# Solving for X in a simple matrix equation system.

I am trying to solve for X in this simple matrix equation system:

$$\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix} - X\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix} = E$$ where $E$ is the identity matrix.

If I multiply $X$ with $\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix}$ I get the following system:

$$\begin{bmatrix}5x_1 & -1x_2\\6x_3 & -4x_4\\\end{bmatrix}$$

By subtracting this from $\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix}$ I get $\begin{bmatrix}7 - 5x_1 & 7 + 1x_2\\2 - 6x_3 & 4 + 4x_4\\\end{bmatrix} = \begin{bmatrix}1 & 0\\0 & 1\\\end{bmatrix}$

Which gives me:

$7-5x_1 = 1$

$7+1x_2 = 0$

$2-6x_3 = 0$

$4+4x_4 = 1$

These are not the correct answers, can anyone help me out here?

Thank you!

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What is your $E$? – awllower Jul 1 '13 at 14:18
Sorry, That is the identity Matrix. – Lukas Arvidsson Jul 1 '13 at 14:19
I think you have made a mistake in the matrix multiplication, it appears you have multiplied element by element which is not the definition of matrix multiplication.$$\left[ \begin {array}{cc} X_{{1,1}}&X_{{1,2}}\\ X_{ {2,1}}&X_{{2,2}}\end {array} \right] \left[ \begin {array}{cc} 5&-1\\ 6&-4\end {array} \right]= \left[ \begin {array}{cc} 5\,X_{{1,1}}+6\,X_{{1,2}}&-X_{{1,1}}-4\,X_{ {1,2}}\\ 5\,X_{{2,1}}+6\,X_{{2,2}}&-X_{{2,1}}-4\,X_{ {2,2}}\end {array} \right]$$ – Graham Hesketh Jul 1 '13 at 14:42
Thank you @GrahamHesketh, that seems to be the problem here! – Lukas Arvidsson Jul 1 '13 at 14:50

$$\begin{bmatrix}7 & 7\\2 & 4\\\end{bmatrix} - X\begin{bmatrix}5 & -1\\6 & -4\\\end{bmatrix} = I$$ where $I$ is the identity matrix.

Hint: reconsider what multiplication by $X$ will look like: $X$ will be a $2\times 2$ matrix, if matrix multiplication and addition is to be defined for this equation.

So if $X = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}$, then $$X\begin{bmatrix}5 & -1\\6 & -4 \end{bmatrix} = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & -4\end{bmatrix} = \begin{bmatrix}5x_1+6x_2&-x_1-4x_2\\5x_3+6x_4&-x_3-4x_4\end{bmatrix}$$

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Thank you for pointing out my error amWhy! That helped a lot :) – Lukas Arvidsson Jul 1 '13 at 14:59
You're welcome, Lukas! Your post was exactly the sort of post we like: showing your work helps us "hone in" on where the mistake(s) was made. +1 for your question. – amWhy Jul 1 '13 at 15:00
Very nice Amy $\langle +\rangle=\{+,+^+,+^{2+},\cdots\}$ – Babak S. Jul 1 '13 at 15:14
I like the "group theoretic" generator $+$, @Babak! ;-) – amWhy Jul 1 '13 at 15:15
I am glad I could make you smiley just for moments. :-) – Babak S. Jul 1 '13 at 15:16

Since $\begin{pmatrix}7&7\\2&4\end{pmatrix}-X\begin{pmatrix}5&-1\\6&-4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$, we obtain:
$\begin{pmatrix}6&7\\2&3\end{pmatrix}=\begin{pmatrix}5x_1+6x_2&-x_1-4x_2\\5x_3+6x_4&-x_3-4x_4\end{pmatrix}$, where $X=\begin{pmatrix}x_1&x_2\\x_3&x_4\end{pmatrix}$.
Now you can multiply both sides of the equation by $\frac{1}{-14}\begin{pmatrix}-4&1\\-6&5\end{pmatrix}$ =(inverse of $\begin{pmatrix}5&-1\\6&-4\end{pmatrix}$), to find:
$X=\frac{1}{-14}\begin{pmatrix}6&7\\2&3\end{pmatrix}\begin{pmatrix}-4&1\\-6&5\end{pmatrix}=\frac{1}{-14}\begin{pmatrix}-66&41\\-26&17\end{pmatrix}$.
Hope this helps.

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Thank you for your answer awllower. It is not correct but i suspect that might be due to a simple calculation error, it helped me solve the question anyhow. – Lukas Arvidsson Jul 1 '13 at 14:52
And what is the correct answer? – awllower Jul 1 '13 at 14:58
The correct answer is $x_1 = \frac{66}{14}, x_2 = \frac{-41}{14}, x_3 = \frac{26}{14}, x_4 = \frac{-17}{14}$ – Lukas Arvidsson Jul 1 '13 at 15:01
Finally there are no more errors. Hope you like it. :D – awllower Jul 1 '13 at 15:18
Thank you @awllower! – Lukas Arvidsson Jul 3 '13 at 11:33

Lets set your equation to look like $C-XA = E$, where $C$ is the constant matrix and $A$ is your coefficient matrix.

first, isolate your matrix $AX$ by moving the constant matrix to the other side of he equation. manipulate the equation to give you $AX$ as positive. then multiply the matrix $A$ by its inverse. Notice that this is a right hand multiplication. Now you are left with your $X$ vector times the identity on the left.

Now you have $$EX = (C-E)A^{-1}$$

You can combine $C$ and $E$ first, or distribute $A^{-1}$ through, it doesn't matter. Just remember that you are doing a right hand multiplication, not left hand. This is important because multiplication is not communicative in matrices.

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Thank you for your answer Tyler, much appreciated! – Lukas Arvidsson Jul 1 '13 at 14:51
using LaTeX helps a lot – gt6989b Jul 1 '13 at 14:51