Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
3 & 5 & 5 & 2 & 0\\
1 & 0 & 2 & 2& 1 \\
1 & 1 & 1 & -2 & -2 \\
-2 & 0 & -4 & -2 & -2
\end{bmatrix}$
We obtain rowspace $A=\begin{bmatrix}
3 & 5 & 5 & 2 & 0\\
1 & 0 & 2 & 2& 1 \\
1 & 1 & 1 & -2 & -2 \\
-2 & 0 & -4 & -2 & -2
\end{bmatrix} \approx \begin{bmatrix}
3 & 5 & 5 & 2 & 0\\
0 & 0 & 0 & 2& 0 \\
0 & 0 & 8 & 38 & 24 \\
0 & 10 & -2 & -2 & -6
\end{bmatrix} \approx \begin{bmatrix}
3 & 5 & 5 & 2 & 0\\
0 & 10 & -2 & -2& -6 \\
0 & 0 & 8 & 38 & 24 \\
0 & 0 & 0 & 2 & 0
\end{bmatrix} \\
rowspace (A)=\{(3,5,5,2,0),(0,10,-2,-2,-6),(0,0,8,38,24),(0,0,0,2,0\}$
And colspace $A= \begin{bmatrix}
3 & 5 & 5 & 2 & 0\\
0 & 10 & -2 & -2& -6 \\
0 & 0 & 8 & 38 & 24 \\
0 & 0 & 0 & 2 & 0
\end{bmatrix} \approx \begin{bmatrix}
2& 0 & 0 & 0 & 0\\
-2 & 10 & -2 & 0& -6 \\
38 & 0 & 8 & 0 & 24 \\
2 & 5 & 5 & 3 & 0
\end{bmatrix} \approx \begin{bmatrix}
2& 0 & 0 & 0 & 0\\
-2 & 10 & 0 & 0& 0 \\
38 & 0 & 40 & 0 & 0\\
2 & 5 & 30 & 3 & 15
\end{bmatrix} \approx \begin{bmatrix}
2& 0 & 0 & 0 & 0\\
-2 & 10 & 0 & 0& 0 \\
38 & 0 & 40 & 0 & 0\\
2 & 5 & 30 & 3 & 0
\end{bmatrix} \\
colspace (A)=\{(2,-2,38,2),(0,10,5,5),(0,0,40,30),(0,0,0,3)\}$
The matrix is in form $4 \times 5$ and $rank(A)=4$, the solution of the system $Ax=0$ has a uniquely solution.
Consequently, the null space of the given matrix $A$ is $\{(0,0,0,0)\}$
