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:
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
ADD these to get -6
| 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
Add these to get 22
| 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:
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 |
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 |
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:
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.
|
