# Math Help - How do I prove if a vector is a solution to a linear system?

1. ## How do I prove if a vector is a solution to a linear system?

(I do not study in America so forgive me if I don't translate the terms correctly)

I am having trouble solving this question:

Vectors "u" (-2, 4, 4, 2) and "v" (4, -2, -2, 4) are solutions to a linear system (M) with 4 unknowns. And it is known that (2, 2, 2, 2) is not a solution.

a) Prove that this system is not homogenous (that the b vector does not equal 0. I don't know if this is the correct translation)

b) Prove that (0, 2, 2, 0) is a solution.

"a" I was able to answer: there is a rule that says, in a homogenous matrix, with a1,...,ak answers, that every linear combination (again, don't know if this is correct translation) of a1,...,ak is also a solution of the system:
Meaning, if I prove (2, 2, 2, 2) is a linear combination of "u" and "v" then that proves that this system is not homogenous. (Since, otherwise, it should be homogenous, according to the law)

-2 4 2 1 -2 -1 1 0 1
4 -2 2 = 2 -1 1 = 0 1 1
4 -2 2 2 -1 1 0 0 0
-2 4 2 1 -2 -1 0 0 0

1 ( -2, 4, 4, -2) + 1 (4, -2, -2, 4) = (2, 2, 2, 2)

This proves that (2, 2, 2, 2) is a linear combination of "u" and "v". And since it is a given that (2, 2, 2, 2) is not a solution, then (M) is not homogenous.

b) I believe i answered "a" correctly, but I do not know how to answer this second part. I have tried many things. If someone can help me that would be great! thank you so much.

2. Originally Posted by jayshizwiz
(I do not study in America so forgive me if I don't translate the terms correctly)

I am having trouble solving this question:

Vectors "u" (-2, 4, 4, 2) and "v" (4, -2, -2, 4) are solutions to a linear system (M) with 4 unknowns. And it is known that (2, 2, 2, 2) is not a solution.

a) Prove that this system is not homogenous (that the b vector does not equal 0. I don't know if this is the correct translation)

b) Prove that (0, 2, 2, 0) is a solution.

"a" I was able to answer: there is a rule that says, in a homogenous matrix, with a1,...,ak answers, that every linear combination (again, don't know if this is correct translation) of a1,...,ak is also a solution of the system:
Meaning, if I prove (2, 2, 2, 2) is a linear combination of "u" and "v" then that proves that this system is not homogenous. (Since, otherwise, it should be homogenous, according to the law)

-2 4 2 1 -2 -1 1 0 1
4 -2 2 = 2 -1 1 = 0 1 1
4 -2 2 2 -1 1 0 0 0
-2 4 2 1 -2 -1 0 0 0

1 ( -2, 4, 4, -2) + 1 (4, -2, -2, 4) = (2, 2, 2, 2)

This proves that (2, 2, 2, 2) is a linear combination of "u" and "v". And since it is a given that (2, 2, 2, 2) is not a solution, then (M) is not homogenous.

b) I believe i answered "a" correctly, but I do not know how to answer this second part. I have tried many things. If someone can help me that would be great! thank you so much.

Let $ax_1+bx+2+cx_3+dx_4=K\neq 0$ be one of the equations in the (non-homogeneous, you proved this correctly) system, then

$I\,:\, -2a_1+4a_2+4a_3-2a_4=K$

$II\,:\,4a_1-2a_2-2a_3+4a_4=K$ .

Now, multiply eq. I by 2, add this to eq. II and then divide by 3...

Tonio

3. Let $

ax_1+bx_2+cx_3+dx_4=K\neq 0
$
be one of the equations in the (non-homogeneous, you proved this correctly) system, then

$

I\,:\, -2a_1+4a_2+4a_3-2a_4=K
$

$

II\,:\,4a_1-2a_2-2a_3+4a_4=K
$

Now, multiply eq. I by 2, add this to eq. II and then divide by 3...

Tonio
Thanks. I am still pretty new at this stuff so I have a few last questions. Since both equations = K, can't I also write

$-2a_1+4a_2+4a_3-2a_4=4a_1-2a_2-2a_3+4a_4$

which would make

$6a_1-6a_2-6a_3+6a_4=0$

So what does that mean exactly? Is it legal to do that?

Also, is it possible to get the result

$2a_1+2a_2+2a_3+2a_4=K$ using your method of combining equations??