\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 $$ A = \begin{pmatrix} 4 & -3 \ 2 & -1 \end{pmatrix} $$, we need to find the eigenvalues and eigenvectors of $$ A $$.

### Step 1: Find the Eigenvalues

The eigenvalues $$ \lambda $$ are found by solving the characteristic polynomial, which is given by:

$$
\det(A - \lambda I) = 0
$$

where $$ I $$ is the identity matrix. We compute $$ A - \lambda I $$:

$$
A - \lambda I = \begin{pmatrix} 4 - \lambda & -3 \ 2 & -1 - \lambda \end{pmatrix}
$$

Now, we calculate the determinant:

$$
\det(A - \lambda I) = (4 - \lambda)(-1 - \lambda) - (-3)(2)
$$

Calculating this gives:

$$
= (4 - \lambda)(-1 - \lambda) + 6
$$
$$
= -4 - 4\lambda + \lambda + \lambda^2 + 6
$$
$$
= \lambda^2 - 3\lambda + 2
$$

Setting the characteristic polynomial to zero:

$$
\lambda^2 - 3\lambda + 2 = 0
$$

Factoring the polynomial:

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

Thus, the eigenvalues are:

$$
\lambda_1 = 1, \quad \lambda_2 = 2
$$

### Step 2: Find the Eigenvectors

**For $$ \lambda_1 = 1 $$:**

We solve $$ (A - I)v = 0 $$:

$$
A - I = \begin{pmatrix} 3 & -3 \ 2 & -2 \end{pmatrix}
$$

Setting up the equation:

$$
\begin{pmatrix} 3 & -3 \ 2 & -2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}
$$

This simplifies to:

$$
3x_1 - 3x_2 = 0 \quad \Rightarrow \quad x_1 = x_2
$$

Let $$ x_2 = t $$, then $$ x_1 = t $$. The eigenvector corresponding to $$ \lambda_1 = 1 $$ is:

$$
v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}
$$

**For $$ \lambda_2 = 2 $$:**

We solve $$ (A - 2I)v = 0 $$:

$$
A - 2I = \begin{pmatrix} 2 & -3 \ 2 & -3 \end{pmatrix}
$$

Setting up the equation:

$$
\begin{pmatrix} 2 & -3 \ 2 & -3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}
$$

This simplifies to:

$$
2x_1 - 3x_2 = 0 \quad \Rightarrow \quad 2x_1 = 3x_2 \quad \Rightarrow \quad x_1 = \frac{3}{2} x_2
$$

Let $$ x_2 = s $$, then $$ x_1 = \frac{3}{2}s $$. The eigenvector corresponding to $$ \lambda_2 = 2 $$ is:

$$
v_2 = \begin{pmatrix} \frac{3}{2} \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \end{pmatrix} \text{ (by choosing } s = 2\text{)}
$$

### Step 3: Form Matrices $$ P $$ and $$ D $$

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

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

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

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

### Step 4: Verify $$ A = PDP^{-1} $$

To find $$ P^{-1} $$, we calculate the inverse of $$ P $$:

$$
P^{-1} = \frac{1}{\det(P)} \text{adj}(P)
$$

Calculating the determinant of $$ P $$:

$$
\det(P) = (1)(2) - (1)(3) = 2 - 3 = -1
$$

The adjugate of $$ P $$ is:

$$
\text{adj}(P) = \begin{pmatrix} 2 & -3 \ -1 & 1 \end{pmatrix}
$$

Thus,

$$
P^{-1} = -\begin{pmatrix} 2 & -3 \ -1 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix}
$$

Finally, we can verify:

$$
A = P D P^{-1}
$$

Calculating $$ P D $$:

$$
P D = \begin{pmatrix} 1 & 3 \ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 6 \ 1 & 4 \end{pmatrix}
$$

Now calculating $$ A = P D P^{-1} $$:

$$
A = \begin{pmatrix} 1 & 6 \ 1 & 4 \end{pmatrix} \begin{pmatrix} -2 & 3 \ 1 & -1 \end{pmatrix} = \begin{pmatrix} 4 & -3 \ 2 & -1 \end{pmatrix}
$$

Thus, we have confirmed that:

$$
A = P D P^{-1}
$$

### Final Result

The matrices are:

$$
P = \begin{pmatrix} 1 & 3 \ 1 & 2 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \ 0 & 2 \end{pmatrix}
$$
The matrices $$ P $$ and $$ D $$ for the diagonalization of matrix $$ A $$ are: $$ P = \begin{pmatrix} 1 & 3 \ 1 & 2 \end{pmatrix}, \quad D = \begin{pmatrix} 1 & 0 \ 0 & 2 \end{pmatrix} $$

\begin{tabular}{l} \#3.) Diagonalize the matrix A , by finding matrix D and matrix P , where \\ \( \qquad A=\left[\begin{array}{cc}4 & -3 \\ 2 & -1\end{array}\right] \) \\ \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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