# Help determining the parameterized solution to a system of linear equation.

I am quite new to this area so please bear with me if I am overlooking something glaringly obvious here :)

I am trying to solve the following equation system:

$$\begin{array}{lcl} x + y + z & = & 30 \\ x + 2y & = & 25 \\ 2x + 3y +x& = & 55 \end{array}$$

I am using the matrix notation and write the equation system as (i do not know how to typeset the divisor line between the equality part, sorry about that):

$$\begin{pmatrix} 1 & 1 & 1 & 30 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix}$$

This matrix can be written as this: (by multiplying row $2$ with $-1$ and adding it to row $1$, and multiplying the new row $1$ with $-1$ and adding it to row $3$, and finally by multiplying row $2$ with $-1$ and adding it to row $3$).

$$\begin{pmatrix} 1 & 1 & 1 & 30 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix}\longleftrightarrow\begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\2 & 3 & 1 & 55\end{pmatrix} \longleftrightarrow \begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\2 & 4 & 0 & 50\end{pmatrix} \longleftrightarrow \begin{pmatrix} 0 & -1 & 1 & 5 \\ 1 & 2 & 0 & 25\\0 & 0 & 0 & 0\end{pmatrix}$$

This can be written as: $$\begin{array}{lcl} -y + z & = & 5 \\ x + 2y & = & 25 \\ 0& = & 0 \end{array}$$

according to my textbook $z$ is a parameter in this equation system. But from the development above, could not $y$ just as well be the parameter?

By solving the equation system differently you could get the final result as:

$$\begin{pmatrix} 1 & 0 & 2 & 35 \\0 & 1 & -1 & -5\\0 & 0 & 0 & 0\end{pmatrix}\longleftrightarrow \begin{array}{lcl} x + 2z & = & 35 \\ y + -z & = & -5 \\ 0& = & 0 \end{array}$$

From this development is is much more clear that $z$ should be the parameter since $x, y$ can be expressed as linear functions of $z$?

Thank you kindly!

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btw, you did nicely with formatting; we appreciate any efforts made to format. +1 – amWhy Mar 8 '13 at 20:24

From this development is is much more clear that z should be the parameter since x,y can be expressed as linear functions of z?

Precisely. $z$ is a "free" variable (as you can tell by the last row of zeros in both approaches) on which the precise values of $x, y$ depend. Once a value for $z$ is chosen (among infinitely many possible values for $z$, then we can determine $x, y$.

So we can write the solutions as follows: \begin{pmatrix} 35 - 2z \\ z - 5 \\ z \end{pmatrix}

You'll often see the following: Given any assignment $z = \alpha$, then the solution can be expressed: \begin{pmatrix} 35 - 2\alpha \\ \alpha - 5 \\ \alpha \end{pmatrix}

I used the latter solution to the system of equations as it is in the preferred reduced row echelon form. But the first approach will yield the same parametrization, which you can see when you write $y$ as a function of $z$, since then $x$ is also a function of $z$.

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So when doing developments like this the first approach is not correct and / or should be avoided? – Lukas Arvidsson Mar 8 '13 at 19:50
Thank you for your updated answer, I cannot understand that in the first solution $y$ looks like it could just as well be the parameter? Why is that so? – Lukas Arvidsson Mar 8 '13 at 20:02
You could still use the first approach, but note that you're first line should be $-y + z = 5$. Note that z is still the free variable: the fact that $x$ in the first case can be defined as a function of y hides the fact that $-y + z = 5 \implies y = z - 5$, hence, from the second equation (first approach) $x + 2y = 25 \implies x = 25 - 2y \implies x = 25 - 2(z - 5) \implies x = 35 - 2z$ – amWhy Mar 8 '13 at 20:03
Thank you! Could you please update your answer with how to use the first approch as well? That would be very kind of you and I think it would help me out a bit here. – Lukas Arvidsson Mar 8 '13 at 20:06
Perfect! I see that now! Thank you very much!!! – Lukas Arvidsson Mar 8 '13 at 20:08