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So I need to solve $\begin{pmatrix} 1 &2\\ 1 &2 \end{pmatrix}X=\begin{pmatrix} -1 & -5\\ -1 &-5 \end{pmatrix}$. I took $X=\begin{pmatrix} a &b \\ c& d \end{pmatrix}$ and I got that $a+2c=-1$ and $b+2d=-5$. What do I do now or do you have any method to solve this and find values for $a,b,c,d$?

P.S. Determinants are $0$ so you can not use the inverse.

4 Answers 4

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Solve this System of equations and you will get the values $a,b$ in dependence of $c,d$. Therefore this Matrix equation has infinitely many Solutions; it is solved by finding Solutions in dependence on Parameters.

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    So I do not need to find any exact solution right? – 2017-01-03
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    Yes, you can solve by arbitrary variable $a,b,c$ or $d$. Only if Matrix that is bevor the variable Matrix $X$ has nonvanishing determinant you have an unique solution. Here Solutions are expressed in Terms of Parameters. – 2017-01-03
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    So i need to write b depending on d and a depending on c and that's it? – 2017-01-03
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    You can write the solution in the form $X = X_0 + dD+cC$ with matrices $X_0,D$ and $C$. – 2017-01-03
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If you got $a+2c=-1$ and $b+2d=-5$, then the set if all solutions is $$ \{\begin{pmatrix}-1-2c&-5-2d\\c&d\end{pmatrix}:c,d\in\mathbb{R}\} $$

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You're correct that $a + 2c = -1$ and that $b + 2d = -5$. This gives 2 (independent) equations in 4 unknowns, so we can expect $4 - 2 = 2$ degrees of freedom in the final answer.

Note then that $a + 2c = -1 \iff a = -1-2c \,$ and also $\,b + 2d = -5 \iff b = -5-2d$, so: $$\begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} X = \begin{pmatrix}-1 & -5 \\ -1 & -5 \end{pmatrix} \iff X = \begin{pmatrix} -1 - 2c & - 5 - 2d \\ c & d \end{pmatrix}$$

That is, this equation will be solved by any such $X$ in this form.

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You can view this as multiplying the left matrix by $X$

$\begin{pmatrix} 1 &2\\ 1 &2 \end{pmatrix} \cdot \begin{pmatrix} a &b \\ c& d \end{pmatrix} = \begin{pmatrix} a+2c&b+2d \\ a+2c & b+2d \end{pmatrix}$

So $\begin{pmatrix} a+2c&b+2d \\ a+2c & b+2d \end{pmatrix}$ and $\begin{pmatrix} -1 & -5\\ -1 &-5 \end{pmatrix}$ are equivalent. Then express solutions in dependence of $c$ and $d$:

$a=-1-2c$ and $b=-5-2d$ then this system has infinite solutions. This means that variables $a$ and $b$ will change everytime you choose values for $c$ and $d$.