MHB Linear Combination - missing data ?

[Yankel]

Dear all,

I am trying to solve a question, and I think that something is missing.

It is given that the vectors u and v are solutions to the non-homogeneous system of equations Ax=b.

If the vector ku-3v is a solution to the same system, then:

a) k = 4
b) k = 3
c) k = 0

The correct solution is apparently k = 4. How can you tell ?? I can't figure it out.

Thank you in advance.

 

[castor28]

Dear all,

I am trying to solve a question, and I think that something is missing.

It is given that the vectors u and v are solutions to the non-homogeneous system of equations Ax=b.

If the vector ku-3v is a solution to the same system, then:

a) k = 4
b) k = 3
c) k = 0

The correct solution is apparently k = 4. How can you tell ?? I can't figure it out.

Thank you in advance.

Hi Yankel,

You have:
$$
A(ku-3v) = kAu - 3Av = (k-3)b
$$
On the other hand, as $ku-3v$ is also a solution of the system, you have:
$$
A(ku-3v) = b
$$
Since $b\ne0$ by hypothesis, the conclusion follows.

 

[Yankel]

So are you saying that k can't be 3 ?

 

[castor28]

So are you saying that k can't be 3 ?

You must have $(k-3)b = b$; since $b\ne0$, this implies $k-3 = 1$ and $k=4$. This is the solution you mentioned as correct.