# Algebra: find all the solutions of each system of equations?

• Sep 22nd 2010, 08:20 AM
eleahy
Algebra: find all the solutions of each system of equations?
Algebra: find all the solutions of each system of equations?

X^1 - 2X^2 + X^3 = 1
X^2 - 2X^3 + X^4 = 0
X^3 - 2X^4 + X^5 = 2
X^4 - 2X^5 + X^6 = 0
X^5 - 2X^6 + X^7 = 1

im not sure how to start such a big system of equations ....do you have to isolate all the different x's(x^1,x^2....)?
any help much appreciated!!!!
• Sep 22nd 2010, 09:10 AM
Ackbeet
Question: does x^2 mean (as it usually does) $x^{2}?$ Or does it mean (as I suspect it does) $x_{2}?$
• Sep 22nd 2010, 09:23 AM
eleahy
ya your right its the second one!!! you said sorry i dont know how to do that!!!
• Sep 22nd 2010, 09:33 AM
Ackbeet
Interesting. Your system looks like a discretization of a DE. Anyway. Here's your system in matrix form:

$\begin{bmatrix}
1 &-2 &1 &0 &0 &0 &0\\
0 &1 &-2 &0 &0 &0 &0\\
0 &0 &1 &-2 &1 &0 &0\\
0 &0 &0 &1 &-2 &1 &0\\
0 &0 &0 &0 &1 &-2 &1
\end{bmatrix}
\begin{bmatrix}
x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\\ x_{6}\\ x_{7}
\end{bmatrix}=\begin{bmatrix}
1 \\0 \\2 \\0 \\1
\end{bmatrix}.$

Correct? I would start doing ERO's on the augmented matrix

$\left[\begin{matrix}
1 &-2 &1 &0 &0 &0 &0\\
0 &1 &-2 &0 &0 &0 &0\\
0 &0 &1 &-2 &1 &0 &0\\
0 &0 &0 &1 &-2 &1 &0\\
0 &0 &0 &0 &1 &-2 &1
\end{matrix}\left|\begin{matrix}
1 \\0 \\2 \\0 \\1
\end{matrix}\right].$

What do you get?

[EDIT] Actually, you can do the back substitution immediately, because your matrix is upper triangular. I think you'll need two parameters to describe all of the solutions.
• Sep 22nd 2010, 10:14 AM
eleahy
im confused how you could do back substitution here i thought you had to have a solution for one value of x immediately to do this??
• Sep 22nd 2010, 10:17 AM
Ackbeet
Sorry. See the edit in Post # 4.
• Sep 22nd 2010, 10:21 AM
eleahy
im confused how you could do back substitution here i thought you had to have a solution for one value of x immediately to do this??
• Sep 22nd 2010, 10:35 AM
Ackbeet
Well, you don't have enough equations to get a unique solution. You're going to have a two-parameter family of solutions (I get two from the fact that you have 7 unknowns and 5 equations. 7 - 5 = 2.) What I would do is use two parameters $s=x_{6}$ and $t=x_{7}.$ Then you take the last equation there and solve for $x_{5}$ as follows:

$x_{5}-2x_{6}+x_{7}=1,$ so

$x_{5}=1+2x_{6}-x_{7}=1+2s-t.$ Now that you have $x_{5},$ you can proceed to obtain $x_{4}$, and so on. Make sense? Your final answer should look like this:

$\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\\ x_{6}\\x_{7}\end{bmatrix}=\begin{bmatrix}a\\b\\c\\ d\\e\\f\\g\end{bmatrix}+s\begin{bmatrix}h\\i\\j\\k \\l\\m\\n\end{bmatrix}+t\begin{bmatrix}o\\p\\q\\r\ \u\\v\\w\end{bmatrix},$

where $a$ through $r$ are known, and $u$ through $w$ are known. You've already got the following:

$\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\\ x_{6}\\x_{7}\end{bmatrix}=\begin{bmatrix}a\\b\\c\\ d\\1\\0\\0\end{bmatrix}+s\begin{bmatrix}h\\i\\j\\k \\2\\1\\0\end{bmatrix}+t\begin{bmatrix}o\\p\\q\\r\ \-1\\0\\1\end{bmatrix}.$
• Sep 27th 2010, 07:26 AM
eleahy
Sorry about the delay I have all the values of x in terms of s and t
X1=12+6s-5t
X2=8+5s-4t
X3=5+4s-3t
X4=2+3s-2t
X5=2s-t+1
x6=s
X7=t

But I'm not sure how you got your final answer with this or is it ok to leave it in the form I have it
• Sep 27th 2010, 07:30 AM
Ackbeet
Your answer is equivalent in form to the form that I gave. Which one is preferable depends on your teacher. If I were teaching, I'd accept either. To convert from one to the other, just read off the coefficients. For example:

$\begin{bmatrix}a\\b\\c\\d\\e\\f\\g\end{bmatrix}=\b egin{bmatrix}12\\8\\5\\2\\1\\0\\0\end{bmatrix},$

which is the constant vector. Make sense?