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.