# Solve linear system using augmented matrix

• merzperson
In summary, rock.freak667 found that clearing the first column by dividing by 3, then subtracting 2x that row from the second row, then subtracting 5x the new row from the third row, gets rid of all the zeroes in the first column.

## Homework Statement

Solve using an augmented matrix:

3r-2s+4t=6
2r+3s-5t=-8
5r-4s+3t=7

2. The attempt at a solution

[3 -2 4 | 6] => [1 0 0 | x]
[2 3 -5 |-8] => [0 1 0 | y]
[5 -4 3 | 7] => [0 0 1 | z]

I have successfully used 2x2 augmented matrices, but with 3x3 it seems like there are a bunch of different options and each one just messes up everything I did to get a zero or one in the right spot. I'm not sure if there is a default starting point with these or you are looking for something to cancel to a zero or what. Thanks!

Use the main diagonal as your pivot points i.e. use that row to make the other rows zero. Ex. take the 3 in the first row to make the other two elements below it zero. Repeat with the 3 in the second row and so on.

Thanks rock.freak667, I have followed your directions and that helped a lot. However, I seem to be going wrong somewhere. I solved the system of equations using software and got r=1.376E-14 , s=-1 , t=1 . When I solved using the matrix I get some fractional numbers (z=63/23). I was extra careful and double-checked all my numbers. Am I wrong?

merzperson said:
Thanks rock.freak667, I have followed your directions and that helped a lot. However, I seem to be going wrong somewhere. I solved the system of equations using software and got r=1.376E-14 , s=-1 , t=1 . When I solved using the matrix I get some fractional numbers (z=63/23). I was extra careful and double-checked all my numbers. Am I wrong?

Could you post all your working with the augmented matrix?

Work one column at a time, left to right. For example, to "clear" the first column, divide the entire row by "3", to get a 1 in the first column, first row.
Then subtract 2 times that new first row from the second row to get a "0" in the first column second row. Finally, subtract 5 times the new first row from the third row to get a "0" in the first column third row.

Then look at the second row. Since the "pivot" of the second row is in the second column, divide that row by whatever number is in second row, second column to get a "1" there. Because that number is now "1" to get a "0" in the second column of another row, just subtract the new second row times what ever number is in the second column of that row. Notice that, because you will always have "0" to the left of the pivot, you will never change the "1"s and "0" you already have there.