# Solve for $X$ in a simple $2\times2$ equation system.

I posted a similar question recently but I still have problem with this problem and would appreciate any help!

$$\left[ \begin{array}{cc} 9 & -3\\ 5 & -5\end{array} \right] - X \left[ \begin{array}{cc} -9 & -2\\ 8 & 5\end{array} \right] = E$$ With $E$ i pressume they mean the identity matrix $\left[ \begin{array}{cc} 1 & 0\\ 0 & 1\end{array} \right]$.

How should I go on and solve this for the $2\times2$ matrix $X$? a full development so I can follow your solution would be very much appreciated!

Thank you kindly for you help!

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You should specify in your question whether $X$ is a scalar or a matrix. – user1551 Mar 15 '13 at 8:55
Sorry for that. $X$ is a $2\times2$ matrix. – Lukas Arvidsson Mar 15 '13 at 8:58

I will express it as equations, as I think that is easier (at least for beginners).
With $$X=\begin{pmatrix} x_{11} & x_{12}\\ x_{21} & x_{22} \\ \end{pmatrix}$$ At first we make the multiplication \begin{align*} X \cdot \begin{pmatrix} -9 & -2\\ 8 & 3 \end{pmatrix}&= \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \cdot \begin{pmatrix} -9 & -2 \\ 8 & 3 \end{pmatrix}\\ &= \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\ \end{pmatrix} \end{align*} So our equation is $$\begin{pmatrix} 9 & -3 \\ 5 & -5 \end{pmatrix} - \begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\ \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$ Now we write it as a system of equations: \begin{align*} 1&=9- (-9 x_{11} +8x_{12})\\ 0&= -3-(-2x_{11} +3x_{12}) \\ 0&=5-(-9x_{21} +8 x_{21}) \\ 1&=-5 - (-2x_{21}+3x_{22}) \\ \end{align*} If you need help solving this system tell me.

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Thank you very much for your excellent answer! – Lukas Arvidsson Mar 15 '13 at 9:27
You are welcome :) – Dominic Michaelis Mar 15 '13 at 9:28

First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution.

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Denote $B=\begin{pmatrix}-9&-2\\8&5\end{pmatrix}$. Then we have: $$\begin{pmatrix}9&-3\\5&-5\end{pmatrix}-E=XB$$ Observe that $\det B=-45+16\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have: $$X=\left(\begin{pmatrix}9&-3\\5&-5\end{pmatrix}-E\right)B^{-1}$$ Do you know how to find $B^{-1}$?

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Thanks for your answer! I get the final result to be: $\left[ \begin{array}{cc} \frac{-64}{29} & \frac{-43}{29}\\ \frac{-48}{29} & \frac{-54}{29}\end{array} \right]$ Is that correct? – Lukas Arvidsson Mar 15 '13 at 9:13
Appearantly, the two above is correct but not the two below... – Lukas Arvidsson Mar 15 '13 at 9:16
Yes, you have a computation error somewhere. You should get $X=\frac1{29}\begin{pmatrix}-64&-43\\-73&-64\end{pmatrix}$ – Dennis Gulko Mar 15 '13 at 9:19
Yes that is correct. Thank you for your help! – Lukas Arvidsson Mar 15 '13 at 9:26