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

• Mar 29th 2010, 06:40 AM
jayshizwiz
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.
• Mar 29th 2010, 08:06 AM
tonio
Quote:

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 \$\displaystyle 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

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

\$\displaystyle 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...(Wink)

Tonio
• Mar 30th 2010, 09:36 AM
jayshizwiz
Quote:

Let \$\displaystyle

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

\$\displaystyle

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

\$\displaystyle

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

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

which would make

\$\displaystyle 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

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