# Thread: Help with Matrix equation

1. ## Help with Matrix equation

Hello,

I'm asked to solve for X

I have; AX = 2X + C

$\displaystyle A = \begin{matrix} 3 & -1 \\ 1 & 2 \\ \end{matrix}$

$\displaystyle C = \begin{matrix} 1 & 2 & 0 \\ 0 &-1 & 1 \\ \end{matrix}$

I have an answer but it looks wrong (not very attractive numbers)

What I've done is solve for X

X = C /(A-2) or X=-0.5 (A^-1)(C)

but I think here is where I am going wrong, I also calculated the determinant of A and adj(A) to find the inverse but my numbers are like 0.286 0.143 etc

any help would be appreciated

---

EDIT: Sorry the matrix messed up in Latex but its still visible

2. ## Re: Help with Matrix equation

Hey entrepreneurforum.co.uk.

Given you have AX = 2X + C you can get (A-2I)X = C which implies X = (A-2I)^(-1)*C.

Try calculating (A-2I)^(-1) and see what you get for that.

Note that you can't do what you did (divide by a matrix): you can only pre-multiply or post-multiply and in this particular instance, you need to post multiply since Y^(-1)*Y = I.

3. ## Re: Help with Matrix equation

An alternative to the matrix inversion method is to see that X is 2 by 3, calling its elements a,b,c,d,e,f, say, substituting in on both sides of the equation, simplifying, and equating corresponding elements.
The resulting equations solve easily.

4. ## Re: Help with Matrix equation

Originally Posted by entrepreneurforum.co.uk
A = \begin{matrix} 3 & -1 \\ 1 & 2 \\
\end{matrix}

C =
\begin{matrix}
1 & 2 & 0 \\
0 &-1 & 1 \\
\end{matrix}
A quick LaTeX tip: You are using |...the format is \ for the coding. Similarly use \\ between lines of the matrix. It is also a good idea to put everything on one line.
For example:
$$A = \left ( \begin{matrix} 3 & -1 \\ 1 & 2 \end{matrix} \right )$$
gives
$\displaystyle A = \left ( \begin{matrix} 3 & -1 \\ 1 & 2 \end{matrix} \right )$

Likewise the matrix for C is:
$\displaystyle C = \left ( \begin{matrix} 1 & 2 & 0 \\ 0 & -1 & 1 \end{matrix} \right )$

-Dan