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Parametrization from Matrix

Jundoe

New member
Nov 23, 2013
10
I'm facing some doubts regarding the parametrization of a given matrix.

Let's say, the following matrix is reduced.

From:
$\begin{bmatrix}0 & 2 & -8\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}$

To:
$\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}$

To Parametrize that I would do the following:

x2=0, x3=0

$\begin{bmatrix}x1\\x2\\x3\end{bmatrix}$= $\begin{bmatrix}0\\1\\1\end{bmatrix}$

But that doesn't seem right. For some reason when the matrix is bigger with more integers I can do it simply with chosen variables r, s, t... But with only zeroes like this I get super confused.

I would usually proceed with assigned variables, which may yield:

$\begin{bmatrix}x1\\x2\\x3\end{bmatrix}$= r $\begin{bmatrix}0\\1\\0\end{bmatrix}$ + s $\begin{bmatrix}0\\0\\1\end{bmatrix}$

But even this feels odd, seeing as I'm assigning a variable to a pivot.

Can someone please clarify this for me.
Thank You.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I'm facing some doubts regarding the parametrization of a given matrix.

Let's say, the following matrix is reduced.

From:
$\begin{bmatrix}0 & 2 & -8\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}$

To:
$\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}$

To Parametrize that I would do the following:

x2=0, x3=0

$\begin{bmatrix}x1\\x2\\x3\end{bmatrix}$= $\begin{bmatrix}0\\1\\1\end{bmatrix}$

But that doesn't seem right. For some reason when the matrix is bigger with more integers I can do it simply with chosen variables r, s, t... But with only zeroes like this I get super confused.

I would usually proceed with assigned variables, which may yield:

$\begin{bmatrix}x1\\x2\\x3\end{bmatrix}$= r $\begin{bmatrix}0\\1\\0\end{bmatrix}$ + s $\begin{bmatrix}0\\0\\1\end{bmatrix}$

But even this feels odd, seeing as I'm assigning a variable to a pivot.

Can someone please clarify this for me.
Thank You.
Welcome to MHB, Jundoe! :)

I think you are trying to solve $x1$, $x2$, and $x3$ from:
$$\begin{bmatrix}0 & 2 & -8\\0 & 2 & 0\\0 & 0 & 2\end{bmatrix}
\begin{bmatrix}x1\\x2\\x3\end{bmatrix}
= \begin{bmatrix}0\\0\\0\end{bmatrix}$$
Let me know if I am misunderstanding.

Row reduction turns this into:
$$\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix}
\begin{bmatrix}x1\\x2\\x3\end{bmatrix}
= \begin{bmatrix}0\\0\\0\end{bmatrix}$$
You correctly deduced that $x2=0$ and $x3=0$.

However, after that you seem to give them a non-zero value, which can't be right.
What you do have, is that $x1$ has an unspecified value. Let's call it $r$. So $x1 = r$.
Then your solution should be:
$$\begin{bmatrix}x1\\x2\\x3\end{bmatrix}= r \begin{bmatrix}1\\0\\0\end{bmatrix}$$
 

Jundoe

New member
Nov 23, 2013
10
However, after that you seem to give them a non-zero value, which can't be right.
What you do have, is that $x1$ has an unspecified value. Let's call it $r$. So $x1 = r$.
Then your solution should be:
$$\begin{bmatrix}x1\\x2\\x3\end{bmatrix}= r \begin{bmatrix}1\\0\\0\end{bmatrix}$$
Much clearer, thank you. I was aware of assigning free variables yet, for some odd reason, thought the zeroes were pretty much obsolete, and would ignore them instead of assigning them a variable.