solve systems of equations by use of determinants?

Question

2x + y - z=4
4x - 3y - 2z= -2
8x - 2y - 3z =3

Answer 1

See if I can remember how to do this.

The denominator will be:

| +02 +01 -01 |
| +04 -03 -02 |
| +08 -02 -03 |

Which should be:

2((-3)(-3) - (-2)(-2)) - 1((4)(-3) - (8)(2)) - 1((4)(2) - (8)(-2)) =
2(9 - 4) - (-12 + 16) - (8 + 16) =
10 + 4 - 24 = -10

And the numerator for x would be:

| +04 +01 -01 |
| -02 -03 -02 |
| +03 -02 -03 |

4(9 - 4) - (6 + 6) - (4 + 9) = 20 - 12 - 13 = -5

For y:

| +02 +04 -01 |
| +04 -02 -02 |
| +08 +03 -03 |

2(6 + 6) - 4(-12 + 16) - (12 + 16) = 24 - 16 - 28 = -30

For z:

| +02 +01 +04 |
| +04 -03 -02 |
| +08 -02 +03 |

2(-9 - 4) - (12 + 16) + 4(-8 + 24) = -26 - 28 + 64 = 10

Therefore:

x = 5/10=1/2, y = -30/10 = -3, z = 10/10 = 1

Answer 2

The easiest way to solve this problem is with a matix.

list the coefficents. varibles, and sums.
[ 2 1 -1 ] [x ] [4]
[4 -3 -2 ] [y] = [-2]
[ 8 -2 -3 ] [z] [3]

[ 2 1 -1 ] -1 [4] [x]
[4 -3 -2 ] [-2] = [y]
[ 8 -2 -3 ] [3] [z]

find the determinat of the 3x3 matrix, put it under 1, multiply the matrix with 1/det and then mulitply the product with the 3x1 matrix to get the answer for x, y, and z.

Answer 3

x = |18 . 3| divided by ability of |4 . 3| .. .. |19 . 5|... .. .. .. ...|3 . 5|   = (18*5 - 19*3)/(4*5 - 3*3)   = 33/11   = 3 y = |4 . 18| divided by ability of |4 . 3| .. .. |3 . 19|... .. .. .. ...|3 . 5|   = (4*19 - 18*3)/11   = 22/11   = 2