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Let's solve the problem for the matrix \( A = \begin{pmatrix} 8 & -5 \\ 10 & -7 \end{pmatrix} \) to find the diagonal matrix \( D \) such that \( A = PDP^{-1} \), where \( P \) is an invertible matrix.

### Step 1: Find the Eigenvalues
The diagonal elements of \( D \) will be the eigenvalues of \( A \). To find them, we solve the characteristic equation:
\[
\text{det}(A - \lambda I) = 0
\]
where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. The matrix \( A - \lambda I \) is:
\[
A - \lambda I = \begin{pmatrix} 8 & -5 \\ 10 & -7 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 8 - \lambda & -5 \\ 10 & -7 - \lambda \end{pmatrix}
\]
The determinant of this matrix is:
\[
\text{det}\left(\begin{pmatrix} 8 - \lambda & -5 \\ 10 & -7 - \lambda \end{pmatrix}\right) = (8 - \lambda)(-7 - \lambda) - (-5)(10)
\]
Expanding this determinant:
\[
(8 - \lambda)(-7 - \lambda) + 50 = \lambda^2 - \lambda - 56 + 50 = \lambda^2 - \lambda - 6
\]
So, the characteristic equation is:
\[
\lambda^2 - \lambda - 6 = 0
\]

### Step 2: Solve the Quadratic Equation
We solve the quadratic equation:
\[
\lambda^2 - \lambda - 6 = 0
\]
Using the quadratic formula:
\[
\lambda = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2}
\]
The solutions are:
\[
\lambda_1 = \frac{6}{2} = 3 \quad \text{and} \quad \lambda_2 = \frac{-4}{2} = -2
\]

### Step 3: Form the Diagonal Matrix \(D\)
The diagonal matrix \( D \) will have these eigenvalues on its diagonal:
\[
D = \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}
\]
Thus:
- \( a = 3 \)
- \( d = -2 \)

### Final Answer
- \( a = 3 \)
- \( d = -2 \)

Would you like more details on any part of this solution? Here are some related questions you might find useful:

1. How do you compute the eigenvectors for this matrix \(A\)?
2. How does the matrix \(P\) relate to the eigenvectors?
3. What properties must a matrix have to be diagonalizable?
4. Can you always find a diagonal matrix for any square matrix?
5. How is diagonalization used in solving differential equations?

**Tip:** When working with eigenvalues and eigenvectors, it is helpful to check if the matrix is symmetric, as symmetric matrices have real eigenvalues and are always diagonalizable.

Learn how to diagonalize the matrix A = [[8, -5], [10, -7]] using eigenvalues and eigenvectors. This problem covers matrix algebra concepts, including characteristic equations and diagonalization theorems. Suitable for undergraduate students.

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