# Parametrize (as a subset of R5) the solution space of the system of equations

I am just wondering how I would parametrize the system of equations from the augmented matrix

I know the following:
v = -2 - 4w -3z
w = 0
x = 0
y = 8 - 5z
z = 0

$$\begin{bmatrix} 1 & 4 & -2 & 0 & -3& -2\\ 0 & 0& 0& 1& 5 &8 \end{bmatrix}$$

Please note that the last two numbers are supposed to have a line.
|-2
|8

• Your matrix doesn’t correspond to the system of equations that you’ve written here. If you back-substitute for $w$, $x$ and $z$ into the other two equations, you’ll find that the solution is unique—there’s nothing to parameterize. – amd Apr 20 at 18:48

## 1 Answer

Considering your equations the matrix should be: $$A|\vec{b}=\left ( \begin{array}{cc} 1 & 4 &0 & 0 &3 \\ 0 & 0 &0 &1&5 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&0&1 \end{array} \right | \left. \begin{array}{c} -2 \\ 8 \\ 0 \\ 0 \\ 0 \end{array}\right )$$ with $$A \cdot \begin{pmatrix} v\\ w\\ x\\ y\\ z \end{pmatrix} = \vec{b}$$ In this case the only fitting solution would be $$\vec{x} = \begin{pmatrix} v\\ w\\ x\\ y\\ z \end{pmatrix} = \begin{pmatrix} -2\\ 0\\ 0\\ 8\\ 0 \end{pmatrix}$$
In case you misinterpreted the matrice, the parametrization will look as follows: $$A|\vec{b}=\left ( \begin{array}{cc} 1 & 4 &-2 & 0 &-3 \\ 0 & 0 &0 &1&5 \\ \end{array} \right | \left. \begin{array}{c} -2 \\ 8 \\ \end{array}\right )$$ We now use the Gaussian Elimination (Wikipedia Entry) . Since the matrix is already in reduced row echolon form (Wikipedia Entry), we can do the following:
We set every variable in a row as a free parameter $$\lambda_{i}\in \mathbb{R}$$ , except for one variable (usually the one with coefficient 1) :
$$w=\lambda_{1}; x=\lambda_{2};z=\lambda_{3}$$
First row: $$v= -2 \,-4\lambda_{1}+2\lambda_{2}+3\lambda_{3}$$
Second row: $$y= 8- 5\lambda_{3}$$
Now we can rewrite this in vectorform, which gives us the parametrization: $$\begin{pmatrix} v\\ w\\ x\\ y\\ z \end{pmatrix} = \begin{pmatrix} -2 \,-4\lambda_{1}+2\lambda_{2}+3\lambda_{3} \\ \lambda_{1}\\ \lambda_{2}\\ 8- 5\lambda_{3}\\ \lambda_{3} \end{pmatrix} =\begin{pmatrix} -2\\ 0\\ 0\\ 8\\ 0 \end{pmatrix} + \lambda_{1} \cdot \begin{pmatrix} -4\\ 1\\ 0\\ 0\\ 0 \end{pmatrix}+\lambda_{2} \cdot \begin{pmatrix} 2\\ 0\\ 1\\ 0\\ 0 \end{pmatrix}+\lambda_{3} \cdot \begin{pmatrix} 3\\ 0\\ 0\\ -5\\ 1 \end{pmatrix}\\$$

• Thank you , your final conclusion is what I was looking for exactly! – KyleMcCann Apr 21 at 18:52