Gaussian elimination (pivoting)

[fonseh]

Homework Statement

In my book , the author stated that when we do the pivoting , we need to make the element below the leading element = 0 . For example for 3x3 matrix , for the first column , we have to make a11 = max , and make sure that a21 and a31 less than a11 , for the second column , we have to make the a22 = max , make sure that the a32 less than a22 ...

But , according to my lecturer ( in second photo)( I have reduced it into row echelon formed in first matrix , for the 2nd matrix , my lecturer just swap the row) , he make the a11 max compared to the other element( a12 and a13) in the first row ...

For row 2 , he make the a22 max compared to a21 and a23 in the same row

For row 2 , he make the a33 max compared to other element (a31 and a32) ...

Homework Equations

The Attempt at a Solution

I don't think the lecturer is correct . Can someone explain about it ? I have read a lot of online resource , i can't find them similar to the way that the lecturer read [/B]

 

[BvU]

Not clear to me what your question is. I see a piece of text in hhh.jpg and a worked out pivoting example in 140116.jpg and 140126
Is 134955.jpg what you call 'the second photo' ? It only shows a swap of row 1 and 2 of a different matrix.

What is it you think the lecturer does incorrectly ? placing the -3 in position 11 is according to the text.

 

[fonseh]

Not clear to me what your question is. I see a piece of text in hhh.jpg and a worked out pivoting example in 140116.jpg and 140126
Is 134955.jpg what you call 'the second photo' ? It only shows a swap of row 1 and 2 of a different matrix.

What is it you think the lecturer does incorrectly ? placing the -3 in position 11 is according to the text.

140116.jpg and 140126 are the textbook question . So , it follows the rule that stated in hhh.jpg

 

[fonseh]

Not clear to me what your question is. I see a piece of text in hhh.jpg and a worked out pivoting example in 140116.jpg and 140126
Is 134955.jpg what you call 'the second photo' ? It only shows a swap of row 1 and 2 of a different matrix.

What is it you think the lecturer does incorrectly ? placing the -3 in position 11 is according to the text.

134955,jpg is the question that the lecturer showed and solved ... I didnt posted the full solution for 134955,jpg because i am just having problem with the pivotisation ( I doubt the way that the lecturer switching row is correct ) . After the pivotisation , it's just the normal gaussian elimination , followed by backward substituition to solve the question...

 

[fonseh]

Not clear to me what your question is. I see a piece of text in hhh.jpg and a worked out pivoting example in 140116.jpg and 140126
Is 134955.jpg what you call 'the second photo' ? It only shows a swap of row 1 and 2 of a different matrix.

What is it you think the lecturer does incorrectly ? placing the -3 in position 11 is according to the text.

Perhaps you can refer to here for more info

https://www.physicsforums.com/threads/pivoting-gaussian-elimination.905148/

 

[BvU]

It's getting to be a bit labyrinthic ..

What is it you think the lecturer does incorrectly ? placing the -3 in position 11 is according to the text.

I doubt the way that the lecturer switching row is correct

Why ? He picks the biggest $a_{i1}$ and moves that row to row 1

 

[fonseh]

It's getting to be a bit labyrinthic ..
Why ? He picks the biggest $a_{i1}$ and moves that row to row 1

140116.jpg and 140126 shows that the author get the max a11 first , then he make the element below a11 = 0 , get max a22 , make the element below it = 0 ...

but , the lecturer gpot all the max a11 , a22 and a33 , then he only do the gauss elimination (make the element below a11 = 0 , make the element below a22 = 0 ... ) ...

Both method are correct ?

 

[BvU]

but , the lecturer gpot all the max a11 , a22 and a33 , then he only do the gauss elimination (make the element below a11 = 0 , make the element below a22 = 0 ... ) ...

Where do I see what he did ? ($\LaTeX$ please )

 

[Mark44]

@fonseh, please start using LaTeX to represent your matrices instead of posting images of them. In post #1 of this thread you attached four images. That alone is enough for many helpers to refuse to provide help. What's worse is that your descriptions of what's in the images, such as "in second photo", are confusing, since it's not clear which one is the second photo.

In your other thread, @BvU also made this request. I am copying hs comment from that post, but have made a very minor change -- I like matrices to be inside brackets (i.e., bmatrix) rather than parentheses (pmatrix).

