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Math  /  Algebra

QuestionConsider the sequence defined recursively by a1=−1,a2=1,an+1=−10an−1+7ana_{1}=-1, a_{2}=1, a_{n+1}=-10 a_{n-1}+7 a_{n}. We can use matrix diagonalization to find an explicit formula for ana_{n}. a. Find a matrix that satisfies [anan+1]=M[an−1an]\left[\begin{array}{c}a_{n} \\ a_{n+1}\end{array}\right]=M\left[\begin{array}{c}a_{n-1} \\ a_{n}\end{array}\right] b. Find the appropriate exponent kk such that [anan+1]=Mk[a1a2]k=□ 媌 \begin{array}{l} {\left[\begin{array}{c} a_{n} \\ a_{n+1} \end{array}\right]=M^{k}\left[\begin{array}{l} a_{1} \\ a_{2} \end{array}\right]} \\ k=\square \text { 媌 } \end{array} c. Find a diagonal matrix DD and an invertible matrix PP such that M=PDP−1M=P D P^{-1}. d. Find P−1P^{-1}. e. Find M5=PD5P−1M^{5}=P D^{5} P^{-1}. f. Use parts b. and e. to find a6a_{6}. a6=a_{6}= □\square g. Develop an explicit formula for ana_{n} using part b. and a formula for Mk=PDkP−1M^{k}=P D^{k} P^{-1}.

Studdy Solution

STEP 1

1. The sequence is defined recursively by a1=−1 a_1 = -1 , a2=1 a_2 = 1 , and an+1=−10an−1+7an a_{n+1} = -10a_{n-1} + 7a_n .
2. We can use matrix diagonalization to find an explicit formula for an a_n .

STEP 2

1. Find the matrix M M that represents the recursive relationship.
2. Determine the exponent k k for expressing the sequence in terms of matrix powers.
3. Diagonalize the matrix M M by finding matrices D D and P P .
4. Compute the inverse of matrix P P .
5. Calculate M5 M^5 using the diagonalization.
6. Use the results to find a6 a_6 .
7. Develop an explicit formula for an a_n .

STEP 3

To find the matrix M M , we use the recursive relation an+1=−10an−1+7an a_{n+1} = -10a_{n-1} + 7a_n .
This can be rewritten in matrix form as: [anan+1]=[01−107][an−1an]\begin{bmatrix} a_{n} \\ a_{n+1} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -10 & 7 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n} \end{bmatrix}
Thus, the matrix M M is: M=[01−107]M = \begin{bmatrix} 0 & 1 \\ -10 & 7 \end{bmatrix}

STEP 4

To express [anan+1] \begin{bmatrix} a_{n} \\ a_{n+1} \end{bmatrix} in terms of Mk M^k , we need to find the exponent k k such that: [anan+1]=Mn−1[a1a2]\begin{bmatrix} a_{n} \\ a_{n+1} \end{bmatrix} = M^{n-1} \begin{bmatrix} a_{1} \\ a_{2} \end{bmatrix}
Therefore, k=n−1 k = n-1 .

STEP 5

To diagonalize M M , we find the eigenvalues and eigenvectors.
The characteristic equation is given by: det⁡(M−λI)=det⁡[−λ1−107−λ]=λ2−7λ+10=0\det(M - \lambda I) = \det \begin{bmatrix} -\lambda & 1 \\ -10 & 7-\lambda \end{bmatrix} = \lambda^2 - 7\lambda + 10 = 0
Solving this quadratic equation gives the eigenvalues λ1=5 \lambda_1 = 5 and λ2=2 \lambda_2 = 2 .
Next, we find the eigenvectors for each eigenvalue.
For λ1=5 \lambda_1 = 5 : (M−5I)v=[−51−102][xy]=[00](M - 5I) \mathbf{v} = \begin{bmatrix} -5 & 1 \\ -10 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} Solving gives v1=[15] \mathbf{v}_1 = \begin{bmatrix} 1 \\ 5 \end{bmatrix} .
For λ2=2 \lambda_2 = 2 : (M−2I)v=[−21−105][xy]=[00](M - 2I) \mathbf{v} = \begin{bmatrix} -2 & 1 \\ -10 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} Solving gives v2=[12] \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} .
Thus, the matrix P P is: P=[1152]P = \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}
And the diagonal matrix D D is: D=[5002]D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}

