| 1 | 1 | 1 | 1 | 4 |
| 1 | 3 | -2 | a | 3 |
| 2 | 2a-2 | -a-2 | 3a-1 | 1 |
| 3 | a+2 | -3 | 2a+1 | 6 |
I'm failing to convert this matrix into reduced row echelon form, can anyone help me? I managed to do reduce the first two roles but considering how I'm getting stuck afterwards I assume I'm doing something wrong.
----
Already solved It.
I don't know what you mean by "roles", but I'll certainly agree that doing reduction with variables can be a royal pain in the... um, keister. The process is the same as always. But I would
strongly recommend
lots of scratch paper! :shock:
As an example for others who may read your post later, here are a few steps in the process:
Ones are easy to deal with, so I'll start by using Row 1 (R1) to clear out the R1-entries in columns 2, 3, and 4. To do this, I'll multiply R1 by -1 (for R2), by -2 (for R3), and by -3 (for R4). The first multiplication, -1*R1, will give me:
. . . . .-1
. . .-1
. . .-1
. . .-1
. . .-4
...to add to:
. . . . .1
. . .3
. . .-2
. . .a
. . .3
Doing the addition, I get:
. . . . . 1
. . .. 3
. . .-2
. . .. a
. . ..3
. . . . .-1
. . .-1
. . .-1
. . . -1
. . .-4
. . . . . 0
. . . 2
. . ..-3
. . ..a-1
. . .-1
Doing similar calculations with the other two rows, I get:
-2*R1 + R3:
. . . . .-2
. . .-2
. . . . .-2
. . . . .-2
. . .-8
. . . . ..2
. . .2a-2
. . .-a-2
. . .3a-1
. . .6
. . . . .0
. . .2a-4
. . .-a-4
. . .3a-3
. . .-7
-3*R1 + R4:
. . . . .-3
. . .-3
. . . .-3
. . ..-3
. . ..-12
. . . . ..3
. . .a+2
. . .-3
. . .2a+1
. . .6
. . . . ..0
. . .a-1
. . .-6
. . .2a-2
. . .-6
Inserting these revised rows back into the matrix, I get:
| 1 | 1 | 1 | 1 | 4 |
| 0 | 2 | -3 | a - 1 | -1 |
| 0 | 2a - 4 | -a - 4 | 3a - 3 | -7 |
| 0 | a - 1 | -6 | 2a - 2 | -6 |
Now the "fun" starts: I have to clear entries with variables. Ugh! Okay, the steps are the same; the method remains the same. It's just that it's messier now....
I've cleared out all the first-column entries, other than for just the one row. So (1) I don't want to use that row any more, since it would mess up the other first-column entries if I did and (2) I need to pick one of the remaining rows to use for clearing out the second-column entries in other other rows.
Because R2 leads with a simple number (just a "2") after the leading zero, I'll reduce this to a leading 1 by multiplying through by 1/2:
(1/2)*R2:
. . . . .0
. . .1
. . .-3/2
. . .a/2 - 1/2
. . .-1/2
So now my matrix is:
| 1 | 1 | 1 | 1 | 4 |
| 0 | 1 | -3/2 | a/2 - 1/2 | -1/2 |
| 0 | 2a - 4 | -a - 4 | 3a - 3 | -7 |
| 0 | a - 1 | -6 | 2a - 2 | -6 |
To clear the leading entries in R3 and R4, I'll need to multiply R2 by -(2a - 4) and -(a - 1), respectively:
-(2a - 4)*R2 + R3:
. . . . .0
. . .-2a+4
. . .3a-6
. . .-a
2+3a-2
. . .a-2
. . . . .0
. . .+2a-4
. . .-a-4
. . . . .3a-3
. . . .-7
. . . . .0
. . . . .0
. . .2a-10
. . .-a
2+6a-5
. . .a-9
-(a - 1)*R2 + R4:
. . . . .0
. . .-a+1
. . . .3a/2 - 3/2
. . . . . .-a
2/2 - 1/2
. . . . . . . .a/2 - 1/2
. . . . .0
. . .+a-1
. . . . . . .-6
. . . . . . . . . .2a - 2
. . . . . . . . . . . .-6
. . . . .0
. . . . .0
. . .(1/2)(3a - 15)
. . .(1/2)(-a
2 + 4a - 5)
. . .(1/2)(a - 13)
Now my matrix is:
| 1 | 1 | 1 | 1 | 4 |
| 0 | 1 | -3/2 | a/2 - 1/2 | -1/2 |
| 0 | 0 | 2a - 10 | -a2 + 6a - 5 | a - 9 |
| 0 | 0 | (1/2)(3a - 15) | (1/2)(-a2 + 2a - 5) | (1/2)(a - 13) |
It is very important to notice that if you work ON a row (like multiplying through by 1/2), you can work ONLY on THAT row. Make the change; rewrite the matrix. Only
then do you try a different step. If you are working WITH a row ON a DIFFERENT row (like adding -1R1 to R2), then you can only make changes to that OTHER row (in the example, only to R2). The row you're working WITH stays the same in the matrix!
Continuing...
