Thread: 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

Hello, Kim Nu!

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

Now I understand that if you let: .
Another possible solution, according to the book is: .

The matrix equation gives this system: .

. . which is just one equation: .


A solution would be: .


Therefore: . . for any values of

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.

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: and are both free variables and are independent from each other. Also note that Soroban's final answer should be .