Parametrize solution

Question

Consider the matrix $$G=\begin{bmatrix}1&2&-2&1&0\\ 0&0&1&-2&0\\ 0&0&0&1&-2\\ 0&0&0&0&0\\ 0&0&0&0&0\end{bmatrix}$$

$$G\begin{bmatrix}v\\w\\x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\1\\1\\0\\0\end{bmatrix}.$$ By which sets of variables can $V$ be parametrised?

My attempt at a solution was to find the reduced row echelon form of the augmented matrix, which I found to be: $$\left[\begin{array}{ccccc|c} 1&2&0&0&-6&6\\ 0&0&1&0&-4&3\\ 0&0&0&1&-2&1\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\end{array}\right]$$ Written as a system of equations, we have: $$v+2w-6z=6$$ $$x-4z=3$$ $$y-2z=1$$ Thus I determined the sets of variables able to parametrize the solution space to be:

$vxy$ (x can be anything, y anything), $wxy$ (x can be anything, y can be anything), $zxy$ (x anything, y anything), $xvwy$ (v, w, and y can be anything), $zvwy$ (v, w, and y can be anything), and $zxvw$ (x, v, and w can by anything).

Thus my answer is: $vxy$, $wxy$, $zxy$, $xvwy$, $zvwy$, $zxvw$. Is this correct or am I totally incorrect in my thinking? Thank you.

Answer 1

Your thinking and answer are wrong. Rather:

• If $z$ is a parameter, it determines $x,y$.
• If $z$ is not a parameter, one of $x,y$ can be a parameter, determining $z$ and hence the non-parameter in $x,y$.
• Either way, we may choose one of $v,w$ as parameter, and this determines the other.

Thus the variable sets that parametrise $V$ are $vz,wz,xv,xw,yv,yw$. All these have cardinality 2, consistent with $G$'s nullity being 2.