SOLUTION: Solve using Gaussian elimination 5x − 3y − z = 2 x − 6y + z = 7 2x + 6y − 2z = −8

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Question 1092353: Solve using Gaussian elimination
5x − 3y − z = 2
x − 6y + z = 7
2x + 6y − 2z = −8
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

There are countless different paths you can take to the final answer any time you are solving a system of equations using Gaussian elimination. I will show you one path, along with some comments about why I chose the path I did.

All of the steps in Gaussian elimination are straightforward; but it is annoyingly easy to make silly arithmetic errors along the way.

Since the process is prone to arithmetic errors, one thing I do whenever possible is avoid using fractions, since careless errors are easier to make with fractions than with whole numbers.

matrix%283%2C4%2C5%2C-3%2C-1%2C2%2C1%2C-6%2C1%2C7%2C2%2C6%2C-2%2C-8%29

The first thing I would do is factor out any common factor in a row if it doesn't introduce fractions. The third row has a common factor of 2. So

matrix%283%2C4%2C5%2C-3%2C-1%2C2%2C1%2C-6%2C1%2C7%2C1%2C3%2C-1%2C-4%29

The first step is to get a 1 in row 1, column 1. By far the easiest way to do that is move either the second or third row to the first row.

matrix%283%2C4%2C1%2C3%2C-1%2C-4%2C1%2C-6%2C1%2C7%2C5%2C-3%2C-1%2C2%29

Next we use the 1 in row 1 column 1 to get 0's in the rest of column 1. We replace row 2 with (row 2 - row 1); we replace row 3 with (row 3 - 5*row 1).
matrix%283%2C4%2C1%2C3%2C-1%2C-4%2C0%2C9%2C-2%2C-11%2C0%2C-18%2C4%2C22%29

Again we see where we can factor a common factor out of a row, in row 3:
matrix%283%2C4%2C1%2C3%2C-1%2C-4%2C0%2C9%2C-2%2C-11%2C0%2C-9%2C2%2C11%29

Looking at this matrix, we see that row 3 and row 2 are opposites of each other, so when we add row 2 and row 3 we get a row of all 0's. This tells us that we are going to get an infinite family of solutions, rather than a single solution.
matrix%283%2C4%2C1%2C3%2C-1%2C-4%2C0%2C9%2C-2%2C-11%2C0%2C0%2C0%2C0%29

Next we need to get a 1 in row 2 column 2 -- without changing column 1. Unfortunately our only option is to divide the second row by 9, introducing fractions.
matrix%283%2C4%2C1%2C3%2C-1%2C-4%2C0%2C1%2C-2%2F9%2C-11%2F9%2C0%2C0%2C0%2C0%29

This matrix tells us that z can be any number; then y can be found using the equation
y-%282%2F9%29z+=+-11%2F9 or
y+=+%282%2F9%29z+-+11%2F9
and then x could be found using the equation
x%2B3y-z+=+-4 or
x+=+-3y%2Bz-4

But it is better to take the Gaussian elimination one step further so that we can express both x and y in terms of the single parameter z. So we use the "-2/9" in row 2 column 3 to get a 0 in row 1 column 3.
matrix%283%2C4%2C1%2C0%2C-1%2F3%2C-1%2F3%2C0%2C1%2C-2%2F9%2C-11%2F9%2C0%2C0%2C0%2C0%29

This is as far as we can go. Our final matrix tells us that z can be any number, and then
x+-+%281%2F3%29z+=+-1%2F3 or
x+=+%281%2F3%29z+-+1%2F3
and
y+-+%282%2F9%29z+=+-11%2F9 or
y+=+%282%2F9%29z+-+11%2F9

So our final solution to the system of equation is the infinite set of solutions defined by
z = z;
x = (1/3)z - 1/3; and
y = (2/9)z - 11/9