I was doing my homework and stumbled upon this question where I have a system of 3 equations with 5 unknowns/variables. I got stuck trying to make it into an echelon form of this matrix.
The problem is stated as follows:
Find which values of c does the system have a solution? For each c, find all solutions as linear combinations of certain vectors.
In the image below I tried to solve it but got stuck and I'm not sure how to continue this equation..
Best Answer
Set $x_5 = c $ and $x_6 = c^2 $, then the system becomes
$\begin{bmatrix} 2&&1&&5&&4&&0&&-1 \\ 1 && -1&& 1 &&-1&&-4&&0 \\ 1 && 0 && 2 && 1 && -1 && 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 3 \end{bmatrix} $
Solving using Gauss-Jordan elimination, the solutions are
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} 5 \\ -8\\ 0 \\ 0 \\ 2 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t_3 \begin{bmatrix} -1 \\ 3 \\ 0 \\ 0 \\ -1 \\ 1 \end{bmatrix} $
From which it follows that
$x_5 = c = 2 - t_3 $ and $ x_6 = c^2 = t_3 $
Hence, we must have $2 - c = c^2 $, i.e. $c^2 + c - 2 = 0$ which can be factored as $ (c + 2)(c - 1) = 0 $. Therefore, the two values of $c$ for which there is a solution are $c = -2 $ and $c = 1$.
For $c = -2$, we get $t_3 = 4 $ and for $c = 1$ we get $t_3 = 1 $
Therefore, there are two corresponding sets of solutions:
Case I: for $c = -2$ we get,
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ 0 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \end{bmatrix} $
Case II: for $c = 1$, we get,
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \\ 0 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \end{bmatrix} $