# Determining Which Variables Are Free

1. Feb 22, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
Find the column of the matrix in Exercise 39 that can be deleted and yet have the remaining matrix columns span $\mathbb{R}^4$

2. Relevant equations
The matrix from problem 39 is

$\begin{bmatrix} 10 & -7 & 1 & 4 & 6 \\ -8 & 4 & -6 & -10 & -3 \\ -7 & 11 & -5 & -1 & -8 \\ 3 & -1 & 10 & 12 & 12 \\ \end{bmatrix}$

3. The attempt at a solution

I had matlab compute the row-reduced echelon form of the matrix augmented with the arbitrary point/vector $\mathbf{b}$, where $\mathbf{b} \in \mathbb{R}^4$. The result is given as an attachment.

I was wondering, does this augmented matrix contain any free variables. I believe it does, and allow me to explain:

The augmented matrix can be written as

$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 & f_1(b) \\ 0 & 1 & 0 & 1 & 0 & f_2(b) \\ 0 & 0 & 1 & 1 & 0 & f_3(b) \\ 0 & 0 & 0 & 0 & 1 & f_4(b) \\ \end{bmatrix}$

Writing out the system of linear equations, that this augmented matrix represents, gives

$\begin{array} \\ x_1 & + & x_2 & + & x_3 & + & x_4 & = & f_1(b) \\ & & x_2 & + & x_3 & & & = & f_2(b) \\ & & & & x_3 & + & x_4 & = & f_3(b) \\ & & & & & & x_4 & = & f_4(b) \\ \end{array}$

If I understand what a free variable is, it is one of which can be written in terms of the others. In this example, we have the other variables can be written as a function of the variables x_4. So, we the variable x_4 is free because you can choose any value of it, and find the corresponding b_i's will be; then from these b_i's, x_3, x_2, and x_1 can be determined.

Am I wrong?

#### Attached Files:

• ###### Capture.PNG
File size:
1.5 KB
Views:
56
Last edited: Feb 22, 2014
2. Feb 22, 2014

### Ray Vickson

I your augmented form is right then your answer is also right.

3. Feb 22, 2014

### Bashyboy

So, then x_4 is the only free variable?

4. Feb 22, 2014

### Ray Vickson

It is the free variable for the particular row and column ordering you have selected. If you permute the columns you might come up with a different free variable.

I don't think looking at free variables is the way to go in this question. You want to know which four columns of the matrix form a basis, so the question you need to answer is this: if you omit one of the columns, do you have an invertible matrix? Your final echelon form has the identity matrix in columns 1,2,3,5, so if you leave out column 4 you get the identity matrix. That means that columns 1,2,3,5 form a non-singular matrix.

5. Feb 22, 2014

### Bashyboy

No, I realize that determining which variables are free won't aid in my solving the actual problem. I was simply curious. Thank you.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted