TITLE: Find a basis and the dimension of the solution space W of the following homogeneous system
QUESTION [0 upvotes]: Good morning, I need help with this problem.
Find a basis and the dimension of the solution space $W $of the following homogeneous system
$\begin{cases}
x+2y-2z+2s-t=0\\
x+2y-z+3s-2t=0\\
2x+4y-7z+s+t=0
\end{cases}
 $

REPLY [1 votes]: Note that $W$ is the null space of the matrix
$$
A=
\left[\begin{array}{rrrrr}
1 & 2 & -2 & 2 & -1 \\
1 & 2 & -1 & 3 & -2 \\
2 & 4 & -7 & 1 & 1
\end{array}\right]
$$
Recall that $A$ and the reduced row-echelon form of $A$ have the same null space. We can compute $\DeclareMathOperator{rref}{rref}\rref A$ using row-reductions. In our case we have
$$
\rref{A}=
\left[\begin{array}{rrrrr}
1 & 2 & 0 & 4 & -3 \\
0 & 0 & 1 & 1 & -1 \\
0 & 0 & 0 & 0 & 0
\end{array}\right]
$$
Here, $\rref A$ has two pivot columns, so the rank-nullity theorem tells us that $W$ has dimension
$$
\dim W=(\#\text{ columns of }A)-(\#\text{ pivot columns of }\rref A)=5-2=3
$$
Can you find three linearly independent vectors in the null space of $\rref A$?
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