# Null space

• robertjford80

## Homework Statement ## The Attempt at a Solution

I don't understand how they get the numbers on the right. This is a null space problem so the 3x3 matrix = 0. By my reckoning

(1/3)x3 = 0 so x3 = 0. So then I try the second row.

2x3 = (3/2)x2

Divide both sides by two and x2 = 3/4

Plug that into the first row and we have

2x1 - 3 [(3/4)x3] - 2x3 = 0

Simplify

x1 = (11/8)x3

It's now clear that I'm in disagreement with the book.

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It's not a null space problem, it's a system where the unknown is a vector C and known parameters are a vector X.
In other words they are solving $$\textbf{AC=X}$$

well, i still don't know how they got the numbers on the right.

They just performed Gauss moves about the rows.

First step is: R2 -> R2-(1/2) R1
Second step: R3 -> R3+(2/3) R2

They're doing very simple calculations, you are probably missing the point here, I can't get your confusion.
Which is then the final purpose of this system, it's before on your textbook.

I understand how they get

[2 -3 -2]
[0 -3/2 2]
[0 0 1/3]

I do not understand how they get

x1
-.5x1 + x2
-.33x1 + .672 + x3

The first matrix in your problem is an augmented matrix. You should have been able to recognize its form. This means the original form of the matrix is: $AC=X$ where the matrix C is 3x1: (c1, c2, c3)^T, and the matrix X is another 3x1: (x1, x2, x3)^T.

hold on

After you have reduced the augmented matrix to its triangular form, write it again into its separate matrices $UC=X'$. Then, try matrix multiplication on the left-hand side. Finally, equate both sides.

Note: $UC=X'$ is in similar form to the original $AC=X$ but after reducing the augmented matrix, $A|X$, then $U$ simply represents the upper echelon form of the original matrix $A$. Likewise, $X'$ represents the elements of the matrix ##X## after row reduction.

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I tried multiplying

[2 -3 -2]
[0 -1.5 2]
[0 0 .33]

by the original matrix and I got

[1 1 5]
[-1.5 6.5 -3.5]
[0 .33 -.33]

Not the right answer. The I tried multiplying

[2 -3 -2]
[0 -1.5 2]
[0 0 .33]

by

[1 0 0]
[-.5 1 0]
[-.33 .67 1]

And I got approximately

[4 -4 -2]
[.1 -.1 2]
[-.1 .2 .33]

I also tried A = LU decomposition and that yielded no results similar to the book.

$$U\times C= \displaystyle\left[ {\begin{array}{*{20}{c}} 2&-3&-2 \\ 0&-3/2&2 \\ 0&0&1/3 \\ \end{array}} \right]\times \displaystyle\left[ {\begin{array}{*{20}{c}} c_1 \\ c_2 \\ c_3 \\ \end{array}} \right]$$What do you get for this multiplication?

I appreciate your help and I apologize for being so stubbornly stupid about this problem, but I don't know what the values for c1 c2 and c3 are.

Also please don't respond in laTex, my computer can't read it. I'm trying to solve that problem but it's hard.

I appreciate your help and I apologize for being so stubbornly stupid about this problem, but I don't know what the values for c1 c2 and c3 are.
Exactly. You need to find those values. The answer should be in terms of ##x_1##, ##x_2## and ##x_3##.
Also please don't respond in laTex, my computer can't read it. I'm trying to solve that problem but it's hard.
Use a new browser to see LaTeX, which makes the matrix and equations look much clearer and hence more easily understood. Download the latest Firefox. It's free and way better than other browsers. http://www.mozilla.org/en-US/firefox/new/

The only method I know of to find the values is the following and it didn't work

By my reckoning

(1/3)x3 = 0 so x3 = 0. So then I try the second row.

2x3 = (3/2)x2

Divide both sides by two and x2 = 3/4

Plug that into the first row and we have

2x1 - 3 [(3/4)x3] - 2x3 = 0

Simplify

x1 = (11/8)x3

Read my post #10 again. The matrix form is $UC=X'$
You are not multiplying with the correct matrix. Leave matrix ##X'## out of any multiplication, as it's alone on the right-hand side.

I can't really read post 10 because it's in laTex. I have firefox and I can read laTex on this browser but for some reason it doesn't work on this site. I'm trying to find a website where I can just copy the code into some place and the website will translate it.

In any case if it's

UC = X

And if X and C are unknown, then I can't solve for C. I don't see where there are any values for X with which I can't solve for C

I can't really read post 10 because it's in laTex. I have firefox and I can read laTex on this browser but for some reason it doesn't work on this site. I'm trying to find a website where I can just copy the code into some place and the website will translate it.

What version of Firefox are you using? The latest is 12. On what operating system? To know what version you are using, in the browser top menu, click on Help, then About. Upgrade your browser, if necessary.

If you're still having problem, you could try with Google Chrome or Opera. I just tested with both and the LaTeX code displays properly. Always use the latest version.

Ultimately, if your problem persists (in that case, it has to be a problem with your operating system), then your last option would be to use an online LaTeX equation editor: http://www.sciweavers.org/free-online-latex-equation-editor

I'm out the door, so I will look at this later.

ok, I can read post 10 now, but I don't understand what I'm supposed to do to

[2 -3 -2]
[0 -1.5 2]
[0 0 .33]

in order to get c1 c2 c3

I can't multiply it by c1 c2 c2 because they're unknown and I've already tried multiplying it to the original matrix and got the wrong answer.

In the original matrix $AC=X$, you are dealing with three matrices, but two of these matrices, C and X, contain unknowns.

Following post #10, here is what you get:

row1 of matrix U x column of matrix C
row2 of matrix U x column of matrix C
row3 of matrix U x column of matrix C
$$2c_1-3c_2-2c_3 \\0c_1-(3/2)c_2+2c_3 \\0c_1+0c_2+(1/3)c_3$$
Now, writing this multiplication result in its proper matrix form, you will get a single column matrix.
$$\displaystyle\left[ {\begin{array}{*{20}{c}} 2c_1-3c_2-2c_3 \\ -(3/2)c_2+2c_3 \\ (1/3)c_3 \\ \end{array}} \right]$$
Then, equate this column matrix with the (right-hand side) column matrix that you previously obtained after reducing the augmented matrix to its triangular form, in other words, ##X'##.

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