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) (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.
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
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.
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).
Again we see where we can factor a common factor out of a row, in row 3:
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.
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.
This matrix tells us that z can be any number; then y can be found using the equation
or
and then x could be found using the equation
or
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.
This is as far as we can go. Our final matrix tells us that z can be any number, and then
or
and
or
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
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