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$A= \begin{pmatrix}-1&3\\ -2&6\end{pmatrix}$

Find two $2\times 2$ matrices $B$ and $C$ such that $AB=AC$ but $B\ne C$.

I have tried to do some row operations along with multiplication but I keep getting the wrong answer.

Any help?

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    $\begingroup$ What do you mean by wrong answers? There are infinite solutions to the given problem. $\endgroup$ – Math Lover Oct 1 '17 at 19:23
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$B=\left( \begin{array}{ll} 3 & 0 \\ 2 & 1 \\ \end{array} \right);\;C=\left( \begin{array}{ll} -6 & 9 \\ -1 & 4 \\ \end{array} \right)$

$AB=AC=\left( \begin{array}{ll} 3 & 3 \\ 6 & 6 \\ \end{array} \right)$

I did in this way

Called $B=\left( \begin{array}{ll} a & b \\ c & d \\ \end{array} \right);\;C=\left( \begin{array}{ll} e & f \\ g & h \\ \end{array} \right)$

I computed $AB$ and $AC$ and wrote an undeterminate system such that they were equal

I got

$g= -\dfrac{a}{3}+c+\dfrac{e}{3},\;h= -\dfrac{b}{3}+d+\dfrac{f}{3}$

And all the other unknowns arbitrary values.

Hope this can be useful

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We have $AB=0$ for all $B$ of the form $$ B=\begin{pmatrix} 3\alpha & 3\beta \cr \alpha & \beta \end{pmatrix} $$

On the other hand, $AC=0$ for $C=0$. Now choose $\alpha,\beta$ non-zero. Then we have $AB=AC$ but $B\neq C$.

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    $\begingroup$ You can also add the proposed matrix to an arbitrary $B$ to get a matrix $C$ with $AB = AC$. $\endgroup$ – student Oct 1 '17 at 19:46
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There's no unique answer.

But the task of finding some valid answer is easy.

Since the determinant of $A$ is zero, there exists a nonzero vector $v$ such that $Av = 0$.

Find such a vector $v$, and let $B$ be the $2\,\times\,2$ matrix with both columns equal to $v$. It follows that $AB=0$.

Then just let $C$ be the zero matrix.

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