# Inhomogeneous Linear System Solutions

1. Aug 11, 2014

### GregJ

Hi all, I have been searching around and cannot seem to find an answer. I am doing a past paper, and have the answers but do not understand one part. I hope someone can help.

The question is:

$A \in M_{3 \times 3}(\mathbb{R})$, $\vec{b} \in \mathbb{R}^{3}$. The general solution to the equation $A\vec{x} = \vec{b}$ is given by:

$\vec{x} = \lambda \left(\begin{array}{rrr|r}-1\\3\\2\end{array}\right) + \left(\begin{array}{rrr|r}0\\1\\2\end{array}\right)$

Let: $\vec{y} = \left(\begin{array}{rrr|r}-1\\5\\6\end{array}\right)$

1) Calculate $\mu_{1} , \mu_{2}$ such that:

$\vec{y} = \mu_{1} \left(\begin{array}{rrr|r}-1\\3\\2\end{array}\right) + \mu_{2}\left(\begin{array}{rrr|r}0\\1\\2\end{array}\right)$

2) Express $A \vec{y}$ in terms of $\vec{b}$

I can do 1) without problem. However it is 2) that I am struggling with. Now I understand that the general solution $\vec{x}$ is made up of homogeneous + inhomogeneous solutions. However I am unable to tell which is which.

I know that the solution to the homogeneous part may be multiplied by a scalar and still be a solution. Would this automatically imply that the homogeneous part is $\lambda \left(\begin{array}{rrr|r}-1\\3\\2\end{array}\right)$? Or is there another reason that I am missing?

2. Aug 11, 2014

### mathman

Your implication is correct. Since the answer for 1) is μ_1 = 1 and μ_2 = 2, then x = y/2, so Ay = 2b.