# Reducing 2x4 matrix to row-echelon form

• Feb 15th 2010, 03:02 PM
yen yen
Reducing 2x4 matrix to row-echelon form
the 2x4 matrix is

[2+i 2+i 5 6+i
1-2i 1-2i -2+i 2-i ]

the answer for row-echelon form is
[4-3i 4-3i 5-10i 8-11i
0 0 -10+10i -3+11i]

and for reduced row echelon form is
[1 1 0 1.6+0.7i
0 0 1 0.7-0.4i]

i cant figure out what operations to use to get the answers. the exercise says to use an algorithm. By the way i came across this problem in my problem sheet book.
• Feb 15th 2010, 09:14 PM
snaes
hope this helps...
having never used "i" in row reductions before here is what i would do (even though its kind of a pain).

First step...(other steps are of a similar for)
$R_1(1-2i)+-(2+i)R_2$ "replace -->" $R_2$

For the first column this should give $(2+i)(1-2i)+-(2+i)(1-2i)$ which should simplify to $(4-3i)+-(4-3i)$ which is $0$

Just "swap" places with their coefficients and multiply by the other row and put a negative sign in front of the second row. Notice how this works with fractions too (you have done the same thing with fractions before i would guess just didnt realize this applies here). By multiplying this way you get things to cancel nicely. This should work, but there may be a better way. Hope this helps and good luck.
• Feb 16th 2010, 02:16 AM
HallsofIvy
The simplest "algorithm", by the way, is this:

Work one column at a time, working from left to right.

Make the "pivot", the number in that column on the main diagonal, equal to 1 by dividing the entire row by whatever is in that place. (The case that is it 0 is a problem- then you will have to swap the pivot row with a lower row.)

Make each number in that column below the pivot 0 by subtracting the pivot row multiplied by the number in that new row from the new row.

To find the the "reduced" row echelon form, do the same with number above the pivot.