Solve the linear system Ax = B I apologize about the formatting but I'm stuck on this problem where I have to solve the linear system $Ax=b$, the vector is a 4×1 while the system is a 4×3. Is this type of question even solvable becaue I tried doing it after watching a video which featured a 3×1 vector and a 3×3 system.
$$A=\begin{bmatrix}
1&0&-2\\2&1&-3\\0&-2&1\\4&1&-2
\end{bmatrix}\qquad b=\begin{bmatrix}6\\9\\0\\11\end{bmatrix}$$


I've added the actual question from the sheet
 A: Remember that the clause $A \vec{x} = \vec{b}$ means that $\vec{x}$ is a vector that can be dot-multiplied on the right of $A$, so is a three component vector, $\vec{x} = (x_1, x_2, x_3)$.  Then the clause means \begin{align*}
1\cdot x_1 + 0 \cdot x_2 -2 \cdot x_3 &= 6 \\
2\cdot x_1 + 1 \cdot x_2 -3 \cdot x_3 &= 9 \\
0\cdot x_1 - 2 \cdot x_2 +1 \cdot x_3 &= 0 \\
4\cdot x_1 + 1 \cdot x_2 -2 \cdot x_3 &= 11 \\
\end{align*}
So you have four equations in three unknowns.  It's possible that this system is overdetermined (i.e., has no solution).
But really, you should just try Gaussian elimination, using the first equation to eliminate the $x_1$s from the other three equations, then the reduced second equation to eliminate the $x_2$s from the following two reduced equations, then scale the third to remove the x_3 from the twice reduced last equation.  Either the last equation becomes trivial ($0+0+0=0$) and the system has a solution which you can get by backsubstitution or the last equation is not trivial and no solution exists.
