Linear Equations Problem. Matrices. Help Please?

Question

Hello, I've done a few of these problems myself, but this one stumps me. Can someone complete the problem step by step so I'll know how to do this type of problem? Thank you very very much.

-x + y - z = 7
5x - 4z = 8
- 4x - y - 3z = 1

Answer 1

Are you allowed to use a calculator?

Because this is a 3 x 3 matrix and those matrices take a while.

I don't know how to do it out by hand unless its 2x2 matrix.

But if you're allowed to use a calculator then here's how you do it (TX 83 or 84):

Press 2nd (upper left corner) and then X^-1. You should see the word "MATRIX".

Press enter.

Go to "EDIT". Then press enter (doesn't matter which one you want):

Enter the numbers like this: (Make sure it's 3x3 box, you can edit it)

[ -1 1 -1

5 0 -4

-4 -1 -3]

Once you're done press 2nd + QUIT.

Go back to MATRIX. Then go back to EDIT and choose a different letter than you chose last time.

Then press enter. (Make sure this time its a 3 x 1 box)

[7

8

1]

2nd + QUIT again.

Then go back to MATRIX and this time don't go over to EDIT.

Go press on the box you entered in for the FIRST matrix u entered in.

Then do the X^-1 button so it should look like this on your calc: (If you entered ur numbers on A)

[A]^-1. (You're finding it's inverse)

Then press enter.

Then go back to MATRIX and times it by the original A^-1 answer you got.

Then your answer is:

[0

5

-2]

Meaning X = 0

Y=5

Z=-2



Hoped I helped.

Or you can use substitution but calculator is faster :P.

Answer 2

want to know a simple trick? Use Matrices.



-1 1 -1 | 7

5 0 -4 | 8

-4 -1 -3 | 1



Then use Gauss-Jordan elimination to put the matrix in reduced row-echelon formation



5(row 1) + row 2 >>row2; 4(row 1) - row 3 >> row3



-1 1 -1 | 7

0 5 -9 | 43

0 5 -1 | 27



then clear the second column

row 2 - 5 (row 1) >> row 1 ; row 2 - row 3 >> row 3



-1 0 -4 | 8

0 5 -9 | 43

0 0 -8 | 16



third column

row 3 - 2(row 1) >> row 1 ; 9(row 3) - 8 (row 2) >> row 2



-1 0 0 | 0

0 5 0 | -200

0 0 -8 |16

use the columns

-x = 0 ; 5y = -200 ; -8z = 16

x = 0

y = -40 (doesn't work. 5 is the real value)

z = -2



Okay, I'm not sure where my math was wrong in the elimination, but it was wrong somewhere, and I've checked multiple times. The values are correct, but the equations aren't working.

Answer 3

-x + y - z = 7 eq 1

5x - 4z = 8 eq 2

- 4x - y - 3z = 1 eq 3



-5x-4z=8 eq 4=eq 1 +eq 3

-8z=16 eq 4= eq 4+eq 2

z=16/-8

z=-2

5x=8+4z

5x=0

5x=0

x=0

y=7+x+z

y=5