And if we are going to do much more of this, you should learn a bit of $\LaTeX$ instead of making photos. It's fun and pretty easy.

$$\begin{bmatrix}
1 & 2& 0 & 9 \\
6 & 6 & -8 & 1 \\
0 & -3 & 1 & 0
\end {bmatrix} $$

gives you
$$\begin{bmatrix}
1 & 2& 0 & 9 \\
6 & 6 & -8 & 1 \\
0 & -3 & 1 & 0
\end {bmatrix} $$

It's OK to post an image of the problem statement, but work on matrices should be posted using LaTeX, as in the example above. If you don't make an effort to comply with this I will close the thread.

 

[Ray Vickson]

140116.jpg and 140126 shows that the author get the max a11 first , then he make the element below a11 = 0 , get max a22 , make the element below it = 0 ...

but , the lecturer gpot all the max a11 , a22 and a33 , then he only do the gauss elimination (make the element below a11 = 0 , make the element below a22 = 0 ... ) ...

Both method are correct ?

I agree with Mark44 about getting you to stop posting images if you want to continue using this forum. If you want, you can use a tabular form instead of matrices; in LaTeX you do that using an "array", allowing you to have vertical and horizontal separating lines and labelled rows if you want, like this:
$$\begin{array}{c|rrr|c}
\text{row}&x_1 & x_2 & x_3 & \text{r.h.s.}\\ \hline
1 &5 & -5 & 1 & 2\\
2 &-3 & 1 & 1 & 1 \\
3 &3 & 2 & -7 & 1 \\ \hline
4 &-3 & 1 & 1 & 1\\
5 &5 & -5 & 1 & 2\\
6 &3 & 2 & -7 & 1 \\ \hline
\vdots &\vdots & \vdots & \vdots & \vdots\\
\hline
\end{array}
$$
To see how this was done, right-click on the image and choose "show math as tex commands" from the displayed menu.

 

[fonseh]

$$\begin{bmatrix}
2 & -5& 1 & 2 \\
3 & 2 & -7 & 1 \\
-3 & 1 & 1 & 1
\end {bmatrix} $$

The author then make a11 , a22 and a33 max all at ONCE , , hence , become
$$\begin{bmatrix}
-3 & 1 & 1 & 1\\
2 & -5& 1 & 2 \\
3 & 2 & -7 & 1
\end {bmatrix} $$In the 2nd matrix , we can see that all the a11 , a22 and a33 is max

Is the author's working of making all the a11 , a22 and a33 = max correct ?

 

[Mark44]

$$\begin{bmatrix}
2 & -5& 1 & 2 \\
3 & 2 & -7 & 1 \\
-3 & 1 & 1 & 1
\end {bmatrix} $$

The author then make a11 , a22 and a33 max all at ONCE , , hence , become
$$\begin{bmatrix}
-3 & 1 & 1 & 1\\
2 & -5& 1 & 2 \\
3 & 2 & -7 & 1
\end {bmatrix} $$In the 2nd matrix , we can see that all the a11 , a22 and a33 is max

Is the author's working of making all the a11 , a22 and a33 = max correct ?

I don't think so. In the 2nd matrix, as far as I can see, it just happens that a11, a22, a33 have the largest values (absolute values).

There are only three row operations that you can apply to get a new matrix with the same solution set as the one you started with:
1) Swapping two rows (R_i <--> R_j)
2) Replacing a row by a nonzero multiple of itself (R_i <-- k*R_i)
3) Replacing a row by the sum of it and a nonzero multiple of another row (R_i <-- R_i + k * R_j)

For the two matrices above, it looks like there were two row operations performed:
Rows 1 and 2 were swapped, resulting in
$\begin{bmatrix} 3 & 2 & -7 & 1 \\ 2 & -5& 1 & 2 \\ -3 & 1 & 1 & 1\end{bmatrix}$
and in this new matrix, rows 1 and 3 were swapped, resulting in

$\begin{bmatrix} -3 & 1 & 1 & 1 \\ 2 & -5& 1 & 2 \\ 3 & 2 & -7 & 1\end{bmatrix}$

The most important thing is that whichever row you're going to pivot on, the leading entry in that column should not be zero. You then use that entry to eliminate all the entries in the columns below that entry, with the goal being a more-or-less diagonal matrix, with entries of zero below the diagonal.

Keep in mind that all this matrix business is just a shorthand way to solve a system of equations. The same row operations apply; namely a) you can interchange any two equations; b) you can replace an equation with a nonzero multiple of itself; c) you can replace any equation with the sum of it and a nonzero multiple of another equation.

