Linear Algebra exercise problem: intersection of two subspaces I’m trying to solve a linear algebra exercise, but my solution does not match the solution given on the exercise book. This is the exercise:
Given $U = \langle(1,0,1,2), (1,1,2,3)\rangle$, a subspace of $\Bbb R^4$, and $S$, solution of the system 
$$\begin{cases}
x+z-t=0 \\ 
y=1
\end{cases}
$$
Find the intersection $U\cap S$.
First, I determined the form of the generic vector $\mathbf u\in U$, using scalars $a$ and $b$, thus obtaining $u=(a+b,b,a+2b,2a+3b)$. Then, I solved the system using a matrix in canonical form, 
$$
    \begin{bmatrix}
    1 & 0 & 1 & -1 & 0 \\
    0 & 1 & 0 &  0 & 1 \\
    \end{bmatrix}
$$
obtaining $\mathbf s\in S$, $\mathbf s=(t-z,1,z,t)$.
To find the intersection, I equalled the two generic vectors, $\mathbf u=\mathbf s$, getting $(a+b, b, a+2b, 2a+3b) = (t-z, 1, z, t)$. I solved this equation by setting  up a system:
$$\begin{cases}
a+b=t-z \\ 
b=1 \\ 
a+2b=z \\
2a+3b=t
\end{cases}
$$
With the corresponding matrix:
$$
    \begin{bmatrix}
    1 & 0 & -1 & 1 & -1 \\
    0 & 1 & 0 &  0 & 1 \\
    1 & 0 & 0 & -1 & -2 \\
    2 & 0 & -1 & 0 & 3
    \end{bmatrix}
$$
Solving this matrix gave me a solution in form of a linear variety, $U\cap S=(-2,1,-1,0)+\langle(1,0,2,1)\rangle$, but this answer differs from the one on the textbook, which is $U\cap S=(1,1,2,3)+\langle(1,0,1,2)\rangle$. Can you help me understand where I made a mistake?
Thank you in advance for your help!
P. S. Sorry if I have misspelled some entity, but English is not my native language!

The first row of the $4\times5$ matrix, the last entry should be $0$ rather than $-1$.
The last two rows of the matrix correspond to equations $x-t=-2$ and $2x-z=3$. How do these equations come from the given equations? There seems to be some confusion.
Here is how the problem should be worked:
$$\begin{pmatrix}
x\\y\\z\\t
\end{pmatrix}=a\begin{pmatrix}
1\\0\\1\\2
\end{pmatrix}+b\begin{pmatrix}
1\\1\\2\\3
\end{pmatrix}=\begin{pmatrix}
a+b\\b\\a+2b\\2a+3b
\end{pmatrix}$$
Since $y=1$ we have
$$\begin{pmatrix}
x\\y\\z\\t
\end{pmatrix}=\begin{pmatrix}
a+b\\b\\a+2b\\2a+3b
\end{pmatrix}
=\begin{pmatrix}
a+1\\1\\a+2\\2a+3
\end{pmatrix}
$$
Then
\begin{equation}
x+z-t=
(a+1)+(a+2)-(2a+3)=0
\end{equation}
So the intersection is the line
$$\begin{pmatrix}
x\\y\\z\\t
\end{pmatrix}=a\begin{pmatrix}
1\\0\\1\\2
\end{pmatrix}+\begin{pmatrix}
1\\1\\2\\3
\end{pmatrix}
$$
Notice, also, that $S$ is not a subspace of $\mathbb{R}^4$ since it does not contain the zero vector.
