Respuesta :
[tex]\begin{cases}a_1=-2\\a_2=2\\a_{n+1}=-2a_n+8a_{n-1}&\text{for }n\ge2\end{cases}[/tex]
From the given starting values, we have
[tex]\begin{bmatrix}a_3\\a_2\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]
and more generally, the [tex](n+1,n)[/tex]-th term pair is captured by
[tex]\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}[/tex]
so that
[tex]\mathbf M=\begin{bmatrix}-2&8\\1&0\end{bmatrix}[/tex]
We notice that
[tex]\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\mathbf M\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\mathbf M^2\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\cdots=\mathbf M^{n-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]
which is to say that the explicit formula for [tex]a_n[/tex] is given by the dot product of the second row of [tex]\mathbf M^{n-1}[/tex] with [tex]\begin{bmatrix}a_2&a_1\end{bmatrix}^\top[/tex].
First we diagonalize [tex]\mathbf M[/tex]. It has eigenvalues [tex]\lambda[/tex], where
[tex]\det(\mathbf M-\lambda\mathbf I)=\begin{vmatrix}-2-\lambda&8\\1&-\lambda\end{vmatrix}=0\implies\lambda_1=2,\lambda_2=-4[/tex]
with corresponding eigenvectors [tex]\vec\eta[/tex], where
[tex](\mathbf M-2\mathbf I)\vec\eta_1=\mathbf0\implies\vec\eta_1=\begin{bmatrix}2\\1\end{bmatrix}[/tex]
[tex](\mathbf M+4\mathbf I)\vec\eta_2=\mathbf0\implies\vec\eta_2=\begin{bmatrix}-4\\1\end{bmatrix}[/tex]
Then we have
[tex]\mathbf M=\mathbf S^{-1}\mathbf{DS}[/tex]
where [tex]\mathbf S[/tex] is the matrix whose columns are [tex]\vec\eta_1[/tex] and [tex]\vec\eta_2[/tex] and [tex]\mathbf D[/tex] is the diagonal matrix whose non-zero entries are the eigenvalues of [tex]\mathbf M[/tex]. So
[tex]\mathbf M=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2&0\\0&-4\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}[/tex]
which gives
[tex]\mathbf M^{n-1}=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2^{n-1}&0\\0&(-4)^{n-1}\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}[/tex]
From here, you would just need to determine the bottom row of the final matrix product.
From the given starting values, we have
[tex]\begin{bmatrix}a_3\\a_2\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]
and more generally, the [tex](n+1,n)[/tex]-th term pair is captured by
[tex]\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\begin{bmatrix}-2&8\\1&0\end{bmatrix}\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}[/tex]
so that
[tex]\mathbf M=\begin{bmatrix}-2&8\\1&0\end{bmatrix}[/tex]
We notice that
[tex]\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}=\mathbf M\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\mathbf M^2\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\cdots=\mathbf M^{n-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}[/tex]
which is to say that the explicit formula for [tex]a_n[/tex] is given by the dot product of the second row of [tex]\mathbf M^{n-1}[/tex] with [tex]\begin{bmatrix}a_2&a_1\end{bmatrix}^\top[/tex].
First we diagonalize [tex]\mathbf M[/tex]. It has eigenvalues [tex]\lambda[/tex], where
[tex]\det(\mathbf M-\lambda\mathbf I)=\begin{vmatrix}-2-\lambda&8\\1&-\lambda\end{vmatrix}=0\implies\lambda_1=2,\lambda_2=-4[/tex]
with corresponding eigenvectors [tex]\vec\eta[/tex], where
[tex](\mathbf M-2\mathbf I)\vec\eta_1=\mathbf0\implies\vec\eta_1=\begin{bmatrix}2\\1\end{bmatrix}[/tex]
[tex](\mathbf M+4\mathbf I)\vec\eta_2=\mathbf0\implies\vec\eta_2=\begin{bmatrix}-4\\1\end{bmatrix}[/tex]
Then we have
[tex]\mathbf M=\mathbf S^{-1}\mathbf{DS}[/tex]
where [tex]\mathbf S[/tex] is the matrix whose columns are [tex]\vec\eta_1[/tex] and [tex]\vec\eta_2[/tex] and [tex]\mathbf D[/tex] is the diagonal matrix whose non-zero entries are the eigenvalues of [tex]\mathbf M[/tex]. So
[tex]\mathbf M=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2&0\\0&-4\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}[/tex]
which gives
[tex]\mathbf M^{n-1}=\begin{bmatrix}2&-4\\1&1\end{bmatrix}^{-1}\begin{bmatrix}2^{n-1}&0\\0&(-4)^{n-1}\end{bmatrix}\begin{bmatrix}2&-4\\1&1\end{bmatrix}[/tex]
From here, you would just need to determine the bottom row of the final matrix product.
The matrix satisfy the [tex]\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =M \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right][/tex] is [tex]M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right][/tex]
Step-by-step explanation:
Given: Consider the sequence defined recursively by [tex]a_1=-2,\;a_2=2,\;a_{n+1}=-2a_n+8a_{n-1}\;\rm{for},\; n\geq2[/tex].
According to question,
From the given values: [tex]\left[\begin{array}{ccc}a_3\\a_2\\\end{array}\right] =\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right] \left[\begin{array}{ccc}a_2\\a_1\end{array}\right][/tex]
And generally we can write for [tex](n+1,\;n)[/tex]: [tex]\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right] \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right][/tex]
So, here
[tex]M=\left[\begin{array}{ccc}-2&8\\1&0\end{array}\right][/tex]
So using diagonalise equation of the matrix using eigen values we can write: [tex]M=S^{-1}DS[/tex]
[tex]M=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2&0\\0&-4\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right][/tex]
Which
[tex]M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right][/tex]
Therefore, the matrix satisfy the [tex]\left[\begin{array}{ccc}a_{n+1}\\a_n\\\end{array}\right] =M \left[\begin{array}{ccc}a_n\\a_{n-1}\end{array}\right][/tex] is [tex]M^{n-1}=\left[\begin{array}{ccc}2&-4\\1&0\end{array}\right]^{-1}\left[\begin{array}{ccc}2^{n-1}&0\\0&(-1)^{n-1}\end{array}\right]\left[\begin{array}{ccc}2&-4\\1&1\end{array}\right][/tex].
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