# Solving Matrix equstions of the type: AX = C (FIND The X) - (NOT HW)

1. Jan 13, 2012

### rugh

Hi,
Last day I learned about matrix equations like this:
AX = C
A–1AX = A–1C
IX = A–1C
X = A–1C

The objective of this is to Find X.

Here they done it by right multiplication of the Inverse matrix (A^-1).

My Questions are:

1) Where can I get more of these exercises and in higher level? (This exercise was fairly simple, but what about something like p^Tx^-1P^-1PX=APX)
Do you have some book that also shows in understandable way how to solve this?
Because when it's get harder I don't know how to approach it.
Do you have any tips? (Or good book)

2) The same question, but for the system of linear equations like this:

X - ky + z = 2 + kz
kx - 4y - 6z = k^2 + 2k - 9
2x + 4y + 8z = kx+2k(y+z-1) + 13 - k^2

Again, How to approach , and maybe some "Rule of thumbs" in general in solving equations.
For example I don't multiply with "-" and all the solution goes wrong…

Thanks a lot :)

2. Jan 13, 2012

### chiro

Hey rugh and welcome to the forums.

What environment are you learning this in? Is it self-study or are you taking a course in this?

In terms of finding the inverse, there is actually more than one method. The first way to learn this is by studying Gaussian Elimination. From that you can go on to other methods, but I would suggest learning that first since that gives you an intuitive idea of how we calculate the inverse from first principles and will help you in other ways like with LU decomposition.

In terms of rearranging matrix equations, you can do either add/subtract a matrix from each side, scalar multiply both matrices on both sides, premultiply a matrix or postmultiply a matrix. Note that if you multiply one side on the left, then you have to do it on the same side on the right. For example if AXA^-1 = C then A^-1 x AXA^-1 = A^-1 x C and not C x A^-1: these are not the same thing.

To approach the 2nd question it will help if you understand Gaussian elimination. Once you understand this, you will be able to understand all these kinds of problems.

3. Jan 13, 2012

### HallsofIvy

If you are attempting problems like this, then presumably you know how to do things like "combining like terms". For example, if, in the first equation, x- ky+ z= 2+ kz, subtracting kz from both sides gives x- ky + (1- k)z= 2 so that we have combined the z terms. The second equation has no x, y, or z on the right so we can leave that. The third equation is the same as
2x+ 4y+ 8z= kx+ 2ky+ 2kz- 2k+ 13- k^2 (distributive law). If we subtract kx, 2ky, and 2kz from both sides, then we have (2- k)x+ (4- 2k)y+ (8- 2k)z= -2k+ 13- k^2.

That is, the three equations are
x- ky+ (1- k)z= 2
kx- 4y- 6z= k^2+ 2k- 9
(2- k)x+ (4- 2k)y+ (8- 2k)z= -2k+ 13- k^2.

We can write that as the matrix equation
$$\begin{bmatrix}1 & -k & 1- k \\ k & -4 & -6 \\ 2- k & 4- 2k & 4- 2k\end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix}2 \\ k^2+ 2k- 9\\ -2k+ 13- k^2\end{bmatrix}$$

Now you can think of that as "Ax= b" where A is the matrix
$$\begin{bmatrix}1 & -k & 1- k \\ k & -4 & -6 \\ 2- k & 4- 2k & 4- 2k\end{bmatrix}$$
and the solution is, as you say, $x= A^{-1}b$.

Of course, whether the matrix A has an inverse and what it is depends upon k.

Last edited by a moderator: Jan 13, 2012
4. Jan 14, 2012

### rugh

Thank all of you for your replay,
I know the Gauss Elimination, I just searching for more exercises and ways of solving the equations I wrote in the 2 Sections.
I just want more exercises, are there any recommended books or web pages?

10X