In all cases here, you end up with a system of equations that looks different, but has exactly the same solution set.

 

[fonseh]

I don't think so. In the 2nd matrix, as far as I can see, it just happens that a11, a22, a33 have the largest values (absolute values).

There are only three row operations that you can apply to get a new matrix with the same solution set as the one you started with:
1) Swapping two rows (R_i <--> R_j)
2) Replacing a row by a nonzero multiple of itself (R_i <-- k*R_i)
3) Replacing a row by the sum of it and a nonzero multiple of another row (R_i <-- R_i + k * R_j)

For the two matrices above, it looks like there were two row operations performed:
Rows 1 and 2 were swapped, resulting in
$\begin{bmatrix} 3 & 2 & -7 & 1 \\ 2 & -5& 1 & 2 \\ -3 & 1 & 1 & 1\end{bmatrix}$
and in this new matrix, rows 1 and 3 were swapped, resulting in

$\begin{bmatrix} -3 & 1 & 1 & 1 \\ 2 & -5& 1 & 2 \\ 3 & 2 & -7 & 1\end{bmatrix}$

The most important thing is that whichever row you're going to pivot on, the leading entry in that column should not be zero. You then use that entry to eliminate all the entries in the columns below that entry, with the goal being a more-or-less diagonal matrix, with entries of zero below the diagonal.

Keep in mind that all this matrix business is just a shorthand way to solve a system of equations. The same row operations apply; namely a) you can interchange any two equations; b) you can replace an equation with a nonzero multiple of itself; c) you can replace any equation with the sum of it and a nonzero multiple of another equation.

In all cases here, you end up with a system of equations that looks different, but has exactly the same solution set.

Do you mean
instead of doing these steps ,
1) interchanging rows to get the the max a11 ,
2) make the element below the a11 (a21 and a31) equal to 0
3) Interchanging the rows to get max a22
4) make the element below a22 = 0We can also interchanging the rows first , to get max a11 and a22 , followed by making the element below a11 and a22 = 0 ?

 

[Ray Vickson]

$$\begin{bmatrix}
2 & -5& 1 & 2 \\
3 & 2 & -7 & 1 \\
-3 & 1 & 1 & 1
\end {bmatrix} $$

The author then make a11 , a22 and a33 max all at ONCE , , hence , become
$$\begin{bmatrix}
-3 & 1 & 1 & 1\\
2 & -5& 1 & 2 \\
3 & 2 & -7 & 1
\end {bmatrix} $$In the 2nd matrix , we can see that all the a11 , a22 and a33 is max

Is the author's working of making all the a11 , a22 and a33 = max correct ?

That was just an accident of the data in this problem.

Generally, we start in column 1 and move the numerically largest value in that column to position (1,1), perhaps by swapping some row i and row 1. Then we do row operations to zero out the remaining elements in column 1.

At that point the rest of the matrix changes (because the row operations affect all the columns, not just column 1.) So, what WAS a largest element before the row operations may no longer be the largest in the new matrix. Therefore, we cannot predict what further row-interchanges will be needed; maybe $a_{22}$ was the largest element in column 2 before row operations, but then maybe the new value of $a_{42}$ is the largest after.

I am trying to get you to understand that you need to re-examine the matrix (below the current pivoting row) that you get after the row operations. Basically, forget about what was largest in the original matrix and look instead what is now largest in the new matrix. It may have moved to a new location.

 

[fonseh]

That was just an accident of the data in this problem.

Generally, we start in column 1 and move the numerically largest value in that column to position (1,1), perhaps by swapping some row i and row 1. Then we do row operations to zero out the remaining elements in column 1.

At that point the rest of the matrix changes (because the row operations affect all the columns, not just column 1.) So, what WAS a largest element before the row operations may no longer be the largest in the new matrix. Therefore, we cannot predict what further row-interchanges will be needed; maybe $a_{22}$ was the largest element in column 2 before row operations, but then maybe the new value of $a_{42}$ is the largest after.

I am trying to get you to understand that you need to re-examine the matrix (below the current pivoting row) that you get after the row operations. Basically, forget about what was largest in the original matrix and look instead what is now largest in the new matrix. It may have moved to a new location.

Yes , i understand what you said ...But , is the way of swapping the row all at once firstly to get max a11 and a22 , then , followed by the zero out the element below a11 and a22 correct ?

