To find the formula for the entries of $A^k$ (the $k$-th power of $A$), where $A$ is diagonalizable, we proceed as follows:

---

Step 1: Diagonalization of $A$

The matrix $A$ can be diagonalized if it can be written as: $A = P D P^{-1}$ where:

• $D$ is the diagonal matrix containing the eigenvalues of $A$.
• $P$ is the matrix whose columns are the eigenvectors of $A$.

Then: $A^k = P D^k P^{-1}$

---

Step 2: Find the Eigenvalues of $A$

The eigenvalues of $A$ are the roots of the characteristic polynomial: $\det(A - \lambda I) = 0$ For $A = \begin{bmatrix} 6 & -8 \\ 4 & -6 \end{bmatrix}$, we compute: $\det\begin{bmatrix} 6-\lambda & -8 \\ 4 & -6-\lambda \end{bmatrix} = (6-\lambda)(-6-\lambda) - (-8)(4)$ $= \lambda^2 - 0\lambda - 100$ $= \lambda^2 - 100$ The eigenvalues are: $\lambda = 10, \quad \lambda = -10$

---

Step 3: Find $D^k$

The diagonal matrix $D$ is: $D = \begin{bmatrix} 10 & 0 \\ 0 & -10 \end{bmatrix}$ Raising $D$ to the $k$-th power is straightforward: $D^k = \begin{bmatrix} 10^k & 0 \\ 0 & (-10)^k \end{bmatrix}$

---

Step 4: Find $P$ and $P^{-1}$

The eigenvectors corresponding to $\lambda = 10$ and $\lambda = -10$ are:

1. For $\lambda = 10$: Solve $(A - 10I)x = 0$: $\begin{bmatrix} -4 & -8 \\ 4 & -16 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$ A solution is $x = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$.

2. For $\lambda = -10$: Solve $(A + 10I)x = 0$: $\begin{bmatrix} 16 & -8 \\ 4 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$ A solution is $x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.

Thus: $P = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}, \quad P^{-1} = \frac{1}{5} \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}$

---

Step 5: Compute $A^k$

Using $A^k = P D^k P^{-1}$, we substitute:

\begin{bmatrix} 10^k & 0 \\ 0 & (-10)^k \end{bmatrix} \frac{1}{5} \begin{bmatrix} 2 & -1 \\ 1 & 2 \end{bmatrix}$$ Perform the multiplications to find the general formula for $$A^k$$. For brevity: $$A^k = \frac{1}{5} \begin{bmatrix} 2 \cdot 10^k + (-10)^k & -10^k + 2 \cdot (-10)^k \\ -2 \cdot 10^k + 10^k & 10^k + 2 \cdot (-10)^k \end{bmatrix}$$ --- ### Final Answer: The entries of $$A^k$$ are: $$A^k = \frac{1}{5} \begin{bmatrix} 2 \cdot 10^k + (-10)^k & -10^k + 2 \cdot (-10)^k \\ -2 \cdot 10^k + 10^k & 10^k + 2 \cdot (-10)^k \end{bmatrix}.$$ Let me know if you'd like to see the detailed computation for the matrix multiplications or have further questions! --- ### Related Questions: 1. What is the significance of diagonalizability in matrix powers? 2. How do eigenvalues affect the behavior of $$A^k$$ as $$k \to \infty$$? 3. Can $$A$$ be diagonalized if its eigenvalues are complex? 4. What happens if $$A$$ is not diagonalizable? How do we compute $$A^k$$? 5. How does the determinant of $$A$$ relate to the product of eigenvalues? --- ### Tip: Diagonalization simplifies the computation of large matrix powers, especially when the eigenvalues have distinct patterns (e.g., repeated or zero eigenvalues).