MHB 311.1.5.14 Use vectors to describe this set as a line in R^4

[karush]

ok, just now looking at some examples of how to do this $x_4$ is just a row with all zeros

 

[karush]

ok don't see any takers on this one but here is a book example that might help, so we have...

$x_1+3x_4, \quad x_2=8+x_4, \quad x_3 =2-5x_4$ with $x_4$ free

$x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}
=\begin{bmatrix}3x_4\\8+x_4\\2-5x_4\\x_4\end{bmatrix}
=\begin{bmatrix}0\\8\\2\\0\end{bmatrix}= ...$

hopefully so far

Spoiler: example from book

 

[HOI]

Frankly it looks to me like you have no idea what you are supposed to be doing!

Yes, since we are told that "[math]x_1= 3x_4[/math], [math]x_2= 8+ x_4[/math], and [math]x_3= 2- 5x_4[/math] we have immediately that [math]\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}= \begin{bmatrix} 3x_4 \\ 8+ x_4 \\ 2- 5x_4 \\ x_4 \end{bmatrix}[/math].

But why then did you set [math]x_4[/math] to 0?? The problem says that [math]x_4[/math] is "free" which means that it can be any number. Saying that a number is "free" certainly does NOT mean that it is 0!

I would say that [math]\begin{bmatrix} 3x_4 \\ 8+ x_4 \\ 2- 5x_4 \\ x_4 \end{bmatrix}[/math] is a perfectly good answer but some people might prefer to replace the "coordinate", [math]x_4[/math] with the "parameter", t:
[math]\begin{bmatrix} 3t \\ 8+ t \\ 2- 5t \\ t \end{bmatrix}[/math].

Some would prefer to write that as
[math]\begin{bmatrix}0 \\ 8 \\ 2 \\ 0 \end{bmatrix}+\begin{bmatrix} 3 \\ 1 \\ -5 \\ 1 \end{bmatrix}t[/math].

 

[karush]

yes it is new material to me

 

[HOI]

Then you need to start by learning the basic definitions!