STEP 6

To find P−1 P^{-1} , we compute the inverse of P P : P−1=1det⁡(P)[2−1−51]P^{-1} = \frac{1}{\det(P)} \begin{bmatrix} 2 & -1 \\ -5 & 1 \end{bmatrix}
The determinant of P P is det⁡(P)=1(2)−1(5)=−3 \det(P) = 1(2) - 1(5) = -3 .
Thus: P−1=1−3[2−1−51]=[−231353−13]P^{-1} = \frac{1}{-3} \begin{bmatrix} 2 & -1 \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & -\frac{1}{3} \end{bmatrix}

STEP 7

To find M5 M^5 , we compute: M5=PD5P−1M^5 = P D^5 P^{-1}
First, compute D5 D^5 : D5=[550025]=[31250032]D^5 = \begin{bmatrix} 5^5 & 0 \\ 0 & 2^5 \end{bmatrix} = \begin{bmatrix} 3125 & 0 \\ 0 & 32 \end{bmatrix}
Then: M5=[1152][31250032][−231353−13]M^5 = \begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} 3125 & 0 \\ 0 & 32 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & -\frac{1}{3} \end{bmatrix}
Calculate the product: M5=[1⋅3125+1⋅01⋅0+1⋅325⋅3125+2⋅05⋅0+2⋅32][−231353−13]M^5 = \begin{bmatrix} 1 \cdot 3125 + 1 \cdot 0 & 1 \cdot 0 + 1 \cdot 32 \\ 5 \cdot 3125 + 2 \cdot 0 & 5 \cdot 0 + 2 \cdot 32 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & -\frac{1}{3} \end{bmatrix}
=[3125321562564][−231353−13]= \begin{bmatrix} 3125 & 32 \\ 15625 & 64 \end{bmatrix} \begin{bmatrix} -\frac{2}{3} & \frac{1}{3} \\ \frac{5}{3} & -\frac{1}{3} \end{bmatrix}
Calculate the final matrix multiplication to find M5 M^5 .

STEP 8

Continue calculating the matrix product: M5=[3125⋅−23+32⋅533125⋅13+32⋅−1315625⋅−23+64⋅5315625⋅13+64⋅−13]M^5 = \begin{bmatrix} 3125 \cdot -\frac{2}{3} + 32 \cdot \frac{5}{3} & 3125 \cdot \frac{1}{3} + 32 \cdot -\frac{1}{3} \\ 15625 \cdot -\frac{2}{3} + 64 \cdot \frac{5}{3} & 15625 \cdot \frac{1}{3} + 64 \cdot -\frac{1}{3} \end{bmatrix}
Simplify each element to find the final form of M5 M^5 .

STEP 9

Using M5 M^5 and the initial vector [a1a2]=[−11] \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} , compute: [a6a7]=M5[−11]\begin{bmatrix} a_6 \\ a_7 \end{bmatrix} = M^5 \begin{bmatrix} -1 \\ 1 \end{bmatrix}
Calculate the resulting vector to find a6 a_6 .

STEP 10

Develop an explicit formula for an a_n using: an=[10]Mn−1[a1a2]a_n = \begin{bmatrix} 1 & 0 \end{bmatrix} M^{n-1} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}
Substitute Mn−1=PDn−1P−1 M^{n-1} = P D^{n-1} P^{-1} to express an a_n in terms of the diagonalized form.
The explicit formula for an a_n is derived using the diagonalization approach.

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