**3. This matrix has repeated eigenvalues for which value(s) of $a$? Show your work.**Given the matrix $M$:\[ M = \begin{bmatrix} a & -1 \\ 4 & 4 \end{bmatrix} \](Show the matrix $M$ above your solution.)---To find the value(s) of $a$ for which the matrix $M$ has repeated eigenvalues, let's follow these steps:1. **Write the characteristic equation of the matrix $M$.** The characteristic equation for a matrix $M$ is given by the determinant of $(M - \lambda I) = 0$, where $\lambda$ is an eigenvalue and $I$ is the identity matrix:\[ \text{det} \left( \begin{bmatrix} a & -1 \\ 4 & 4 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = 0 \]2. **Subtract $\lambda$ times the identity matrix from matrix $M$:**\[ M - \lambda I = \begin{bmatrix} a & -1 \\ 4 & 4 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} a - \lambda & -1 \\ 4 & 4 - \lambda \end{bmatrix} \]3. **Calculate the determinant of the resulting matrix:**\[ \text{det} \left( \begin{bmatrix} a - \lambda & -1 \\ 4 & 4 - \lambda \end{bmatrix} \right) = (a - \lambda)(4 - \lambda) - (-1 \cdot 4) \]4. **Simplify the equation:**\[ (a - \lambda)(4 - \lambda) + 4 = a(4 - \lambda) - \lambda (4 - \lambda) + 4 \]\[ = 4a - a\lambda - 4\lambda + \lambda^2 + 4

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Answered: 3. This matrix has repeated eigenvalues…

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3. This matrix has repeated eigenvalues for which value(s) of a? Show your wor -1] a M 4

Linear Algebra: A Modern Introduction

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ISBN:9781285463247

Author:David Poole

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Chapter2: Systems Of Linear Equations

Section2.2: Direct Methods For Solving Linear Systems

Problem 3CEXP

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Question

**3. This matrix has repeated eigenvalues for which value(s) of $a$? Show your work.**

Given the matrix $M$:

\[
M = \begin{bmatrix}
a & -1 \\
4 & 4
\end{bmatrix}
\]

(Show the matrix $M$ above your solution.)

---

To find the value(s) of $a$ for which the matrix $M$ has repeated eigenvalues, let's follow these steps:

1. **Write the characteristic equation of the matrix $M$.**

The characteristic equation for a matrix $M$ is given by the determinant of $(M - \lambda I) = 0$, where $\lambda$ is an eigenvalue and $I$ is the identity matrix:

\[
\text{det} \left( \begin{bmatrix}
a & -1 \\
4 & 4
\end{bmatrix} - \lambda \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \right) = 0
\]

2. **Subtract $\lambda$ times the identity matrix from matrix $M$:**

\[
M - \lambda I = \begin{bmatrix}
a & -1 \\
4 & 4
\end{bmatrix} - \begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix} = \begin{bmatrix}
a - \lambda & -1 \\
4 & 4 - \lambda
\end{bmatrix}
\]

3. **Calculate the determinant of the resulting matrix:**

\[
\text{det} \left( \begin{bmatrix}
a - \lambda & -1 \\
4 & 4 - \lambda
\end{bmatrix} \right) = (a - \lambda)(4 - \lambda) - (-1 \cdot 4)
\]

4. **Simplify the equation:**

\[
(a - \lambda)(4 - \lambda) + 4 = a(4 - \lambda) - \lambda (4 - \lambda) + 4
\]
\[
= 4a - a\lambda - 4\lambda + \lambda^2 + 4

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