# Simple Linear Algebra Problem

I'm trying to solve for the full solution of $Ax=b$ where $$A = \pmatrix{20 &40 &60\\ 30 &60 &90}, b = \pmatrix{10\\ 15}$$

The answers I am getting are $$\pmatrix{-2\\1\\ 0}x_1 + \pmatrix{-3\\0\\ 1}x_2 + \pmatrix{1/2\\0\\ 0}$$

I'm not exactly sure if this is correct or not as the answers given to me by my professor is different and I don't know why. Can someone offer some insight?

Thanks

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$$A \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} = A \begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

and

$$A \begin{pmatrix} 1 / 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 10 \\ 15 \end{pmatrix} \ .$$

As for the answer given by your professor, you must take into account that those vectors $(-2 ,0 , 1), (-3,0, 1)$ and $(1/2, 0 ,0)$ are not unique. What is unique is the solution set

$$(1/2, 0 ,0) + \ \mathrm{span}\left\{(-2 ,0 , 1), (-3,0, 1) \right\} \ .$$

But there are infinitely many ways to write it. For instance, another way to write the same solution set could be the following:

$$(0, 0, 1/6) +\ \mathrm{span}\left\{(-1 ,0 , 1/2), (-1,0, 1/3) \right\} \ .$$

EDIT. I forgot: in order to fully check that your solution is right, you should also verify that the rank of your matrix $A$ is one (as it is); so the solution set has indeed dimension = number of unknowns - $\mathrm{rank}\ A = 3 -1 = 2$.

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