how do you solve 2x-y-3z=13, x+y+z=2, 2x-2y-3z=11 in matrix?

Answer 1

2 -1 -3 x = 13
1 1 1 y = 2
2 -2 -3 z = 11

The inverse of this 3x3 matrix =

-0.2 0.6 0.4
1 0 -1
-0.8 0.4 0.6

x = -0.2 0.6 0.4 13
y = 1 0 -1 2
z = -0.8 0.4 0.6 11

x = 3, y = 2 z = -3

Answer 2

Okay, in my Algebra 2 class we just learned how to do these.
the number in front of each varable are put in their corresponding spots in the matrix.
first the x, y, then z. D = the determinent.
So D = [2 -1 -3
1 1 1
2 -2 -3] = 5

Then you solve for the dterminent of x: to do this you put the answers of the equations in for the x's
Dx = [13 -1 -3
2 1 1
11 -2 -3] Which equals -15


You do this same thing with the y's and z's.
Dy = [2 13 -3
1 2 1
2 11 -3] = 10
Dz = [2 -1 13
1 1 2
2 -2 11] = -15

We know that x = Dx/D; y = Dy/D; z = Dz/D

So when you solve out the determinents you get x=-3; y=2; and z=-3



Sorry for the way the matrices are, yahoo wouldnt let me write them the way I originally did.