Determining Which Variables Are Free

[Bashyboy]

Homework Statement

Find the column of the matrix in Exercise 39 that can be deleted and yet have the remaining matrix columns span $\mathbb{R}^4$

Homework Equations

The matrix from problem 39 is

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

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?

 

[Ray Vickson]

Homework Statement

Find the column of the matrix in Exercise 39 that can be deleted and yet have the remaining matrix columns span $\mathbb{R}^4$

Homework Equations

The matrix from problem 39 is

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

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

[itex]\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}/itex]

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?

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

 

[Bashyboy]

So, then x_4 is the only free variable?

 

[Ray Vickson]

So, then x_4 is the only free variable?

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.

 

[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.