 

[Ray Vickson]

Yes , i understand what you said ...But , is the way of swapping the row all at once firstly to get max a11 and a22 , then , followed by the zero out the element below a11 and a22 correct ?

It is not incorrect, but it is a waste of time and nobody who writes linear algebra software (for example) would ever do it! The author should not have done that, unless it happened just by accident when he but the largest $|a_{i1}|$ into position (1,1) by swapping two rows. The fact that $a_{22}$ and $a_{33}$ are then also the largest is purely accidental and has no significance whatsoever. It is unimportant! Forget it!

You are trying to read way too much into things that have no importance, and it is getting in the way of your learning the subject.

 

[fonseh]

It is not incorrect, but it is a waste of time and nobody who writes linear algebra software (for example) would ever do it! The author should not have done that, unless it happened just by accident when he but the largest $|a_{i1}|$ into position (1,1) by swapping two rows. The fact that $a_{22}$ and $a_{33}$ are then also the largest is purely accidental and has no significance whatsoever. It is unimportant! Forget it!

You are trying to read way too much into things that have no importance, and it is getting in the way of your learning the subject.

So , just follow these steps will be enough ?
1) interchanging rows to get the the max a11 ,
2) make the element below the a11 (a21 and a31) equal to 0
3) Interchanging the rows to get max a22
4) make the element below a22 = 0

 

[Mark44]

Emphasis added

You are trying to read way too much into things that have no importance, and it is getting in the way of your learning the subject.

I agree completely.

So , just follow these steps will be enough ?
1) interchanging rows to get the the max a11 ,
2) 1) make the element below the a11 (a21 and a31) equal to 0
3) Interchanging the rows to get max a22
4) 2)make the element below a22 = 0

You do the above at each pivot row, moving across the matrix column by column.

Here's a simple system as an example.
x + y + z = 6
2x -z = -1
x + z = 4

As an augmented matrix this system is
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 2 & 0 & -1 & | & -1 \\ 1 & 0 & 1 & | & 4 \end{bmatrix}$

Replace R2 by -2* R1 + R2:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -2 & -3 & | & -13 \\ 1 & 0 & 1 & | & 4 \end{bmatrix}$
Now replace R3 by -1 * R1 + R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -2 & -3 & | & -13 \\ 0 & -1 & 0 & | & -2 \end{bmatrix}$

The pivot was the a_1,1 entry. I used it to eliminate the entries below it in the 2nd and 3rd rows.
To save some time I could swap rows 2 and 3 for a reason I'll explain later.

Swap R2 and R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -1 & 0 & | & -2\\ 0 & -2 & -3 & | & -13 \end{bmatrix}$

Replace R2 by -1 times itself
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & 1 & 0 & | & 2\\ 0 & -2 & -3 & | & -13 \end{bmatrix}$
The second row tells me that y = 2, and this is why I swapped R2 and R3.

Replace R3 by 2 * R2 + R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & 1 & 0 & | & 2\\ 0 & 0 & -3 & | & -9 \end{bmatrix}$

Finally, replace R3 by -1/3 * R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & 1 & 0 & | & 2\\ 0 & 0 & 1 & | & 3 \end{bmatrix}$

Notice that I used only the three row operations I mentioned in my earlier post, nothing more.

This matrix is now in reduced echelon form, with each row having a leading entry of 1, and every row with a leading 1 has 0 entries below it. The second row tells me that y = 2. The third row tells me that z = 3. By substituting these values into the equation represented by the first row (x + y + z = 6), I can determine that x = 1.

The solution to the system of equations is x = 1, y = 2, z = 3, which you can check by substituting into the original system of equations.

 

[fonseh]

Notice that I used only the three row operations I mentioned in my earlier post, nothing more.

why 3 operation in the earlier post you are referring to ?

 

[Mark44]

why 3 operation in the earlier post you are referring to ?

Because those are the only operations that result in a new system with the same solution set.

 

[fonseh]

Because those are the only operations that result in a new system with the same solution set.

Sorry , i mean which 3 operation in the earlier post you are referring to ?

 

[fonseh]

Emphasis added
I agree completely.

You do the above at each pivot row, moving across the matrix column by column.

