Solve the system using Gauss-Jordan elimination

Hello frctl. Are you still working on this exercise?

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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 | 0
0 + 9 + 12 | 9
 
frctl said:
… row2 becomes row2
minus two times row1
-2 + 2 + 4 | 2
2 + 3 + 1 | -2
----------------
0 + 5 + 5 | 4
Click to expand...
1 - 1 - 2 | -1
0 + 5 + 5 | 4
5 + 4 + 2 | 4
…
Click to expand...
… row3 becomes row3
minus five times row1
1 - 1 - 2 | -1
0 + 5 + 5 | 0
0 + 9 + 12 | 9
Click to expand...
Hello. Are you copying and pasting from other posts? Otherwise, why do you continue typing 2-2=4? (I see that you eventually changed that 4 into 0, while working in the third row.)

You're done working with column 1 because you have a pivot in row1 and zeros below it.

Now work with column 2. Get a pivot in row2, and then get a zero below it.

After that, get a pivot in the third column (in row3). You will then have row echelon form (REF). Let's check your work, before continuing to reduced row echelon form (RREF).

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Thank you, I have fixed these errors. Which operation will get me a pivot in row2?

Perhaps -5R2 + R1
 
frctl said:
… Which operation will get me a pivot in row2?
Click to expand...
Remember, we're trying to get the first three columns (i.e., the coefficient matrix) into the form

100
010
001

Those ones are the pivots. The pivot in row2 is in col2.

Rich (BB code): 
1  -1  -2  -1
0   5   5   0
0   9  12   9

In other words, you need to turn that 5 above into a 1. The row operation for that is to multiply row2 by 1/5.

Note: In case you've forgotten, we call 1/5 the reciprocal of 5 (in fraction form: 5/1). We will always get a product of 1, when we multiply a number by its reciprocal.

\(\frac{1}{5}×\frac{5}{1} = 1\)

So, whenever you need to change some matrix element into 1, just multiply that row by the reciprocal of the element. If the element is 3, multiply the row by 1/3. If the element is 5/4, then multiply the row by 4/5. If the element is k, then multiply the row by 1/k, etc.

(As MarkFL mentioned to you earlier, multiplying by 1/5 is the same as dividing by 5.)

After you get the pivot in row2, add the appropriate multiple of row2 to row3 to get a zero below the pivot. Then get the pivot in row3 the same way as you did in row2: Multiply row3 by the fraction that will change 3 into 1 (i.e., multiply by the reciprocal of 3).

?
 
my work:

multiply row 2 by 1/5
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 9 + 12 | 9
multiply row 3 by 1/3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 3 + 4 | 3
multiply row 3 by 1/3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 1 + 4/3 | 3

Is it fine to have the fraction appear?
 
frctl said:
… multiply row 2 by 1/5
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 9 + 12 | 9
[then] multiply row 3 by 1/3 …
Click to expand...
That's correct. But don't start working in col3, for the next step. You're not done with col2, yet. We need to get a zero below the pivot in row2.

Code: 
1  -1  -2  -1
0   1   1   0
0   9  12   9
Now that you have a pivot above 9 in col2, use it to change 9 into zero. What multiple of row2 do you need to add to row3, to change that 9 into 0?

After you've done that, then multiply row3 to get your pivot in the third column.

Is it fine to have the fraction appear?
Click to expand...
Yes! It's okay if some elements are fractions. Some solutions turn out to be fractions, too.

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..multiply row 2 by 1/5
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 9 + 12 | 9
add -9 to row3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 3 | 0
multiply row 3 by 1/3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 0 | 0
 
frctl said:
multiply row 2 by 1/5
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 9 + 12 | 9
add -9 times row2 to row3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 3 | 0
multiply row 3 by 1/3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 0 | 0
Click to expand...
Hi. The values shown in red above need correcting. If those are typos, then please proofread your posts before submitting. It'll save time.

Otherwise, have you been double-checking arithmetic, as you go? I use scratch paper for that, when I do matrix arithmetic by hand. Often, I'll write out each column's arithmetic, if I've been making too many mistakes. For example, I'll write a step like -2*R1+R2 using the numbers from each column:

(-2)(1) + (2) = 0
(-2)(2) + (4) = 0
etc …

That way, it's easy to proofread the numbers I've copied (comparing to the latest matrix) and also to double-check my arithmetic before writing the resulting matrix. Maybe something like that would help you save time. Cheers

?
 
