Solve the system using Gauss-Jordan elimination

I obtain the augmented matrix:
-1 + 1 + 2 | 1
2 + 3 + 1 | -2
5 + 4 + 2 | 4

In order to obtain a 1 instead of a -1 what operation should I take?
 
I obtain:
1 - 1 - 2 | 1
2 + 3 + 1 | -2
5 + 4 + 2 | 4
Now I wish to eliminate the 2 in row2
Which operation should I take?
 
The next two operations I would do is:

[MATH]-2R_1+R_2[/MATH] and [MATH]-5R_1+R_3[/MATH]
Do you see why these will work?
 
frctl said:
Now I wish to eliminate the 2 in row2
Which operation should I take?
Click to expand...
If you want to change 2 into 0 using multiples of 1, then subtract two times 1 from 2 because 2-2(1)=0. That is,

Row2 becomes Row2 minus two times Row1

Likewise, Row3 becomes Row3 minus five times Row1

This is why the first step is to get a leading 1 in the first row -- we use it to zero out the remaining elements in the first column. When you get a 1 in the second column in Row2, you'll use it to zero out the remaining elements in the second column.

?
 
I have tried the first operation, is my reasoning correct?

1 - 1 - 2 | 1
2 + 3 + 1 | -2
5 + 4 + 2 | 4
-2R1 + R2
-2 + 2 + 4 | 1
2 + 3 + 1 | -2
----------------
0 + 5 + 5 | -1
therefore
1 - 1 - 2 | 1
0 + 5 + 5 | -1
5 + 4 + 2 | 4
 
Is

Row2 becomes Row2 minus two times Row1
−2R1+R2
and
Row3 becomes Row3 minus five times Row1
−5R1+R3

the same?
 
Now I must apply
-R1 + R3
to
1 - 1 - 2 | 1
0 + 5 + 5 | -1
5 + 4 + 2 | 4

Correct?
 
frctl said:
Is [R2−2R1 and] -2R1+R2 … the same?
Click to expand...
Yes! Adding the opposite of a number to something is always the same as subtracting that number from the something. 'Adding the opposite' could be an informal (algebraic) definition for subtraction.

-1 + 2 = 3

2 − 1 = 3

:)
 
After previous operation:
1 - 1 - 2 | 1
0 + 5 + 5 | -1
5 + 4 + 2 | 4

My next operation:
-R1 + R3
row3 becomes row3
minus five times row1

5 + 4 + 2 | 4
5 - 5 - 10 | 5
---------------
0 + 9 + 12 | -1
therefore
1 - 1 - 2 | 1
0 + 5 + 5 | -1
0 + 9 + 12 | -1

Am I correct?
 
frctl said:
Which steps remain.
Click to expand...
I gave you the next step, in post #7. After you zero out the elements in the first column, get a 1 in R2C2 and then use it to zero out the remaining elements in the second column. Likewise, finish by getting a 1 in R3C3 and zero out the elements above it.

Your goal is to get ones along the diagonal (in the square matrix, to the left of the augmented column):

100
010
001

frctl said:
…
1 - 1 - 2 | -1
0 + 5 + 5 | -1
0 + 9 + 12 | -1

Am I correct?
Click to expand...
The elements in red above are not correct. Double check your arithmetic in the augmented column.

I've added the missing negative sign to 1 in R1C4. Typo?

\(\;\)
 
Are you doing the same operations within the augmented column? I just noticed that you didn't multiply R1C4 by -1, in your very first row operation (post #5).

When you complete any row operation, you must do the arithmetic in every column (including the augmented column). It is those changes to the values in the augmented column that eventually convert the beginning values into the solutions (at the final row operation).

At this point, you ought to have the following 3×4 matrix.
Code: 
1  -1  -2  -1
0   5   5   0
0   9  12   9

?
 
augmented matrix:
-1 + 1 + 2 | 1
2 + 3 + 1 | -2
5 + 4 + 2 | 4
multiply row1 by -1
1 - 1 - 2 | 1
2 + 3 + 1 | -2
5 + 4 + 2 | 4
-2R1 + R2
row2 becomes row2
minus two times row1
-2 + 2 + 4 | 2
2 + 3 + 1 | -2
----------------
0 + 5 + 5 | 4
therefore
1 - 1 - 2 | 1
0 + 5 + 5 | 4
5 + 4 + 2 | 4
-R1 + R3
row3 becomes row3
minus five times row1
5 + 4 + 2 | 4
5 - 5 - 10 | 5
---------------
0 + 9 + 12 | -1
therefore
1 - 1 - 2 | 1
0 + 5 + 5 | 4
0 + 9 + 12 | -1

Where have I gone wrong so far?
I obtain the wrong value in row3.
 
Thank you, therefore I am presently at:

1 - 1 - 2 | -1
0 + 5 + 5 | 4
0 + 9 + 12 | -1

correct?
 
frctl said:
… multiply row1 by -1

… -2R1 + R2
row2 becomes row2 minus two times row1 …
Click to expand...
It's not necessary to label your steps both ways. You may type something like

R1 -> -R1

R2 -> -2R1 + R2

?
 
frctl said:
Thank you, therefore I am presently at:

1 - 1 - 2 | -1
0 + 5 + 5 | 4
0 + 9 + 12 | -1
Click to expand...
Those values are not correct. (I had given you the correct values in post #16, so that you would have them to check your work.)

Maybe you're still using 1 instead of -1 in the first row (R1C4).

R2 -> -2(R1) + R2

R3 -> -5(R1) + R3

In the augmented column, the arithmetic is:

R2C4 -> (-2)(-1) + (-2)

R3C4 -> (-5)(-1) + (4)

At this point, you ought to have the following 3×4 matrix.
Code: 
1  -1  -2  -1
0   5   5   0
0   9  12   9
\(\;\)
 
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