# Solve the following system of equations. x <mrow class="MJX-TeXAtom-ORD"> n

Solve the following system of equations.
${x}_{n+1}=2{y}_{n}-{z}_{n}$
${y}_{n+1}={y}_{n}$
${z}_{n+1}={x}_{n}-2{y}_{n}+2{z}_{n}$
What is the solution in general for ${x}_{0}$, ${y}_{0}$, ${z}_{0}$ arbitrary?
It is intended to be solved using Jordan Normal Form of the Linear Algebra knowledge. I have no idea how to start, can anyone give a hint?
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hopeloothab9m
Hint. Note that for $n\ge 0$,
$\left(\begin{array}{c}{x}_{n}\\ {y}_{n}\\ {z}_{n}\end{array}\right)=\left(\begin{array}{ccc}0& 2& -1\\ 0& 1& 0\\ 1& -2& 2\end{array}\right)\left(\begin{array}{c}{x}_{n-1}\\ {y}_{n-1}\\ {z}_{n-1}\end{array}\right)={\left(\begin{array}{ccc}0& 2& -1\\ 0& 1& 0\\ 1& -2& 2\end{array}\right)}^{n}\left(\begin{array}{c}{x}_{0}\\ {y}_{0}\\ {z}_{0}\end{array}\right).$
Now find the Jordan normal form $J$ of the matrix
$M:=\left(\begin{array}{ccc}0& 2& -1\\ 0& 1& 0\\ 1& -2& 2\end{array}\right)$
and a matrix $P$ such that $M=PJ{P}^{-1}$. Then ${M}^{n}=P{J}^{n}{P}^{-1}$.
###### Not exactly what you’re looking for?
Jamiya Weber
By the @Robert Z answer the jordan form of matrix $M$ is as follows:
$M=P\phantom{\rule{thinmathspace}{0ex}}J\phantom{\rule{thinmathspace}{0ex}}{P}^{-1}$
where the matices $P$ and $J$ are in the following form;
$P=\left(\begin{array}{ccc}-1& 3& 2\\ 0& 1& 1\\ 1& 0& 0\end{array}\right)\phantom{\rule{1em}{0ex}},\phantom{\rule{1em}{0ex}}J=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
with the induction on $n$ you can prove that the $n$-th power of matrix $J$ is as follows:
${J}^{n}=\left(\begin{array}{ccc}1& n& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
So, the $n$-th power of matrix $M$ is in the following form:
${M}^{n}=P\phantom{\rule{thinmathspace}{0ex}}{J}^{n}\phantom{\rule{thinmathspace}{0ex}}{P}^{-1}=\left(\begin{array}{ccc}1-n& 2\phantom{\rule{thinmathspace}{0ex}}n& -n\\ 0& 1& 0\\ n& -2\phantom{\rule{thinmathspace}{0ex}}n& 1+n\end{array}\right)$