Consider the following matrix.A = 23 12 −36 −19Find the eigenvalues and associated eigenvectors of A. (Arrange the eigenvalues so that 𝜆1 < 𝜆2.)

1. Break Down the Problem

To find the eigenvalues and associated eigenvectors of the matrix $A$, we will:

1. Calculate the characteristic polynomial by finding the determinant of $A - \lambda I$.
2. Solve for the eigenvalues $\lambda$.
3. For each eigenvalue, solve for the corresponding eigenvector.

2. Relevant Concepts

• The characteristic polynomial of a matrix $A$ is given by $\det(A - \lambda I) = 0$.
• Eigenvalues are the solutions $\lambda$ to this determinant equation.
• Eigenvectors $v$ corresponding to each eigenvalue can be obtained by solving $(A - \lambda I)v = 0$.

3. Analysis and Detail

Given the matrix: $A = \begin{pmatrix} 23 & 12 \\ -36 & -19 \end{pmatrix}$

Step 3.1: Find the Characteristic Polynomial

We first find $A - \lambda I$: $A - \lambda I = \begin{pmatrix} 23 - \lambda & 12 \\ -36 & -19 - \lambda \end{pmatrix}$

Now, calculate the determinant: $\det(A - \lambda I) = (23 - \lambda)(-19 - \lambda) - (12)(-36)$

Calculating this gives: $= (\lambda - 23)(\lambda + 19) + 432$ $= \lambda^2 - 4\lambda - 437$

Step 3.2: Solve for Eigenvalues $\lambda$

Set the characteristic polynomial to zero: $\lambda^2 - 4\lambda - 437 = 0$

Using the quadratic formula: $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1, b = -4, c = -437$: $\lambda = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-437)}}{2 \cdot 1}$ $= \frac{4 \pm \sqrt{16 + 1748}}{2}$ $= \frac{4 \pm \sqrt{1764}}{2}$ $= \frac{4 \pm 42}{2}$ Thus, we have two eigenvalues: $\lambda_1 = \frac{46}{2} = 23, \quad \lambda_2 = \frac{-38}{2} = -19$

Step 3.3: Find Eigenvectors

For $\lambda_1 = 23$: $(A - 23I)v = 0 \implies \begin{pmatrix} 0 & 12 \\ -36 & -42 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ From the first row, we get: $12x_2 = 0 \implies x_2 = 0$ We can take $x_1 = 1$ as a free variable. So, an eigenvector corresponding to $\lambda_1$ is: $v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

For $\lambda_2 = -19$: $(A + 19I)v = 0 \implies \begin{pmatrix} 42 & 12 \\ -36 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ From the first row, we can express: $42x_1 + 12x_2 = 0 \implies x_2 = -\frac{42}{12}x_1 = -\frac{7}{2}x_1$ Let $x_1 = 2$: $v_2 = \begin{pmatrix} 2 \\ -7 \end{pmatrix}$

4. Verify and Summarize

We found and confirmed the eigenvalues:

• $\lambda_1 = -19$
• $\lambda_2 = 23$

And their corresponding eigenvectors:

• $v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ for $\lambda_1$
• $v_2 = \begin{pmatrix} 2 \\ -7 \end{pmatrix}$ for $\lambda_2$

Final Answer

Eigenvalues: $\lambda_1 = -19, \quad \lambda_2 = 23$

Eigenvectors: $v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 2 \\ -7 \end{pmatrix}$