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#### karush

##### Well-known member

- Jan 31, 2012

- 3,066

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 3,066

- Thread starter
- #2

- Jan 31, 2012

- 3,066

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

- Jan 30, 2018

- 821

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

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

But why then did you set \(\displaystyle x_4\) to 0?? The problem says that \(\displaystyle x_4\) 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 \(\displaystyle \begin{bmatrix} 3x_4 \\ 8+ x_4 \\ 2- 5x_4 \\ x_4 \end{bmatrix}\) is a perfectly good answer but some people might prefer to replace the "coordinate", \(\displaystyle x_4\) with the "parameter", t:

\(\displaystyle \begin{bmatrix} 3t \\ 8+ t \\ 2- 5t \\ t \end{bmatrix}\).

Some would prefer to write that as

\(\displaystyle \begin{bmatrix}0 \\ 8 \\ 2 \\ 0 \end{bmatrix}+\begin{bmatrix} 3 \\ 1 \\ -5 \\ 1 \end{bmatrix}t\).

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

But why then did you set \(\displaystyle x_4\) to 0?? The problem says that \(\displaystyle x_4\) 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 \(\displaystyle \begin{bmatrix} 3x_4 \\ 8+ x_4 \\ 2- 5x_4 \\ x_4 \end{bmatrix}\) is a perfectly good answer but some people might prefer to replace the "coordinate", \(\displaystyle x_4\) with the "parameter", t:

\(\displaystyle \begin{bmatrix} 3t \\ 8+ t \\ 2- 5t \\ t \end{bmatrix}\).

Some would prefer to write that as

\(\displaystyle \begin{bmatrix}0 \\ 8 \\ 2 \\ 0 \end{bmatrix}+\begin{bmatrix} 3 \\ 1 \\ -5 \\ 1 \end{bmatrix}t\).

Last edited:

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- Jan 31, 2012

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yes it is new material to me

- Jan 30, 2018

- 821

Then you need to start by learning the basic definitions!