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Date: 09/21/98 at 13:39:14
From: Khamisi Easton
Subject: Linear algebra

I know that you have already been asked this question but I didn't
really understand the answer that you gave. I am trying to solve a 3x3
matrix. I know how to solve for a 2x2 matrix. (2 equations 2 unknowns).
I don't know how to solve for three equations three unknowns.

Here is what I have so far. I would appreciate it if you could help me
out. Can you please solve any 3x3 matrix using this format?

A^(-1) * A = I

[3 -1  2]
A = [1  4 -3]
[2  5  3]

So:

[c11 c12 c13]   [3 -1   2]   [1 0 0]
[c21 c22 c23] * [1  4  -3] = [0 1 0]
[c31 c32 c33]   [2  5   3]   [0 0 1]

3c11 + 1c12 + 2c13 = 1
-1c21 + 4c22 + 5c23 = 0
2c31 + -3c32 + 3c33 =0

...and so on.

Thank you,
Khamisi
```

```
Date: 09/21/98 at 17:43:03
From: Doctor Anthony
Subject: Re: Linear algebra

Hello Khamisi,

You certainly don't want to solve 9 simultaneous equations, which your
method suggests. The two ways that are normally used are either row
operations or finding the adjugate matrix. Below is an example of the
latter method.

Use the inverse matrix method to solve the following set of equations.

x + 3y - z = -3
3x - y + 2z = 1
2x -y + z = -1

You require the inverse of the matrix formed by the coefficients of
the equations. A1, A2, B1, C2, etc are the cofactors of the elements
a1, a2, b1, c2 and the way they are formed is described below.

If  M = [1   3   -1|  then:      A1 = 1     -A2 = 2      A3 = 5
|3  -1    2|            -B1 = -1     B2 = 3     -B3 = 5
|2  -1    1]             C1 = -1    -C2 = -7     C3 = -10

labeling the positions of the matrix as:

[a1   b1   c1|
|a2   b2   c2|
|a3   b3   c3]

Note that A1 is the cofactor of a1, A2 is the cofactor of a2, and so
on. To get the cofactors, cross-out the row and column and find the
determinant of the remaining matrix. For example, to find the cofactor
of A1, cross out its corresponding row and column to get the matrix:

[-1    2|
|-1    1]

which has determinant 1. We write the cofactors in the transposed
positions as shown, and use the rule of signs with alternating
+ and - signs.

[A1   A2   A3|  =  [1   -2     5|
|B1   B2   B3|     |1    3    -5|
|C1   C2   C3]     |-1   7   -10]

Finally the inverse matrix is:

so we must get det(M) to complete the calculation.

det(M) = -1+12+3-2+2-9 = 5

and so:

M^(-1)  = (1/5)[1   -2     5|
|1    3    -5|
|-1   7   -10]

We then have:

[x|  = (1/5)[1   -2     5| [-3|
|y|         |1    3    -5| |1 |
|z]         |-1   7   -10] |-1]

The solutions are then found by multiplying out the matrices on the
right:

[x|    [-2|
|y| =  | 1|
|z]    | 4]

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Algebra

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