\begin{tabular}{l} \#3.) Diagonalize the matrix A , by finding matrix D and matrix P , where \ $$ \qquad A=\left[\begin{array}{cc}4 & -3 \ 2 & -1\end{array} ight] $$ \ \hline Note: We want to express A as $$ A=P D P^{-1} \quad $$, where $$ D $$ is a diagonal matrix. \end{tabular}

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\begin{tabular}{l} \#3.) Diagonalize the matrix A , by finding matrix D and matrix P , where \ $$ \qquad A=\left[\begin{array}{cc}4 & -3 \ 2 & -1\end{array}ight] $$ \ \hline Note: We want to express A as $$ A=P D P^{-1} \quad $$, where $$ D $$ is a diagonal matrix. \end{tabular}

UpStudy AI · Accepted Answer

To diagonalize the matrix $$ A $$, we need to find a matrix $$ P $$ whose columns are the eigenvectors of $$ A $$, and a diagonal matrix $$ D $$ whose diagonal elements are the corresponding eigenvalues. The process involves the following steps:

1. Find the eigenvalues of $$ A $$ by solving the characteristic equation $$ \det(A - \lambda I) = 0 $$.
2. For each eigenvalue, find the eigenvectors by solving the equation $$ (A - \lambda I)v = 0 $$.
3. Form the matrix $$ P $$ with the eigenvectors as columns.
4. Form the diagonal matrix $$ D $$ with the eigenvalues on the diagonal.

Let's start with step 1:

### Step 1: Find the eigenvalues

The characteristic equation is given by:

$$
\det(A - \lambda I) = \det\left[\begin{array}{cc}4 & -3 \ 2 & -1\end{array}\right] - \lambda \det\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}\right] = 0
$$

This simplifies to:

$$
\det\left[\begin{array}{cc}4 - \lambda & -3 \ 2 & -1 - \lambda\end{array}\right] = (4 - \lambda)(-1 - \lambda) - (2)(-3) = 0
$$

Expanding the determinant, we get:

$$
\lambda^2 - 3\lambda - 4 + 6 = 0 \implies \lambda^2 - 3\lambda + 2 = 0
$$

This quadratic equation factors as:

$$
(\lambda - 1)(\lambda - 2) = 0
$$

So the eigenvalues are $$ \lambda_1 = 1 $$ and $$ \lambda_2 = 2 $$.

### Step 2: Find the eigenvectors

For $$ \lambda_1 = 1 $$:

$$
(A - \lambda_1 I)v = \left[\begin{array}{cc}4 - 1 & -3 \ 2 & -1 - 1\end{array}\right]v = \left[\begin{array}{cc}3 & -3 \ 2 & -2\end{array}\right]v = 0
$$

This gives us the system of equations:

$$
\begin{cases}
3x - 3y = 0 \
2x - 2y = 0
\end{cases}
$$

Both equations are the same, so we can choose $$ x = 1 $$ and $$ y = 1 $$. Thus, an eigenvector for $$ \lambda_1 = 1 $$ is $$ v_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} $$.

For $$ \lambda_2 = 2 $$:

$$
(A - \lambda_2 I)v = \left[\begin{array}{cc}4 - 2 & -3 \ 2 & -1 - 2\end{array}\right]v = \left[\begin{array}{cc}2 & -3 \ 2 & -3\end{array}\right]v = 0
$$

This gives us the system of equations:

$$
\begin{cases}
2x - 3y = 0 \
2x - 3y = 0
\end{cases}
$$

Again, both equations are the same, so we can choose $$ x = 3 $$ and $$ y = 2 $$. Thus, an eigenvector for $$ \lambda_2 = 2 $$ is $$ v_2 = \begin{bmatrix} 3 \ 2 \end{bmatrix} $$.

### Step 3: Form the matrix $$ P $$

The matrix $$ P $$ is formed by the eigenvectors as columns:

$$
P = \begin{bmatrix} 1 & 3 \ 1 & 2 \end{bmatrix}
$$

### Step 4: Form the diagonal matrix $$ D $$

The diagonal matrix $$ D $$ is formed by the eigenvalues on the diagonal:

$$
D = \begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix}
$$

### Step 5: Find $$ P^{-1} $$

To diagonalize $$ A $$, we need to find $$ P^{-1} $$. The inverse of $$ P $$ is:

\[
P^{-1} = \frac{1}{\text{det}(P)} \begin{bmatrix} 2 & -3 \ -1 & 1 \end{bmatrix
To diagonalize matrix $$ A $$, find the eigenvectors and eigenvalues, form matrix $$ P $$ with eigenvectors, and diagonal matrix $$ D $$ with eigenvalues. Then, find $$ P^{-1} $$.

#3.) Diagonalize the matrix A , by finding matrix D and matrix P , where

Note: We want to express A as , where is a diagonal matrix.

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#3.) Diagonalize the matrix A , by finding matrix D and matrix P , where Note: We want to express A as , where is a diagonal matrix.