Here's a simple system as an example.
x + y + z = 6
2x -z = -1
x + z = 4

As an augmented matrix this system is
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 2 & 0 & -1 & | & -1 \\ 1 & 0 & 1 & | & 4 \end{bmatrix}$

Replace R2 by -2* R1 + R2:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -2 & -3 & | & -13 \\ 1 & 0 & 1 & | & 4 \end{bmatrix}$
Now replace R3 by -1 * R1 + R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -2 & -3 & | & -13 \\ 0 & -1 & 0 & | & -2 \end{bmatrix}$

The pivot was the a_1,1 entry. I used it to eliminate the entries below it in the 2nd and 3rd rows.
To save some time I could swap rows 2 and 3 for a reason I'll explain later.

Swap R2 and R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & -1 & 0 & | & -2\\ 0 & -2 & -3 & | & -13 \end{bmatrix}$

Replace R2 by -1 times itself
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & 1 & 0 & | & 2\\ 0 & -2 & -3 & | & -13 \end{bmatrix}$
The second row tells me that y = 2, and this is why I swapped R2 and R3.

Replace R3 by 2 * R2 + R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & 1 & 0 & | & 2\\ 0 & 0 & -3 & | & -9 \end{bmatrix}$

Finally, replace R3 by -1/3 * R3:
$\begin{bmatrix} 1 & 1 & 1 & | & 6\\ 0 & 1 & 0 & | & 2\\ 0 & 0 & 1 & | & 3 \end{bmatrix}$

Notice that I used only the three row operations I mentioned in my earlier post, nothing more.

This matrix is now in reduced echelon form, with each row having a leading entry of 1, and every row with a leading 1 has 0 entries below it. The second row tells me that y = 2. The third row tells me that z = 3. By substituting these values into the equation represented by the first row (x + y + z = 6), I can determine that x = 1.

The solution to the system of equations is x = 1, y = 2, z = 3, which you can check by substituting into the original system of equations.

I don't understand why you swapped row 2 and row 3 , at here , a22 is already max , why we need to swap row 2 and row 3 ? Swapping row 3 would make the a22 become not maximum ... To pivot 'a22' , we need to make it max , right ? Then , we can only make the element below a22 = 0

 

[fonseh]

Or i misunderstood something ?

 

[BvU]

No you didn't misunderstand. Mark tells you: 'to save time'. Do it both ways and you already have found the explanation Mark gives a few lines later.

Note that swapping rows is for numerical analysis purposes only (one wants to avoid subtracting big numbers from big numbers) and it's no big deal NOT to swap an $|a_{22}| = 2$ of for one with absolute value $1$.

 

[fonseh]

No you didn't misunderstand. Mark tells you: 'to save time'. Do it both ways and you already have found the explanation Mark gives a few lines later.

Note that swapping rows is for numerical analysis purposes only (one wants to avoid subtracting big numbers from big numbers) and it's no big deal NOT to swap an $|a_{22}| = 2$ of for one with absolute value $1$.

sorry , i didnt quite understand the example that Mark gave in post #18...

Can you explain based on the example i gave in post #11 ?

Do you mean that it is actually unnecessary to interchange all the row at ONCE to get max a11 , a22 and a33 ?

We can also change the row to get the max a11 first , then zero out a21 and a31 , after that , change the row to get max a22 ... and so on ?

 

[Mark44]

sorry , i didnt quite understand the example that Mark gave in post #18...

What parts of my example didn't you understand? I explained every operation in detail in terms of the row operations I used.

Can you explain based on the example i gave in post #11 ?

No, we're not going to do your work for you. Why don't you give it a try?

Do you mean that it is actually unnecessary to interchange all the row at ONCE to get max a11 , a22 and a33 ?

Not only that, as Ray said in post #14, it's not usually possible to get the largest values in a11, a22, and a33.

That was just an accident of the data in this problem.

We can also change the row to get the max a11 first , then zero out a21 and a31 , after that , change the row to get max a22 ... and so on ?

It's not necessary to have the leading entry of the pivot row be the maximum of the entries in that column. It's enough that that entry isn't 0.

 

[fonseh]

In this image , I found that the gauss elimination with pivoting is suitable when a11 or a22 is 0 . But , based on the example I uploaded and discussed earlier , the a11 or a22 are not 0 , how could the author use the gauss elimination with pivoting ?
Mod edit: I inserted one of the images inline.

 

[Mark44]

What is so difficult to understand here? In the visible image of your post, they swapped rows 1 and 2. As I mentioned before, this is one of the three possible row operations. The next operation is to replace row 2 by -.02/500 * R1 + R2. This is another of the three row operations I mentioned. The a_{2, 1} entry in the second row will be 0 after this operation.

