To find the characteristic polynomial and eigenvalues of the matrix $A$, we need to follow these steps:

1. Compute the characteristic polynomial: The characteristic polynomial of a matrix $A$ is given by $\det(A - \lambda I)$, where $\lambda$ is a scalar and $I$ is the identity matrix.

2. Solve for the eigenvalues: The eigenvalues are the roots of the characteristic polynomial.

Given the matrix $A = \begin{bmatrix} 2 & -7 \\ -6 & 3 \end{bmatrix}$:

Step 1: Compute the Characteristic Polynomial

The characteristic polynomial is found by calculating the determinant of $A - \lambda I$:

$A - \lambda I = \begin{bmatrix} 2 - \lambda & -7 \\ -6 & 3 - \lambda \end{bmatrix}$

The determinant of this matrix is:

$\det(A - \lambda I) = (2 - \lambda)(3 - \lambda) - (-7)(-6)$

$= (2 - \lambda)(3 - \lambda) - 42$

$= 6 - 5\lambda + \lambda^2 - 42$

$= \lambda^2 - 5\lambda - 36$

So, the characteristic polynomial is:

$\lambda^2 - 5\lambda - 36$

Step 2: Solve for the Eigenvalues

To find the eigenvalues, we solve the characteristic polynomial:

$\lambda^2 - 5\lambda - 36 = 0$

Using the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -5$, and $c = -36$:

$\lambda = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-36)}}{2(1)}$

$= \frac{5 \pm \sqrt{25 + 144}}{2}$

$= \frac{5 \pm \sqrt{169}}{2}$

$= \frac{5 \pm 13}{2}$

Thus, the eigenvalues are:

$\lambda_1 = \frac{5 + 13}{2} = 9$ $\lambda_2 = \frac{5 - 13}{2} = -4$

Eigenvalues

The eigenvalues of the matrix $A$ are $\lambda_1 = 9$ and $\lambda_2 = -4$.

If you need any further details or have questions, feel free to ask!

Here are 8 related questions to explore further:

1. How do you find the eigenvectors corresponding to each eigenvalue?
2. What is the geometric interpretation of eigenvalues and eigenvectors?
3. How can you verify if a given vector is an eigenvector of a matrix?
4. What are the properties of diagonalizable matrices?
5. How do you compute the determinant of a matrix?
6. How does the trace of a matrix relate to its eigenvalues?
7. What is the significance of the characteristic polynomial in linear algebra?
8. How do eigenvalues affect the stability of a system of differential equations?

Tip: To find the eigenvectors, substitute each eigenvalue back into the equation $(A - \lambda I)v = 0$ and solve for the vector $v$.