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

Eden Solomon 2022-06-13 Answered
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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Answers (2)

hopeloothab9m
Answered 2022-06-14 Author has 25 answers
Hint. Note that for n 0,
( x n y n z n ) = ( 0 2 1 0 1 0 1 2 2 ) ( x n 1 y n 1 z n 1 ) = ( 0 2 1 0 1 0 1 2 2 ) n ( x 0 y 0 z 0 ) .
Now find the Jordan normal form J of the matrix
M := ( 0 2 1 0 1 0 1 2 2 )
and a matrix P such that M = P J P 1 . Then M n = P J n P 1 .
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Jamiya Weber
Answered 2022-06-15 Author has 2 answers
By the @Robert Z answer the jordan form of matrix M is as follows:
M = P J P 1
where the matices P and J are in the following form;
P = ( 1 3 2 0 1 1 1 0 0 ) , J = ( 1 1 0 0 1 0 0 0 1 )
with the induction on n you can prove that the n-th power of matrix J is as follows:
J n = ( 1 n 0 0 1 0 0 0 1 )
So, the n-th power of matrix M is in the following form:
M n = P J n P 1 = ( 1 n 2 n n 0 1 0 n 2 n 1 + n )
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