# Explanation of a matrix, linear system

• Jun 11th 2008, 12:39 PM
Kim Nu
Explanation of a matrix, linear system
The matrix is:

[-1 0 -3] [k1] [0]
[1 0 3] [k2] = [0]
[1 0 3] [k3] [0]

Now I understand that if you let k1=3 and k3=-1, then k2=0:

[3 ]
[0 ]
[-1]

Another possible solution, according to the back of the book is

[0]
[1]
[0]

Is it correct for me to think that 1 is an aribtrary number and that it really could be anything? Thanks,

Kim
• Jun 11th 2008, 01:07 PM
Soroban
Hello, Kim Nu!

I'm not sure what your question is.
The matrix equation is rather silly, isn't it?

Quote:

$\displaystyle \begin{bmatrix}\text{-}1 & 0 & \text{-}3 \\ 1 & 0 & 3 \\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} \:=\:\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now I understand that if you let: .$\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}\:=\:\begin{bma trix}3 \\ 0 \\ \text{-}1 \end{bmatrix}$
Another possible solution, according to the book is: .$\displaystyle \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$

The matrix equation gives this system: .$\displaystyle \begin{array}{ccc}\text{-}x-3z &=&0 \\ x+3z &=&0 \\ x+3z &=& 0 \end{array}$

. . which is just one equation: .$\displaystyle x+3z\:=\:0 \quad\Rightarrow\quad x \:=\:-3z$

A solution would be: .$\displaystyle \begin{array}{ccc}z &=& \text{any number} \\ x &=& -\text{3 times z} \\ y &=& \text{any number} \end{array}$

Therefore: .$\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix} \;=\;\begin{bmatrix} 3s \\ t \\ s\end{bmatrix}\quad\hdots$ . for any values of $\displaystyle s\text{ and }t$

• Jun 11th 2008, 03:15 PM
Kim Nu
Soroban,

Thanks for the response!

So under the system you set up, when y=1, does that mean that z has to be 0? I'm guessing not as x and z are the only ones that are dependent on each other.
• Jun 11th 2008, 08:00 PM
Reckoner
Quote:

Originally Posted by Kim Nu
Soroban,

Thanks for the response!

So under the system you set up, when y=1, does that mean that z has to be 0? I'm guessing not as x and z are the only ones that are dependent on each other.

That's correct: $\displaystyle y$ and $\displaystyle z$ are both free variables and are independent from each other. Also note that Soroban's final answer should be $\displaystyle \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}-3s\\t\\s\end{matrix}\right]$.