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?



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 $C$.

  • $\begingroup$ Thank you! This helps a lot for now :) Happy holidays! $\endgroup$ – 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.


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