So I'm currently taking a Linear Algebra class and am stuck on a problem.
I have the equations:
$$\begin{cases}\begin{align}&x + 2y - z + 3t = 3\\ &2x + 4y + 4z + 3t = 9\\ &3x + 6y - z + 8t = 10 \end{align}\end{cases}$$
I'm not sure on how to solve this with the $4$ variables and the $3$ lines. I can do a $3 \times 3$ but am struggling with this one. I'm not asking for someone to solve it, I'm asking for a point in the right direction.
On
Put your system into the corresponding augmented matrix: $$ [A \mid b] = \left[ \begin{array}{cccccc} 1 & 2 & -1 & 3 & \mid & 3 \\ 2 & 4 & 4 & 3 & \mid & 9 \\ 3 & 6 & -1 & 8 & \mid & 10 \\ \end{array} \right] $$ then do Gaussian elimination. That is, row reduce to the RREF form. Then you can read off the solutions (if they exist). A solution will exists if and only if the rank of $A$ equals the rank of $[A \mid b]$.
$$\left[ \begin{array}{cccc|c} 1 & 2 & -1 & 3 & 3 \\ 2 & 4 & 4 & 3 & 9 \\ 3 & 6 & -1 & 8 & 10 \\ \end{array} \right]$$ $$\left[ \begin{array}{cccc|cc} 1 & 2 & -1 & 3 & 3 \\ 0 & 0 & 6 & -3 & 3 \\ 0 & 0 & 2 & -1 & 1 \\ \end{array} \right]$$ $$\left[ \begin{array}{cccc|cc} 1 & 2 & -1 & 3 & 3 \\ 0 & 0 & 2 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]$$ $$\left[ \begin{array}{cccc|cc} 1 & 2 & 0 & 2.5 & 3.5 \\ 0 & 0 & 2 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]$$ $$x+2y+2.5t=3.5\\2z-t=1$$you have infinitely many solutions