Thread: Matrix Row Reduction

1. Matrix Row Reduction

I’m having serious problems in comprehending matrix row reduction. I know how to turn an equation into a matrix and just to clarify I think these are the rules:

1. I can switch any two rows at any given point.
2. I can multiply or divide a row by any number of my choosing.
3. I can add or subtract any two rows with each other. The resulting answer is then replaced into the row that should change (ex: row1+row2 means row 2 is replaced. If row1-row2 then row1 is replaced).
4. I should be concentrating on an outcome of:
1 0 0 x
0 1 0 y
0 0 1 z
5. If I'm able to get x, y or z separately by itself I can use regular algebra to solve for the rest. (Ex: I have 1 0 0 x but the rows of y and z still don't look as they should. I can simply plug in what x is with the linear equations given above to solve for y and z.)

By this point, I don't understand the logical steps that should be taken when solving a problem. In other words, when I see examples I do not get why someone is choosing to multiply a row or subtract a row. Could someone step me through how I would solve:

2x-y+4z=-3
x-4z=5
6x-y+2z=10

Thanks.

2. I can try - the best way to do these is to just practice - you'll start to get a more intuitive sense of what works and what doesn't. You start with:

2 -1 4 -3
1 0 -4 5
6 -1 2 10

If you can get two rows with 0s in the same column, it's often helpful. Since I have a 0 in R2, C2, I'm going to get a 0 in R3, C2. New R3 = R3 - R1. Note that the ONLY row I'm actually changing is R3, so R1 and R2 get copied exactly as they are.

2 -1 4 -3
1 0 -4 5
4 0 -2 13

Now, I can get another zero in R3 using a combination of R2. I'll get rid of the 4 in R3. New R3 = R3 - 4R2. Again, R1 and R2 don't change.

2 -1 4 -3
1 0 -4 5
0 0 14 -7

Let's finish R3 by dividing by 14.

2 -1 4 -3
1 0 -4 5
0 0 1 -.5

Now, using R3, I can easily eliminate anything I want in column 3 without messing anything else up. Start by cleaning up R2. New R2 = R2 + 4R3.

2 -1 4 -3
1 0 0 3
0 0 1 -.5

Now, let's clean up R1. New R1 = R1 - 4R3.

2 -1 0 -1
1 0 0 3
0 0 1 -.5

Clean up the first position in R1. New R1 = R1 - 2R2.

0 -1 0 -7
1 0 0 3
0 0 1 -.5

Rearrange and multiply the current R1 by -1.

1 0 0 3
0 1 0 7
0 0 1 -.5

So x = 3, y = 7, z = -.5

Let's check:

2x - y + 4z = -3

2(3) - 7 - 2 = 3. Check.

x - 4z = 5

3 + 2 = 5 Check.

6x - y + 2z = 10

6(3) - 7 - 1 = 10 Check.

Important things to remember:

Getting two zeros in the same column is incredibly useful because you can combine those rows to get a zero in an additional column without messing up what you've done. Once you've got a row with two zeros in it, you can pretty quickly clean up the rest. So, aim to get a zero in the same position in two different rows.

Also, as I learned while doing this, copy the problem down correctly.

Hope this helps!