}(a)\Rightarrow B_i-C_i=-k_ix,\,i=\{1,2\}$, where $k_i$ is an arbitrary constant, which can be different for each column. Then $C_i=B_i+k_ix,\,i=\{1,2\}$.
This gives your matrix $C$ as
$$
C=\begin{bmatrix}
1-2k_1&5-2k_2\\
k_1&1+k_2
\end{bmatrix}.
$$
The choice $k_1=1,\,k_2=2$ gives the same matrix as Git Gud showed. Any choice for $k_1$ and $k_2$ would satisfy the required condition.
