We have system of linear equations $A\vec{x}=b$ : $A=\begin{pmatrix} 2& 1& -1& -1& 1& \\ 1& -1& 1& 1& -2& \\ 3& 0& 0& 0& -1& \\ 1& -2& 5& 0& 0& \end{pmatrix}$ and $\vec{b}=\begin{pmatrix} a \\ b\\ c \\ d \end{pmatrix}$.
I know the value of $a=1, b=2, c=3 $ and $d=4$.
Now we know $b$ is a $4\times 1$ matrix, and A is a $4\times 5$.
So to get $4\times 1$ by multiplication of the two matrices, $x$ has to be a $5\times 1$ matrix, which is the variables.
So I think it will be:
$$ \begin{pmatrix} 2& 1& -1& -1& 1& \\ 1& -1& 1& 1& -2& \\ 3& 0& 0& 0& -1& \\ 1& -2& 5& 0& 0& \end{pmatrix} \begin{pmatrix} x\\ y\\ z \\ l \\ k\end{pmatrix}=\begin{pmatrix} a \\ b\\ c \\ d \end{pmatrix} $$
Is this good?
If yes than I get:
$2x+y-z-t+k=1$
$x-y+z+t-2k=2$
$3x-k=3$
$x-2y+5z=4$
Can someone tell me if I am on good way? If yes, can someone help me to do this system of $4$ linear equations, because I'm making some mistakes in calculus always.