SOLUTION: if you were to use gauss-jordan elimination on the following linear system, what would the augmented matrix be?
3w-x=2y+z-4
9x-y+z=10
4w+3y-z=7
12x+17=2y-z+6
Algebra.Com
Question 647443: if you were to use gauss-jordan elimination on the following linear system, what would the augmented matrix be?
3w-x=2y+z-4
9x-y+z=10
4w+3y-z=7
12x+17=2y-z+6
Found 2 solutions by Aaniya, MathLover1:
Answer by Aaniya(63) (Show Source): You can put this solution on YOUR website!
{(567/31,-438/31,-663/31,3589/31)}
i used the augmented matrix system
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
first write it form , for constants
|3| -1| -2| -1| -4|
|0| 9| -1| 1| 10|
|4| 0| 3| -1| 7|
|0| 12| -2| 1| -11|
Step 1: Swap row 3 and 1
|4| 0| 3| -1| 7|
|0| 9| -1| 1| 10|
|3| -1| -2| -1| -4|
|0| 12| -2| 1| -11|
Step 2: Divide row 1 by 4
|1| 0| 0.75| -0.25| 1.75|
|0| 9| -1| 1| 10|
|3| -1| -2| -1| -4|
|0| 12| -2| 1| -11|
Step 3: Subtract (3 × row 1) from row 3
|1| 0| 0.75| -0.25| 1.75|
|0| 9| -1| 1| 10|
|0| -1| -4.25| -0.25| -9.25|
|0| 12| -2| 1| -11|
Step 4: Swap row 4 and 2
|1| 0| 0.75| -0.25| 1.75|
|0| 12| -2| 1| -11|
|0| -1| -4.25| -0.25| -9.25|
|0| 9| -1| 1| 10|
Step 5: Divide row 2 by 12
|1| 0| 0.75| -0.25| 1.75|
|0| 1| -0.167| 0.083| -0.917|
|0| -1| -4.25| -0.25| -9.25|
|0| 9| -1| 1| 10|
Step 6: Subtract (-1 × row 2) from row 3
|1| 0| 0.75| -0.25| 1.75|
|0| 1| -0.167| 0.083| -0.917|
|0| 0| -4.417| -0.167| -10.167|
|0| 9| -1| 1| 10|
Step 7: Subtract (9 × row 2) from row 4
|1| 0| 0.75| -0.25| 1.75|
|0| 1| -0.167| 0.083| -0.917|
|0| 0| -4.417| -0.167| -10.167|
|0| 0| 0.5| 0.25| 18.25|
Step 8: Divide row 3 by -4.417
|1| 0| 0.75| -0.25| 1.75|
|0| 1| -0.167| 0.083| -0.917|
|0| 0| 1| 0.038| 2.302|
|0| 0| 0.5| 0.25| 18.25|
Step 9: Subtract (0.5 × row 3) from row 4
|1| 0| 0.75| -0.25| 1.75|
|0| 1| -0.167| 0.083| -0.917|
|0| 0| 1| 0.038| 2.302|
|0| 0| 0| 0.231| 17.099|
Step 10: Divide row 4 by 0.231
|1| 0| 0.75| -0.25| 1.75|
|0| 1| -0.167| 0.083| -0.917|
|0| 0| 1| 0.038| 2.302|
|0| 0| 0| 1| 73.98|
Matrix is now in row echelon form
Step 11: Subtract (-0.25 × row 4) from row 1
|1| 0| 0.75| 0| 20.245|
|0| 1| -0.167| 0.083| -0.917|
|0| 0| 1| 0.038| 2.302|
|0| 0| 0| 1| 73.98|
Step 12: Subtract (0.083 × row 4) from row 2
|1| 0| 0.75| 0| 20.245|
|0| 1| -0.167| 0| -7.082|
|0| 0| 1| 0.038| 2.302|
|0| 0| 0| 1| 73.98|
Step 13: Subtract (0.038 × row 4) from row 3
|1| 0| 0.75| 0| 20.245|
|0| 1| -0.167| 0| -7.082|
|0| 0| 1| 0| -0.49|
|0| 0| 0| 1| 73.98|
Step 14: Subtract (0.75 × row 3) from row 1
|1| 0| 0| 0| 20.612|
|0| 1| -0.167| 0| -7.082|
|0| 0| 1| 0| -0.49|
|0| 0| 0| 1| 73.98|
Step 15: Subtract (-0.167 × row 3) from row 2
|1| 0| 0| 0| 20.612|...
|0| 1| 0| 0| -7.163|...
|0| 0| 1| 0| -0.49|...
|0| 0| 0| 1| 73.98|...
Matrix is now in reduced row echelon form
check one:
3w - x - 2y - z = -4
3* 20.612- (-7.163) - 2(-0.49)- 73.98 = -4
61.836+7.163+0.98- 73.98 = -4
69.979- 73.98=-4
-4.001=-4
-4=-4
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