What kind of class are you taking, frctl? Does your textbook describe echelon forms or provide any worked examples?

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frctl said:
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
Click to expand...
The numbers in red above are not correct!
 
-x + y + 2z = 1
2x + 3y + z = −2
5x + 4y + 2z = 4
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 | 0
therefore
1 - 1 - 2 | -1
0 + 5 + 5 | 0
5 + 4 + 2 | 4
-R1 + R3
row3 becomes row3
minus five times row1
5 + 4 + 2 | 0
5 - 5 - 10 | 5
---------------
0 + 9 + 12 | -1
therefore
1 - 1 - 2 | -1
0 + 5 + 5 | 0
0 + 9 + 12 | 9
multiply row 2 by
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 9 + 12 | 9
add -9 times row2 to row3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 3 | 1
multiply row 3 by
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 1 | 1
 
What you've done looks mostly good. I have one remark and a single error to point out. First, the remark: I think you left off a bit of information. At two of your steps, you say "multiply row 2 by" and "multiply row 3 by" but multiply the row by... what? From context, I can gather you meant to multiply by 1/5 and 1/3 respectively. When it came to actually manipulating the rows, you did the math you just didn't actually write down the step in full.

And that brings me to the small error I noticed. In the very last step, you forgot to also multiply the "answer" column by 1/3. What you had is the same as saying \(3z = 1\). You then (attempted to) divide both sides by 3. Would you agree that the resulting equation of \(z = 1\) makes no sense, given what followed? How can it be true that both \(z\) and \(3z\) are equal to 1 at the same time?
 
frctl said:
…
-R1 + R3
row3 becomes row3 minus five times row1
5 + 4 + 2 | 0
5 - 5 - 10 | 5
---------------
0 + 9 + 12 | -1
Click to expand...
Typos highlighted in red above.

frctl said:
… add -9 times row2 to row3
1 - 1 - 2 | -1
0 + 1 + 1 | 0
0 + 0 + 3 | 1
Click to expand...
That 1 is not correct because (-9)(0)+9 is 9, so the matrix ought to be:
Code: 
1  -1  -2  -1
0   1   1   0
0   0   3   9
​
The next row operation is R3 -> 1/3*R3
Code: 
1  -1  -2  -1
0   1   1   0
0   0   1   3
​
The matrix is now in row echelon form (REF) because there is a pivot in each column (along the diagonal in the coefficient matrix), and all elements below the pivots are zero.

The REF form shows the given system as consistent because (as noted in your other thread) the bottom row does not have all zeros in the coefficient matrix with a non-zero number in the augmented column.

If the bottom row had looked something like 0 0 0 3 instead, then we could stop. The system would be inconsistent.

Since we know there is a solution, we continue to reduced row echelon form (RREF). That is, we get zeros above the pivots, too. RREF form gives us the solution values for x, y and z. The last three row operations are listed below, for you. See if you can finish and post the solution without any more typos or arithmetic mistakes. In other words, please use scratch paper, proofread numbers you have copied and double-check your arithmetic. You should also verify your final solution, by substituting your values for x, y and z into the three given equations, to ensure they work.

R1 -> R2 + R1 (this will change R1C2 from -1 to 0)

R1 -> R3 + R1 (this will change R1C3 from -1 to 0)

R2 -> -R3 + R2 (this will change R2C3 from 1 to 0)

?
 
You're typing a lot more than you need to. Also, some of your descriptions are wrong.

frctl said:
multiply row1 by -1
Click to expand...
R1 -> -R1

-2R1 + R2
row2 becomes row2
minus two times row1
therefore
Click to expand...
R2 -> -2R1 + R2

-R1 + R3
row3 becomes row3
minus five times row1
therefore
Click to expand...
R3 -> -5R1 + R3

multiply row 2 by
Click to expand...
R2 -> (1/5)R2

add -9 times row2 to row3
Click to expand...
R3 -> -9R2 + R3

multiply row 3 by
Click to expand...
R3 -> (1/3)R3

\(\;\)
 
Hello frctl. I gave you the last three row operations, two days ago. Are you done with this exercise? (I would like to know whether I can throw away my notes for this thread.)

As a self-studying student, it might be better for you to finish each exercise before posting new exercises.

?
 
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