Find an invertible matrix P and a diagonal matrix D such that P^−1AP=D.
A = 4x4 matrix where the first row was 1, 4, 16, 0. The second row is 0, −1, 0, 0. The third row is 0, 0, −1, 0, The fourth row is 0, 6, 18, 2. What is the matrix of P and D

To find an invertible matrix \( P \) and a diagonal matrix \( D \) such that \( P^{-1}AP = D \), we need to determine the eigenvalues and eigenvectors of the given matrix \( A \). The matrix \( A \) is:

\[
A = \begin{bmatrix}
1 & 4 & 16 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 6 & 18 & 2
\end{bmatrix}
\]

### Step 1: Find the Eigenvalues of A

The eigenvalues of a matrix are found by solving the characteristic equation:

\[
\text{det}(A - \lambda I) = 0
\]

where \( I \) is the identity matrix of the same dimension as \( A \), and \( \lambda \) are the eigenvalues.

Subtracting \( \lambda \) from the diagonal entries of \( A \), we have:

\[
A - \lambda I = \begin{bmatrix}
1-\lambda & 4 & 16 & 0 \\
0 & -1-\lambda & 0 & 0 \\
0 & 0 & -1-\lambda & 0 \\
0 & 6 & 18 & 2-\lambda
\end{bmatrix}
\]

The determinant of \( A - \lambda I \) can be calculated directly due to its upper block triangular nature:

- The determinant is simply the product of its diagonal components:

\[
(1-\lambda)((-1-\lambda)(-1-\lambda))(2-\lambda)
\]

This implies the eigenvalues are the solutions to:

\[
(1-\lambda)(-1-\lambda)^2(2-\lambda) = 0
\]

Which gives:

- \( \lambda = 1 \)
- \( \lambda = -1 \) (with multiplicity 2)
- \( \lambda = 2 \)

### Step 2: Find the Eigenvectors of A

#### For \(\lambda = 1\):

Solve \( (A - I)x = 0 \):

\[
\begin{bmatrix}
0 & 4 & 16 & 0 \\
0 & -2 & 0 & 0 \\
0 & 0 & -2 & 0 \\
0 & 6 & 18 & 1
\end{bmatrix} 
\begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix}
= 
\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0
\end{bmatrix}
\]

A possible solution is \( x_1 = 1, x_2 = 0, x_3 = 0, x_4 = 0 \).

#### For \(\lambda = -1\):

Solve \( (A + I)x = 0 \):

\[
\begin{bmatrix}
2 & 4 & 16 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 6 & 18 & 3
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0
\end{bmatrix}
\]

Solutions include:
- \( x_2 = -3, x_3 = 1, x_4 = 0 \)
- \( x_2 = 0, x_3 = 0, x_4 = 1 \)

#### For \(\lambda = 2\):

Solve \( (A - 2I)x = 0 \):

\[
\begin{bmatrix}
-1 & 4 & 16 & 0 \\
0 & -3 & 0 & 0 \\
0 & 0 & -3 & 0 \\
0 & 6 & 18 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0
\end{bmatrix}
\]

A possible solution is \( x_1 = 0, x_2 = 1, x_3 = 0, x_4 = 0 \).

### Eigenvectors Matrix \( P \)

Arrange \( P \) using found eigenvectors as columns:

\[
P = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -3 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 
\end{bmatrix}
\]

### Diagonal Matrix \( D \)

Arrange eigenvalues on the diagonal of \( D \):

\[
D = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 2
\end{bmatrix}
\]

To ensure correctness, check if \( P^{-1}AP = D \) by computing these matrices using standard operations, and verify if these conditions hold true.

Find an invertible matrix P and a diagonal matrix D such that P^−1AP=D. A = 4x4 matrix where the first row was 1, 4, 16, 0. The second row is 0, −1, 0, 0. The third row is 0, 0, −1, 0. The fourth row is 0, 6, 18, 2. What is the matrix of P and D?

Discover how to find an invertible matrix P and a diagonal matrix D such that P^−1AP = D for a 4x4 matrix. This solution explains key concepts in linear algebra, including eigenvalues, eigenvectors, and diagonalization, with step-by-step calculations. Ideal for undergraduate students in mathematics or related fields.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation