# inverse matrices

• Jul 15th 2008, 08:17 PM
mankvill
inverse matrices
This is probably the most confusing thing I've ever stumbled across in my academic career. (Doh)

http://img210.imageshack.us/img210/1134/math1qx8.jpg
• Jul 15th 2008, 08:33 PM
Jhevon
Quote:

Originally Posted by mankvill
This is probably the most confusing thing I've ever stumbled across in my academic career. (Doh)

http://img210.imageshack.us/img210/1134/math1qx8.jpg

if you write the augmented matrices for these problems, you will end up with matrix equations of the form: $\displaystyle A \bold{x} = b$

to solve for $\displaystyle \bold{x}$ using the inverse matrix method, we simply multiply both sides by the matrix $\displaystyle A^{-1}$ to obtain $\displaystyle \bold{x} = A^{-1}b$

For example, you first system of equations can be represented by:

$\displaystyle \left( \begin{array}{ccc} 1 & 3 & 0 \\ 0 & 1 & -3 \\ -2 & -1 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 9 \\ 10 \\ -4 \end{array} \right)$
............$\displaystyle \downarrow$..................$\displaystyle \downarrow$...............$\displaystyle \downarrow$
...........$\displaystyle A$..................$\displaystyle \bold{x}$...............$\displaystyle b$

I hope you see how we got this. now, find the inverse of $\displaystyle A$ and multiply the right side by it (on the left)
• Jul 15th 2008, 08:34 PM
mankvill
I don't understand at all. :(

I only have like 20 minutes left to finish this test, could someone just walk me through it? :( I'd really appreciate it.
• Jul 15th 2008, 08:34 PM
mr fantastic
Quote:

Originally Posted by mankvill
This is probably the most confusing thing I've ever stumbled across in my academic career. (Doh)

http://img210.imageshack.us/img210/1134/math1qx8.jpg

Why is this so confusing? You should have examples in your class notes and textbook to follow .....?

The matrix equation that represents the first system of linear equations is $\displaystyle A X = B$ where

$\displaystyle A = \left[ \begin{tabular}{c c c } 1 & 3 & 0 \\ 0 & 1 & -3 \\ -2 & -1 & 3 \\ \end{tabular} \right]$

$\displaystyle X = \left[ \begin{tabular}{c} x \\ y \\ z \\ \end{tabular} \right]$

$\displaystyle B = \left[ \begin{tabular}{c} 9 \\ 10 \\ - 4 \\ \end{tabular} \right]$.

The solution is given by $\displaystyle X = A^{-1} B$ where I leave it to you find $\displaystyle A^{-1}$ (the inverse of matrix A) and do the matrix multiplication.
• Jul 15th 2008, 08:40 PM
mankvill
But what's the inverse of Matrix A?

The textbook gives an odd way of finding it.
• Jul 15th 2008, 08:45 PM
mr fantastic
Quote:

Originally Posted by mankvill
But what's the inverse of Matrix A?

The textbook gives an odd way of finding it.

Read this: Matrix Inversion: Finding the Inverse of a Matrix.

Or this: The inverse of an nxn matrix.

Or this: Inverse of a Matrix.

And of course you can always just use technology such as a TI-89 calculator.

You should already know how to find the inverse of a 3x3 matrix if you're trying to use matrices to solve systems of linear equations.

You need to go back and revise the basic prerequisite material.
• Jul 15th 2008, 08:49 PM
Jhevon
why did you wait till the last minute. of course we can't do the problem for you, but here is some direction
Quote:

Originally Posted by mankvill
But what's the inverse of Matrix A?

The textbook gives an odd way of finding it.

do you know how to bring a matrix to reduced row-echelon form?

set this up:
...........$\displaystyle A$...............$\displaystyle I_n$
...........$\displaystyle \downarrow$................$\displaystyle \downarrow$
$\displaystyle \left( \begin{array}{ccc|ccc} 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 1 & -3 & 0 & 1 & 0 \\ -2 & -1 & 3 & 0 & 0 & 1 \end{array} \right)$

now, perform row operations (on the entire matrix) to get the matrix $\displaystyle A$ to look like $\displaystyle I_n$. once this is done, the matrix in the $\displaystyle I_n$ position will end up being the matrix $\displaystyle A^{-1}$