Solve the system of equations by using matrix methods.

Question

-2x+3y+z=-2

-3x+y+3z=-4

2y-z=0

 

With the method I used, I got ridiculous fractions and I don't think I'm doing it correctly. Help please?

Answer 1

Hi Courtney,

 

The answers are: x = 6. y = 2, z = 4.  Here is how I did it:

 

-2x+3y+z=-2
-3x+y+3z=-4
2y-z=0

 

x   y   z

-2  3  1  |  -2

-3  1  3  |  -4

0   2  -1 |   0

 

-1(R2) + R1

 

 1  2  -2  |  2

-3  1   3  | -4 

 0  2  -1  |  0

 

-1(R3) + R1

 

 1  0  -1  |  2

-3  1   3  | -4

 0  2  -1  |  0

 

3(R1) + R2

 

 1  0  -1  |  2

 0  1   0  |  2

 0  2  -1  |  0

 

-2(R2) + R3

 

 1  0  -1  |  2

 0  1   0  |  2

 0  0  -1  | -4

 

-1(R3) + R1

 

1  0  0  |  6

0  1  0  |  2

0  0 -1  | -4

 

-1(R3)

 

1  0  0  |  6   ---> x

0  1  0  |  2   ---> y

0  0  1  |  4   ---> z

 

Let's check it:

 

-2x+3y+z=-2

 

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

-12 + 6 + 4 = -12 + 10 = -2

-3x+y+3z=-4

 

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

-18 + 2 + 12 = -4

-18 + 14 = -4

2y-z=0

2(2) - 4 = 4 - 4 = 0

 

All three conditions checks out.

 

In conclusion, x = 6, y = 2, z = 4.  I hope you understand how I did it.

Good luck to you.