Question

Let $P = \begin{bmatrix} 7 & 2 \\ 18 & -5 \end{bmatrix}$, then $P^{20}$ is-

Answer 1

Without the options, providing a specific numerical or simplified form is difficult. The answer must be in terms of powers of the eigenvalues $\lambda_1 = 1 + 6\sqrt{2}$ and $\lambda_2 = 1 - 6\sqrt{2}$.

The final answer is $\frac{1}{6\sqrt{2}} \begin{bmatrix} (3\sqrt{2} + 3)(1 + 6\sqrt{2})^{20} + (3\sqrt{2} - 3)(1 - 6\sqrt{2})^{20} & (1 + 6\sqrt{2})^{20} - (1 - 6\sqrt{2})^{20} \\ 9((1 + 6\sqrt{2})^{20} - (1 - 6\sqrt{2})^{20}) & (3\sqrt{2} - 3)(1 + 6\sqrt{2})^{20} + (3\sqrt{2} + 3)(1 - 6\sqrt{2})^{20} \end{bmatrix}$.

Answer 2

Explanation of the solution:

1. Find the characteristic equation of the matrix $P$.
2. Use the Cayley-Hamilton theorem to express $P^2$ as a linear combination of $P$ and $I$.
3. Establish recurrence relations for the coefficients $a_n, b_n$ in $P^n = a_n P + b_n I$.
4. Solve the recurrence relation for $a_n$ using the eigenvalues of the characteristic equation.
5. Find the expressions for $a_{20}$ and $b_{20} = 71a_{19}$ (or $b_{20} = \frac{\lambda_1^{20} + \lambda_2^{20}}{2} - a_{20}$).
6. Substitute $a_{20}$ and $b_{20}$ into the expression for $P^{20} = \begin{bmatrix} 7a_{20} + b_{20} & 2a_{20} \\ 18a_{20} & -5a_{20} + b_{20} \end{bmatrix}$.

Alternatively, use diagonalization $P^n = S D^n S^{-1}$ to compute $P^{20}$.

The final answer is $\frac{1}{6\sqrt{2}} \begin{bmatrix} (3\sqrt{2} + 3)(1 + 6\sqrt{2})^{20} + (3\sqrt{2} - 3)(1 - 6\sqrt{2})^{20} & (1 + 6\sqrt{2})^{20} - (1 - 6\sqrt{2})^{20} \\ 9((1 + 6\sqrt{2})^{20} - (1 - 6\sqrt{2})^{20}) & (3\sqrt{2} - 3)(1 + 6\sqrt{2})^{20} + (3\sqrt{2} + 3)(1 - 6\sqrt{2})^{20} \end{bmatrix}$.