How do you solve a 4x4 Matrix?

Question

1 1 1 2 | 6
1 2 0 1 | -2
1 1 3 -2| 12
1 1 -4 5| -16
What order do I take to solve this problem?

Answer 1

1 1 1 2 | 6 I
1 2 0 1 | -2 II
1 1 3 -2| 12 III
1 1 -4 5| -16 IV

perform II - I, III - I, IV - I

1 1 1 2 | 6
0 1 -1 -1 | -8
0 0 2 -4 | 6
0 0 -5 3 | -22

III / 2

1 1 1 2 | 6
0 1 -1 -1 | -8
0 0 1 -2 | 3
0 0 -5 3 | -22

II + III, IV + 5III

1 1 1 2 | 6
0 1 0 -3 | -5
0 0 1 -2 | 3
0 0 0 -7 | -7

IV / -7

1 1 1 2 | 6
0 1 0 -3 | -5
0 0 1 -2 | 3
0 0 0 1 | -1

II + 3IV, III + 2IV

1 1 1 2 | 6
0 1 0 0 | -8
0 0 1 0 | 1
0 0 0 1 | -1

I - II - III - 2IV

1 0 0 0 | 15
0 1 0 0 | -8
0 0 1 0 | 1
0 0 0 1 | -1


I'd double check my math since I didn't pull out a piece of paper to solve it. I probably made a small mistake somewhere. However, the steps should be correct.

There are many paths you could have taken, however this is what I would have done.

Answer 2

Use Gaussian elimination:

Subtract the second row from the first, then the third row from the first, then the fourth row from the first. You get

1 1 1 2 | 6
0 1 -1 -1 | -8
0 0 2 -4 | 6
0 0 -5 3 | -22

Now subtract -5/2 times the third row from the fourth:

1 1 1 2 | 6
0 1 -1 -1 | -8
0 0 2 -4 | 6
0 0 0 -7 | -7

Now you can use back substitution to obtain the solution, working your way up from the fourth row.

The solution is [ 1 -2 5 1 ].

Answer 3

use Gaussian elimination on the matrix. or if you can use Gauss-Jordan elimination. (get leading ones with 0 in the rest of the column by doing simple row operation) once you have get the system of equations. then solve.

i hope that explained it well, I'll do it for you if you want, just e-mail me.

Answer 4

By taking determinants of all 3/3 matrices and summing up

Answer 5

no