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I need to find $X$ from

$$\begin{pmatrix} 1 & 2\\ -3 &-6 \end{pmatrix} X \begin{pmatrix} 1 &2 \\ -1 &-2 \end{pmatrix}=\begin{pmatrix} 2 &4 \\ -6 & -12 \end{pmatrix}$$

I wrote $X$ as

$$X=\begin{pmatrix} a & b\\ c &d \end{pmatrix}$$

and I got $a+2c-b-2d=2$ but I do not know what to do next. Please help.

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    Hint: look up inverses of matrices and then try using $A^{-1} A = I$ and apply it to something like: $AB = C$ implies $ B = A^{-1} C$ – 2017-01-03
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    @TheMathsGeek Every single matrix here has determinant $0$. – 2017-01-03
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    Determinant is 0 so there is no inverse – 2017-01-03
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    Right you are! I'm blaming the lack of coffee this morning... – 2017-01-03

1 Answers 1

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You are on good way. When you multiply matrices on left side, you finally get $$a-b+2c-2d = 2$$ which is a linear equation. Solutions are given with three parameters, and thus

$$ X = \begin{pmatrix} a & b\\ c & \frac{a-b+2c-2}{2} \end{pmatrix}$$

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    So you let it like that? – 2017-01-03
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    Yes, you have shown that your matrix equation is equivalent to linear equation $a-b+2c-2d=2$ yourself. Thus, solutions for your matrix equation are $3$-dimensional (+ particular solution). – 2017-01-03