Inactive Tutor answered • 11/13/13
Tutor
New to Wyzant
If you left-multiply
[ 2 -1 0 0]
[ -1 2 -1 0]
[ 0 -1 2 -1]
[ 0 0 -1 2]
by
[ 1 0 0 0]
[ 1/2 1 0 0]
[ 0 0 1 0]
[ 0 0 0 1]
, then you get
[ 2 -1 0 0]
[ 0 3/2 -1 0]
[ 0 -1 2 -1]
[ 0 0 -1 2]
, which would eliminate the first variable from all but the first equation of the system
AX = Y
, for whatever given column vector Y. So
[ 1 0 0 0]
[ 1/2 1 0 0]
[ 0 0 1 0]
[ 0 0 0 1]
[ 1/2 1 0 0]
[ 0 0 1 0]
[ 0 0 0 1]
must be E21. Next, you would need to eliminate the second variable from all but the first two equations. Could you come up with the matrix E32 by which to left-multiply E21A?
Finally, find E43 by which to left-multiply E32E21A to get upper-triangular matrix E43E32E21A.
Inactive Tutor
I wrote a comment to answer your comment, but when I clicked "Save comment" it disappeared.
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11/15/13
Inactive Tutor
It disappeared probably because I'm not allowed to include links, and I included a link to the paper I found the idea in. That's University of Utah Math 2270 - Lecture 7: Elimination Using Matrices by Dylan Zwick, Fall 2012. He starts with an example
matrix that subtracts 2 times the first row from the second. Using that idea, I came up with this:
[ 1 0 0 0 ] [ 2 -1 0 0 ] [ 2 -1 0 0 ]
[ 1/2 1 0 0 ] [ -1 2 -1 0 ] = [ 0 3/2 -1 0 ]
[ 0 0 1 0 ] [ 0 -1 2 -1 ] [ 0 -1 2 -1 ]
[ 0 0 0 1 ] [ 0 0 -1 2 ] [ 0 0 -1 2 ]
So the matrix A you gave me is multiplied on the left by the elimination matrix I came up with. That adds 1/2 times the first row of A to the second row of A, getting rid of that first -1.
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11/15/13
Sun K.
11/13/13