Can Gauss-Jordan Elimination simplify a matrix into a reduced row echelon form?

[askor]

By using row operation, how can

$\left( \begin{array}{rrrr|r} 2 & 0 & -1 & 0 & 0 \\ 6 & 0 & 0 & -2 & 0 \\ 0 & 2 & -2 & -1 & 0 \end{array} \right)$

be

$\left( \begin{array}{rrrr|r} 1 & 0 & 0 & -\frac{1}{3} & 0 \\ 0 & 1 & 0 & -\frac{7}{6} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0 \end{array} \right)$

?

I am trying to write my attempt on this problem but I don't know how to write matrix with row operation in Latex. Please show me example.

 

[BvU]

Hi askor,

There's a link to the LateX Tutorial in the homework guidelines and an extensive one in general math.
Is there a speecific question you need answered ?

Your matrix looks rather straightforward to me:
Subtract 1/3 x row 2 from row 1 to get 2 x new row 3.
Add row 1 to new row 3 to get 2 x new row 1
Add 2 x new row 3 to row 3 to get 2 x new row 2

But I expect you already found this ? If so where do you get stuck ?

--

 

[HallsofIvy]

Mod note: Corrected what appeared to be a typo in the second vector, below. I'm almost certain that the corrected version was what was intended.
It should be obvious, seeing $\begin{pmatrix}0 \\ 0 \\ 2 \end{pmatrix}$ in the middle column of the original matrix and $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ in the result you want, that you will want to swap the second and third rows. If you do that as the first step, you get $$\left(\begin{array}{cccc|c} 2 & -1 & 0 & 1 & 0 \\ 0 & 2 & -2 & -1 & 0 \\ 6 & 0 & 0 & -2 & 0 \end{array}\right)$$.

The rest should be easy.

 

[askor]

Subtract 1/3 x row 2 from row 1 to get 2 x new row 3.
Add row 1 to new row 3 to get 2 x new row 1
Add 2 x new row 3 to row 3 to get 2 x new row 2

But I expect you already found this ? If so where do you get stuck ?

--

Could you write it in matrix rather than words? I am confuse.

My attempt is:

$\left( \begin{array}{rrrr|r} 2 & 0 & -1 & 0 & 0 \\ 6 & 0 & 0 & -2 & 0 \\ 0 & 2 & -2 & -1 & 0 \end{array} \right)$

R1 divided by 2, R2 swap to R3, then I get

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 2 & -2 & -1 & 0 \\ 6 & 0 & 0 & -2 & 0 \end{array} \right)$

R2 divided by 2, then I get

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 6 & 0 & 0 & -2 & 0 \end{array} \right)$

R3 - 6R1, then I get lower triangular matrix

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & -2 & 0 \end{array} \right)$

Why the result is different?

 

[BvU]

Could you write it in matrix rather than words? I am confuse.

Which one is confusing you ?

Subtract 1/3 x row 2 from row 1 to get 2 x new row 3 is so clear to me that I don't know how to simplify it!
Perhaps:

Subtract 1/3 x row 2 from row 1 to and divide the resulting row by -2 to get new row 3​

?

 

[askor]

Which one is confusing you ?

Subtract 1/3 x row 2 from row 1 to get 2 x new row 3 is so clear to me that I don't know how to simplify it!
Perhaps:

Subtract 1/3 x row 2 from row 1 to and divide the resulting row by -2 to get new row 3​

?

Could you write it in my style as above.

 

[BvU]

I could, but if it is unclear the way I write it now, you still have a problem. What is it that makes it undecipherable for you ?

 

[askor]

You said "Subtract 1/3 x row 2 from row 1..."

If written by R1 = row 1, R2 = row 2, which one of below is correct according your above statement?

1/3*R2 - R1

or

R1 - 1/3*R2

 

[BvU]

R1 - 1/3*R2

Gives you $(0,0,-1, {2\over 3})$ which is minus the new row 3. (new meaning: as in the final answer)

So not +2 or -2 x new row 3 as I stupidly typed - twice . I fixed the - sign in post #5 and now I fix the 2 that should have been a 1.

Sorry for the mistakes. No wonder you get confused.

 

[askor]

OK, R1 - 1/3*R2 give me

$\left( \begin{array}{rrrr|r} 0 & 0 & -1 & \frac{2}{3} & 0 \\ 6 & 0 & 0 & -2 & 0 \\ 0 & 2 & -2 & -1 & 0 \end{array} \right)$

Now, what is the next step?

 

[BvU]

From what you post now (#10): line 3 minus 2 x line 1 is $(0,2,0,-7/3)$, then divide by 2 to get line 2 in the answer in post #1.

 

[askor]

Why the result is different with the post #4?

 

[Svein]

Why the result is different with the post #4?

In that post you have not finished the computations!

 

[askor]

In that post you have not finished the computations!

How to finish it? Please tell step-by-step.

Here is my step:

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & -2 & 0 \end{array} \right)$

R3 / 3

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0 \end{array} \right)$

The matrix is already in form of Lower triangular matrix, how to obtain the upper triangular matrix?

What next?

 

[Mark44]

How to finish it? Please tell step-by-step.

Here is my step:

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 3 & -2 & 0 \end{array} \right)$

R3 / 3

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0 \end{array} \right)$

The matrix is already in form of Lower triangular matrix, how to obtain the upper triangular matrix?

What next?

Use the bottom row (3rd column entry) to eliminate the entries directly above it.

Some notation I have seen is this:
$R_2$ ← $R_2 + R_3$
$R_1$ ← $R_1 + \frac 1 2 R_3$
The arrow, ←, means "is replaced by".

 

[askor]

Use the bottom row (3rd column entry) to eliminate the entries directly above it.

Some notation I have seen is this:
$R_2$ ← $R_2 + R_3$
$R_1$ ← $R_1 + \frac 1 2 R_3$
The arrow, ←, means "is replaced by".

OK, I rewrite it with matrix

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0 \end{array} \right)$

$R_2$ ← $R_2 + R_3$

$\left( \begin{array}{rrrr|r} 1 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & -\frac{7}{6} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0 \end{array} \right)$

$R_1$ ← $R_1 + \frac 1 2 R_3$

$\left( \begin{array}{rrrr|r} 1 & 0 & 0 & -\frac{2}{6} & 0 \\ 0 & 1 & 0 & -\frac{7}{6} & 0 \\ 0 & 0 & 1 & -\frac{2}{3} & 0 \end{array} \right)$

 

[BvU]

Well done !