 

[fonseh]

What is so difficult to understand here? In the visible image of your post, they swapped rows 1 and 2. As I mentioned before, this is one of the three possible row operations. The next operation is to replace row 2 by -.02/500 * R1 + R2. This is another of the three row operations I mentioned. The a_{2, 1} entry in the second row will be 0 after this operation.

Then， what is the meaning of gauss elimination with pivoting is suitable when a11 or a22 is 0 ?

 

[Mark44]

Then， what is the meaning of gauss elimination with pivoting is suitable when a11 or a22 is 0 ?

The you look at a33 as a potential pivot.

 

[fonseh]

The you look at a33 as a potential pivot.

?? What do you mean ?

 

[Mark44]

?? What do you mean ?

Then you look at a33 as a potential pivot.

 

[Stephen Tashi]

Generally, we start in column 1 and move the numerically largest value in that column to position (1,1), perhaps by swapping some row i and row 1. Then we do row operations to zero out the remaining elements in column 1.

As I recall, some computer implementations of Gaussian elimination that concern themselves with "numerical methods" begin by switching rows and columns so the element with largest absolute value (picked from the all elements in the entire matrix) is in position 1,1. They follow a similar policy to select subsequent pivots. However, for making theoretical conclusions about the algorithm of Gaussian elimination or prescribing how to do hand-calculations it is simpler to consider an algorithm where less freedom-of-choice is allowed in selecting pivots.

It would be challenging exercise in scholarship to determine if there is a "standard" definition of "Gaussian elimination" as a unique algorithm.

 

[Ray Vickson]

As I recall, some computer implementations of Gaussian elimination that concern themselves with "numerical methods" begin by switching rows and columns so the element with largest absolute value (picked from the all elements in the entire matrix) is in position 1,1. They follow a similar policy to select subsequent pivots. However, for making theoretical conclusions about the algorithm of Gaussian elimination or prescribing how to do hand-calculations it is simpler to consider an algorithm where less freedom-of-choice is allowed in selecting pivots.

It would be challenging exercise in scholarship to determine if there is a "standard" definition of "Gaussian elimination" as a unique algorithm.

In my post I did mention that some implementations just seek a pivot element whose absolute value exceeds some threshold, without bothering to look for the largest. Of course, some implementations do not move rows at all, but just store pointers and lists and the like in order to keep track of which rows have already originated pivots and which are still available for new pivots.

And, of course, with exact, rational arithmetic we do not need to worry at all about the magnitude of the pivot element, just so long as it is nonzero. I believe that "pure" (original) Gaussian elimination is like that; only "floating-point" Gaussian elimination needs to, and does exercise care in the pivot choice.

 

[fonseh]

What is so difficult to understand here? In the visible image of your post, they swapped rows 1 and 2. As I mentioned before, this is one of the three possible row operations. The next operation is to replace row 2 by -.02/500 * R1 + R2. This is another of the three row operations I mentioned. The a_{2, 1} entry in the second row will be 0 after this operation.

Or the notes in this picture is misleading ? It mentioned that the a11 or a22 is 0 , not when we pivot the a11 , the a21 and a31 is 0 ...

 

[Ray Vickson]

Or the notes in this picture is misleading ? It mentioned that the a11 or a22 is 0 , not when we pivot the a11 , the a21 and a31 is 0 ...

The note you post is 100% clear, so you must be reading it incorrectly. Remember: after swapping rows 1 and $i$, the old row $i$ becomes the new row 1, and vice-versa. The old $a_{i1}$ becomes the new $a_{11}$, and the old $a_{11}$ becomes the new $a_{i1}$.

 

[fonseh]

The note you post is 100% clear, so you must be reading it incorrectly. Remember: after swapping rows 1 and $i$, the old row $i$ becomes the new row 1, and vice-versa. The old $a_{i1}$ becomes the new $a_{11}$, and the old $a_{11}$ becomes the new $a_{i1}$.

It didnt mention about when after swapping row , the old a11 will become 0

 

[Ray Vickson]

It didnt mention about when after swapping row , the old a11 will become 0

No, you didn't, but the posted note DID, and that is all that counts here. You are trying to understand what the note is all about, and whether the note is correct. It is correct. (The note itself did not say that the old row i becomes the new row 1, etc., but it takes that as understood by the reader.)