The given system expressed as a matrix would be
0 4 -1 -5
1 5 1 -3
2 -1 3 13
To solve this, first, we have to change the first row for the second
1 5 1 -3
0 4 -1 -5
2 -1 3 13
Now, we multiply the first row by 2, then subtract it from the third row.
1 5 1 -3
0 4 -1 -5
0 -11 1 19
Then, we multiple the second row with -11/4 to subtract it from the third row.
1 5 1 -3
0 4 -1 -5
0 0 -7/4 21/4
The resulting system would be

We solve for z

Therefore, z is equal to -3.
We use z to find y in the second equation.

Therefore, y is equal to -2.
We use y and z to find x in the first equation.

Therefore, x is equal to 10.