- #1

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The question is:

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

[itex]\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)[/itex]

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

1) Calculate [itex]\mu_{1} , \mu_{2}[/itex] such that:

[itex]\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)[/itex]

2) Express [itex]A \vec{y}[/itex] in terms of [itex] \vec{b}[/itex]

I can do 1) without problem. However it is 2) that I am struggling with. Now I understand that the general solution [itex]\vec{x}[/itex] 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 [itex]\lambda \left(\begin{array}{rrr|r}-1\\3\\2\end{array}\right)[/itex]? Or is there another reason that I am missing?