SOLUTION: Pivot the system about the element in row 2, column 2. 1 5 –10 0 0 5 –6 –6 0 10 8 –5

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Question 652103: Pivot the system about the element in row 2, column 2.


1 5 –10 0
0 5 –6 –6
0 10 8 –5



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

They want you to make the 1 a pivot element, meaning that the other numbers in it's column are zero. All you do is use row reduction techniques to accomplish this.
Divide row 2 by 5
|1| 5| -10| 0|
|0| 1| -6/5| -6/5|
|0| 10| 8| -5|

Add (-10+%2A+row2) to row3
|1| 5| -10| 0|
|0| 1| -6/5| -6/5|
|0| 0| 20| 7|

Divide row3 by 20
|1| 5| -10| 0|
|0| 1| -6/5| -6/5|
|0| 0| 1| 7/20|

Add (6%2F5+%2A+row3) to row2
|1| 5| -10| 0|
|0| 1| 0| -39/50|
|0 | 0| 1| 7/20|

Add (10+%2A+row3) to row1
|1| 5 | 0| 7/2|
|0| 1| 0| -39/50|
|0| 0| 1| 7/20|

Add (-5+%2A+row2) to row1
|1| 0| 0| 37/5|
|0| 1| 0| -39/50|
|0| 0| 1| 7/20|
so, your solutions are: x=+37%2F5, y=+-39%2F50, and z=+7%2F20