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 Math Help > Sets, Set theory, Number systems > Matrix Math > Cramer's Rule
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On 7/20/00 4:35:24 PM, Kyla Blomgren wrote:

I need help solving matrices.  My professor has taken one day to explain the "old fashion technique" (so to speak) as well as Cramer's Rule.  Unfortunately, I am not comprehending either as well as I'd like to. Our first exam is tomorrow and matrices are to be expected. Any help/tips/techniques for solving 3x3 matrices would be greatly appreciated. Included is a a sample problem; a step by step solution using both methods would be invaluable.  Thanx in advance for any support.

> 3x + 2y + 2z = 3
> x + 2y - z = 5
> 2x - 4y + z = 0
>
> [3 2 2 3]
> [1 2 -1 5]
> [2 -4 1 0]

I want to thank Stan Hanson for the explanation, which I'm copying here.  The only reason I'm repeating it is to improve the formatting.  I hope you like it...

# Find the determinant of the coefficient matrix

First Problem: How to find the determinant of a 3 X 3 matrix.

Example: Find the determinant of the following:

 3 2 2 1 2 -1 2 -4 1

Rewrite as follows:

 3 2 2 3 2 2 1 2 -1 1 2 -1 2 -4 1 2 -4 1

You now perform a series of multiplications, additions, and subtractions as follows:

Three products you are going to ADD:
3 times 2 times 1 = 6
2 times -1 times 2=-4
2 times 1 times -4=-8

 3 2 2 3 2 2 1 2 -1 1 2 -1 2 -4 1 2 -4 1 -6 = 6 -4 -8

Three products you are going to ADD:
2 times 2 times 2 = 8
2 times 1 times 1 = 2
3 times -1 times -4= 12

 3 2 2 3 2 2 1 2 -1 1 2 -1 2 -4 1 2 -4 1 12 2 8 = 22

Now subtract the 22 from the -6 to get -28
This is the determinant of your matrix.

Comment 1: You only need to repeat the first two columns; not all three, and the method still works.

Comment 2: If you see where the products came from in the rectangle of numbers above, then you don't really need to copy any part of the matrix. Note that the 3 times 2 times 1 starts at the upper left corner and moves diagonally down to the right. Note that the 2 times 2 times 2 starts at the upper right corner and moves diagonally down to the left.  Note that the other products follow similar diagonal patterns, with the diagonal "wrapping" from the left side to the right side, and vice-versa.

The determinant we just found is called "the determinant of the coefficient matrix" because the elements in the matrix are the coefficients of "x" "y" and "z".

# Find the x-determinant

Now, Cramer says that if we want to solve for "x" we should replace the "x" coefficients by the constants (3,5,0) as follows:

 3 2 2 5 2 -1 0 -4 1

Then we should find the determinant of this matrix.

So, the "x-determinant" is:
Add: 6 + 0 + -40 = -34
Add: 0 +10 + 12 = 22
Now subtract 22 from -34 to get -56

 3 2 2 3 2 2 5 2 -1 5 2 -1 0 -4 1 0 -4 1 -34 = 6 0 -40
 3 2 2 3 2 2 5 2 -1 5 2 -1 0 -4 1 0 -4 1 12 10 0 = 22
-34 - 22 = -56

Then we should divide that "x" determinant by the coefficient matrix determinant.

Now, divide -56 by -28 to get +2

So, x = +2

# Find the y-determinant

To find "y" replace the "y" coefficients by the constants to get the following matrix, and find its determinant:

 3 3 2 1 5 -1 2 0 1
 3 3 2 3 3 2 1 5 -1 1 5 -1 2 0 1 2 0 1 9 = 15 -6 0
 3 3 2 3 3 2 1 5 -1 1 5 -1 2 0 1 2 0 1 0 3 20 = 23
9 - 23 = -14

The "y" determinant is -14
Divide -14 by -28 to get y = 0.5

# Find the z-determinant

To find "z" replace the "z" coefficients by the constants to get the following matrix, and find its determinant:

 3 2 3 1 2 5 2 -4 0
 3 2 3 3 2 3 1 2 5 1 2 5 2 -4 0 2 -4 0 8 = 0 20 -12
 3 2 3 3 2 3 1 2 5 1 2 5 2 -4 0 2 -4 0 -60 0 12 = -48
8 + 48 = 56

The "z-determinant" is 56
Divide 56 by -28 to get z=-2

So the solution of the system of equations is as follows:

 x = 2,  y = 0.5,  z = -2

Are we done?  No, we should check the results in the original system of equations:

(3)(2) + (2)(0.5) + (2)(-2) = 3
6 + 1 - 4 = 3

(2) + (2)(0.5) - (-2) = 5
2 + 1 + 2 = 5

(2)(2) - (4)(0.5) + (-2) = 0
4 - 2 - 2 = 0

Yes, it checks.

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