Calculate $\dim W+V$ and $W\cap V$ This is a task from an old exam.
Let define:
$V_{t} = \text{lin}((1,2,2,1),(1,1,-1,t))$
$W = \cases{x_1-x_2=0\\x_3-x_4 =0}$ 
Calculate $\dim W+V_{t}$ and $\dim W\cap V_{t}$
Please verify answer below
 A: basis of $W$ = $\text{lin} ((1,1,0,0),(0,0,1,1))$
$ W+V_{t}=\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
1 & 1 & -1 & t\\
1 & 1 & 0 & 0\\
0 & 0 & 1 & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
0 & -1 & -3 & t-1\\
0 & -1 & -2 & -1\\
0 & 0 & 1 & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
0 & 0 & -1 & t\\
0 & -1 & -2 & -1\\
0 & 0 & 1 & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & 2 & 2 & 1\\
0 & -1 & -2 & -1\\
0 & 0 & 1 & 1\\
0 & 0 & 0 & t+1
\end{array}\right]
 $
$\dim W+V_{t}=3$ for $t=-1$ and $4$ in other case.
$V_{t}:$
$\left[\begin{array}{ccc}
1 & 1 & x_{1}\\
2 & 1 & x_{2}\\
2 & -1 & x_{3}\\
1 & t & x_{4}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 1 & x_{1}\\
0 & 0 & x_{2}-2x_{1}\\
0 & -2 & x_{3}-2x_{1}\\
0 & t & x_{4}-2x_{1}
\end{array}\right]\implies V_{1}:\begin{cases}
x_{2}-2x_{1}=0\\
x_{4}-2x_{1}=0
\end{cases}
 $
$W\cap V_{t}:\begin{cases}
x_{1}-x_{2}=0\\
x_{3}-x_{4}=0\\
x_{2}-2x_{1}=0\\
x_{4}-2x_{1}=0
\end{cases}
 $
$\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
 &  & 1 & -1\\
-2 & 1\\
-2 &  &  & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
 &  & 1 & -1\\
0 & -1\\
0 & -2 &  & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
0 & -1\\
 &  & 1 & -1\\
0 & -2 &  & 1
\end{array}\right]\sim\left[\begin{array}{cccc}
1 & -1 & 0 & 0\\
0 & -1\\
 &  & 1 & -1\\
0 & 0 &  & 1
\end{array}\right]
 $
$\dim W\cap V_{t}=0
 $
