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(1) Sequence $ \{a_n\} $ is defined as $ a_{0} = a_1 = a_2 = 1 $ and

$ a_n = 7a_{n-2} + 6a_{n-3}, for. \ n \geq 3$.

Find a general formula for $ a_n $

I have only solved recurrence relations for n >= 2 so far in my class, I am not sure as to how I should extend it to n >= 3?

Edit: P =

(2) $ C = \begin{pmatrix} 2 &2 & 0 \\ -1 & -1 & 0 \\ 2 & 2 & 3 \end{pmatrix}\ $. Diagonalise the matrix in 3 ways so that the diagonal matrices $ D_1, D_2, D_3 $ are different. I found it's eigenvalues to be

$ \lambda_1 = 0, $ corresponding vector is $ (-1,1,0) $.

$ \lambda_2 = 1, $ corresponding vector is $ (-2,1,1) $

$ \lambda_3 = 3, $ corresponding vector is $ (0,0,1) $

Is it correct to say that $ D_1 = \begin{pmatrix} 3 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ $. $ D_2 = \begin{pmatrix} 1 &0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ $ $ D_1 = \begin{pmatrix} 3 &0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\ $?

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  • $\begingroup$ Ask separate questions $\endgroup$
    – lhf
    Nov 6 '20 at 13:47
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Hint for (1): We have $$ \begin{pmatrix} a_n \\ a_{n-1} \\ a_{n-2} \end{pmatrix} = \begin{pmatrix} 0 &7 & 6 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} a_{n-1} \\ a_{n-2} \\ a_{n-3} \end{pmatrix} $$ whose characteristic polynomial is $x^3 - 7 x - 6$ (no surprises here).

Finally, $x^3 - 7 x - 6=(x + 1) (x + 2) (x - 3)$.

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  • $\begingroup$ Thanks, I found the matrix P(edit in the qn) that diagonalizes A, but do I have to find the inverse of P directly in order to find $ a_n $? $\endgroup$
    – foc
    Nov 6 '20 at 14:31
  • $\begingroup$ @foc, not really. Once you know the eigenvalues, you know that $a_n = c_1 \lambda_1^n + c_2 \lambda_2^n + c_3 \lambda_3^n$ and you can find $c_1,c_2,c_3$ from the initial conditions. $\endgroup$
    – lhf
    Nov 6 '20 at 14:43
  • $\begingroup$ @foc, check your work on $P$ at WA $\endgroup$
    – lhf
    Nov 6 '20 at 14:45

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