# Is there a more intuitive way of coming up with matrices for problems like this? E.g. Find two matrices B and C with AB = AC, and B does not equal C

Let A = $$\begin{bmatrix}1 & 0\\1 & 0\end{bmatrix}$$

Find two matrices $$B$$ and $$C$$ with $$AB = AC$$, and $$B$$ does not equal $$C$$.

I always have trouble with problems like this. Here, I know $$\begin{bmatrix}1\\0\end{bmatrix}$$ and $$\begin{bmatrix}1\\1\end{bmatrix}$$ works, but I've always done it with trial and error. This that takes a lot of time during tests.

So, is there a more intuitive way of coming up with matrices to solve this problem or is it really just guess and check if it works?

Thanks!

## 2 Answers

One way of attacking problems like this is to step back and take a big view. What happens when you multiply any matrix by $\begin{bmatrix}1&0\\1&0\end{bmatrix}$? It's a slightly scary step to use letters $\begin{bmatrix}x\\y\end{bmatrix}$ to represent 'any matrix', but: $\begin{bmatrix}1&0\\1&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x\\x\end{bmatrix}$.

Aha! The $y$ gets tossed away, only $x$ ends up in the result. So if you want $AB=AC$, the numbers at the top of $B$ and $C$ have to match, but you can pick any numbers for the bottom of $B$ and $C$.

• Thank you! This helps a lot for now :) Happy holidays! – heyyo Dec 14 '15 at 22:25

Writing $B = (b_1,b_2)^t$ and $C = (c_1,c_2)^t$, the two conditions translate to $(b_1,b_1) = (c_1,c_1)$ and $(b_1,b_2) \neq (c_1,c_2)$, i.e., $b_1 = c_1$ and $b_2 \neq c_2$. This lets you see immediately all the possible answers. You can do similar computations for arbitrary $2 \times n$ matrices.