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

1. ## 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!!!!

2. Question: does x^2 mean (as it usually does) $\displaystyle x^{2}?$ Or does it mean (as I suspect it does) $\displaystyle x_{2}?$

3. ya your right its the second one!!! you said sorry i dont know how to do that!!!

4. Interesting. Your system looks like a discretization of a DE. Anyway. Here's your system in matrix form:

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

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

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

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.

5. 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??

6. Sorry. See the edit in Post # 4.

7. 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??

8. 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 $\displaystyle s=x_{6}$ and $\displaystyle t=x_{7}.$ Then you take the last equation there and solve for $\displaystyle x_{5}$ as follows:

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

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

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

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

$\displaystyle \begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}\\ x_{6}\\x_{7}\end{bmatrix}=\begin{bmatrix}a\\b\\c\\ d\\1\\0\\0\end{bmatrix}+s\begin{bmatrix}h\\i\\j\\k \\2\\1\\0\end{bmatrix}+t\begin{bmatrix}o\\p\\q\\r\ \-1\\0\\1\end{bmatrix}.$

9. 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

10. 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:

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

which is the constant vector. Make sense?