<s>[INST] Show that it is a solution for all values of $t$ Show that $(1-t, 2+3t, 3-2t)$ is a solution for all values of $t$ to following linear system:
$a + b + c = 6$
$a - b - 2c = -7$
$5a + b - c = 4$
I have found that these rows are linearly dependent but I do not know how to continue. Can you help me? Here is after Gaussian Elimination:
$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0  \end{bmatrix} $$
 [/INST] Hint: One approach is to use Gaussian Elimination, you end up with:
$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & -\dfrac{1}{2} \\0 & 1 & \dfrac{3}{2} & \dfrac{13}{2} \\ 0 & 0 & 0 & 0  \end{bmatrix} $$
Can you take it from here?
Update
We have:


*

*$a = -\dfrac{1}{2} +\dfrac{1}{2}c$

*$b = \dfrac{13}{2} -\dfrac{3}{2}c$


So, $c$ is a free variable and from the problem statement, they chose $c = 3-2t$.
Now, what are $a$ and $b$? Also, can you write this in another form?
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