# Help with reduced row echelon form/solving system

• Mar 22nd 2009, 07:48 PM
Murphie
Help with reduced row echelon form/solving system
Im stuck on this problem:

Use reduced row echelon form to solve the system:
2x+3y-12z=1
x-2y+z=4
4x+y-14z=7

Im stuck after many steps here, no idea if this is right so far.
[1 5 -13 -3]
[0 -7 14 7]
[0 -2 3 4]
• Mar 23rd 2009, 04:34 AM
stapel
Quote:

Originally Posted by Murphie
Use reduced row echelon form to solve the system:
2x+3y-12z=1
x-2y+z=4
4x+y-14z=7

Im stuck after many steps here, no idea if this is right so far.

Without knowing the particular steps you took, it's hard to say whether or not you've done them correctly. Sorry! (Blush)

(Note: My calculator returns different results for the "reduced row echelon form" for the original matrix and your work-in-progress, which suggests that there may have been an error made somewhere along the way.)

One sequence of steps might start out as follows:

[html][ 2 3 -12 1 ]
[ 1 -2 1 4 ]
[ 4 1 -14 7 ]

-2R2 + R1 -> R1
-4R2 + R3 -> R3

[ 0 7 -14 -7 ]
[ 1 -2 1 4 ]
[ 0 9 -18 -9 ]

(1/7)R1 -> R1
(1/9)R3 -> R3

[ 0 1 -2 -1 ]
[ 1 -2 1 4 ]
[ 0 1 -2 -1 ]

-1R1 + R3 -> R3[/html]
...and so forth.

Hope that helps! :D
• Mar 23rd 2009, 06:35 AM
HallsofIvy
It looks like Murphie did the following:

1) Swap the first and second rows.
2) Subtract twice the (new) first row from the (new) second row.
3) Subtract four times the (new) first row from the third row.
However that does not give what Murphie shows for the new third row.
4- 4(1)= 0, 1- 4(-2)= 9, -14- 4(1)= -18, and 7- 4(4)= -9 so the new matrix must be $\displaystyle \left[\new{array}1 & -2 & 1 & 4 \\0 & 7 & -14 & -7\\ 0 & 9 & -19 & -9\end{array}\right]$

Now:
1) Divide the second row by 7 in order to get "1" at the "pivot" point- the second row, second column.
2) Add 2 times that new second row to the first row to get a "0" above the 1.
3) Subtract 9 times that new second row to the third row to get a "0" below the 1.

Looks to me like you won't have to worry about the third column!