Systems of Equations

[math magic]

Could someone please help me with this problem. It would be much appreciated. Thanks.


Solve the following system of equation by writing the augmented matrix and using row reduction.

w + x – y + 3z = -1
x + 2y – 3z = -2
w + 2x + 2y + z = 0
2w + 3x + 2y + 7z = 4

[krtxmrtz]

Could someone please help me with this problem. It would be much appreciated. Thanks.


Solve the following system of equation by writing the augmented matrix and using row reduction.

w + x – y + 3z = -1
x + 2y – 3z = -2
w + 2x + 2y + z = 0
2w + 3x + 2y + 7z = 4

If you Google for "row reduction" you'll get lots of places to go. Here , for example you can type in your matrix and solve by various methods.

At any rate, after you write the augmented matrix,
1 1 -1 3 -1
0 1 2 -3 -1
1 2 2 1 0
2 3 2 7 4
what you've got to do is do linear combinations on the rows like this:

(New Row)_i = p*(Row)_j + q*(Row)_k

where p and q are any appropriate numbers and j and k are any 2 rows (including the old row i) as many times as necessary so that you end up with the identity matrix in the first 4 columns. Then, the 5th column is the solution to the system.

For the matrix you've posted the solution is:

1 0 0 0 1
0 1 0 1 -3
0 0 1 0 2
0 0 0 1 1

so that:
w = 1
x = -3
y = 2
z = 1

[krtxmrtz]

By the way, you may stop the row reduction process when you're half way through, i.e. when your matrix is already in echelon form, like this:

Original matrix.
1 1 -1 3 -1
0 1 2 -3 -1
1 2 2 1 0
2 3 2 7 4

Matrix in echelon form:
1 1 -1 3 -1
0 1 2 -3 -2
0 0 1 1 3
0 0 0 2 2

and now apply back substitution, i.e. solve for the last, use the found (z) value to solve the one before last and so on.