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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!

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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 $C$.

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

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