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 ->  Matrices-and-determiminant -> 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       Log On


   

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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
first write it form aw+%2B+bx+%2B+cy+%2B+dz+=+k, for constants

3w+-+x+-+2y+-+z+=+-4
0w+%2B+9x+-+y+%2B+z+=+10
4w+%2B+0x+%2B+3y+-+z+=+7
0w+%2B+12x+-+2y+%2B+z+=+-11



|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|...w
|0| 1| 0| 0| -7.163|...x
|0| 0| 1| 0| -0.49|...y
|0| 0| 0| 1| 73.98|...z


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