Transcription for Educational Website:---**Problem Statement:**Let \(\{x_n\}\) be a sequence such that \[ x_n = 5x_{n-1} + 6x_{n-2}, \]with initial conditions \( x_1 = -2 \) and \( x_0 = 4 \).**Task:**Use the power of a suitable matrix to solve for \( x_n \).---**Detailed Explanation:**The problem involves finding a sequence \(\{x_n\}\) defined by a linear recurrence relation with constant coefficients. The goal is to express \(x_n\) in terms of matrix operations.1. **Define a Matrix Representation:** To use matrices, set up a vector that can express the relation using matrix multiplication. Define the vector: \[ \mathbf{v}_n = \begin{bmatrix} x_n \\ x_{n-1} \end{bmatrix} \] The recurrence relation can be rewritten in matrix form: \[ \mathbf{v}_{n} = \begin{bmatrix} 5 & 6 \\ 1 & 0 \end{bmatrix} \mathbf{v}_{n-1} \] Let \(\mathbf{A} = \begin{bmatrix} 5 & 6 \\ 1 & 0 \end{bmatrix}\).2. **Initial Conditions:** Given: \[ \mathbf{v}_1 = \begin{bmatrix} x_1 \\ x_0 \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \end{bmatrix} \]3. **Solution Using Matrix Powers:** The general solution for \(\mathbf{v}_n\) can be expressed as: \[ \mathbf{v}_n = \mathbf{A}^{n-1} \cdot \mathbf{v}_1 \] Calculate \(\mathbf{A}^{n-1}\) and multiply it by \(\mathbf{v}_1\) to find \(x_n\). This matrix approach allows for efficient computation of terms in the sequence by leveraging the power of matrices and linear algebra.---This allows students to explore the interplay between linear algebra and recurrence relations, highlighting the practical applications of matrices in solving complex

The sequence is xn=5xn-1+6xn-2 and initial terms x0=4 and x1=-2. Define the vector yn=(xn-1, xn). Then, we also know that yn-1=(xn-2, xn−1) As we have, x2=5x1+6x0=5-2+64=14x3=5x2+6x1=514+6-2=58x4=5x3+6x2=558+614=374 y1=(x0, x1)=4, -2, y2=(x1, x2)=-2, 14, y3=(x2, x3)=14, 58, y4=(x3, x4)=58, 374. Then, we find a matrix A such that yn=Ayn-1xn-1xn=a11a12a21a22xn-2xn-1=a11xn-2+a12xn-1a21xn-2+a22xn-1 For each row, we have xn-1=a11xn-2+a12xn-1 ⇒ a11=0 and a12=1 on comparingand xn=a21xn-2+a22xn-1 ⇒a21=6 and a22=5 on comparing with xn=5xn-1+6xn-2 Thus, the matrix is A=a11a12a21a22=0165. Therefore, xn-1xn=0165xn-2xn-1 Now, the eigenvalues are λ1=6, λ2=-1. Hence, the general solution is of the form yn=c1k1n6+c2k2n-1where the eigenvectors are k1=16, k2=-11. Hence, yn=c116n6+c2-11n-1. Use initial conditions: n=1, x0x1=c116+c2-11=4-2c1-c2=46c1+c2=-2⇒c1=27 and c2=-267 Therefore, the solution is xn-1xn=2716n6-267-11n-1