find the eigenvalues and eigenvectors of the matrix

A=[3105−2−3−4357],A = \begin{bmatrix} 3 & 10 & 5 \\ -2 & -3 & -4 \\ 3 & 5 & 7 \end{bmatrix}, A=​3−23​10−35​5−47​​,

Let's find the eigenvalues and eigenvectors of the matrix \( A = \begin{bmatrix} 3 & 10 & 5 \\ -2 & -3 & -4 \\ 3 & 5 & 7 \end{bmatrix} \).

### Step 1: Find the Eigenvalues
The eigenvalues \(\lambda\) are found by solving the characteristic equation, which is obtained from the determinant of the matrix \( A - \lambda I \), where \( I \) is the identity matrix.

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

So, let's calculate \( A - \lambda I \):

\[
A - \lambda I = \begin{bmatrix} 3 - \lambda & 10 & 5 \\ -2 & -3 - \lambda & -4 \\ 3 & 5 & 7 - \lambda \end{bmatrix}
\]

Now, find the determinant of this matrix:

\[
\text{det}(A - \lambda I) = \begin{vmatrix} 3 - \lambda & 10 & 5 \\ -2 & -3 - \lambda & -4 \\ 3 & 5 & 7 - \lambda \end{vmatrix}
\]

### Step 2: Solve the Determinant
To find the determinant, we'll expand along the first row:

\[
\text{det}(A - \lambda I) = (3-\lambda) \begin{vmatrix} -3-\lambda & -4 \\ 5 & 7-\lambda \end{vmatrix} - 10 \begin{vmatrix} -2 & -4 \\ 3 & 7-\lambda \end{vmatrix} + 5 \begin{vmatrix} -2 & -3-\lambda \\ 3 & 5 \end{vmatrix}
\]

Let's calculate these minors:

1. \(\begin{vmatrix} -3-\lambda & -4 \\ 5 & 7-\lambda \end{vmatrix} = (-3-\lambda)(7-\lambda) - (-4)(5)\)

2. \(\begin{vmatrix} -2 & -4 \\ 3 & 7-\lambda \end{vmatrix} = (-2)(7-\lambda) - (-4)(3)\)

3. \(\begin{vmatrix} -2 & -3-\lambda \\ 3 & 5 \end{vmatrix} = (-2)(5) - (-3-\lambda)(3)\)

After calculating the determinant and setting it to zero, we'll solve the resulting polynomial equation to find the eigenvalues \(\lambda\).

### Step 3: Find the Eigenvectors
For each eigenvalue \(\lambda\), solve the equation \((A - \lambda I)\mathbf{v} = \mathbf{0}\) to find the corresponding eigenvector \(\mathbf{v}\).

I'll calculate this step by step.import sympy as sp

# Define the matrix and the variable lambda
lambda_ = sp.symbols('lambda')
A = sp.Matrix([[3, 10, 5], [-2, -3, -4], [3, 5, 7]])
I = sp.eye(3)  # Identity matrix

# Compute the characteristic equation: det(A - lambda*I) = 0
char_eq = (A - lambda_ * I).det()

# Solve the characteristic equation for eigenvalues
eigenvalues = sp.solve(char_eq, lambda_)

# Now, let's compute the eigenvectors for each eigenvalue
eigenvectors = [A - eig * I for eig in eigenvalues]

# Solve for the eigenvectors corresponding to each eigenvalue
eigenvectors_solutions = [eig.nullspace() for eig in eigenvectors]

eigenvalues, eigenvectors_solutionsThe eigenvalues and corresponding eigenvectors of the matrix \( A = \begin{bmatrix} 3 & 10 & 5 \\ -2 & -3 & -4 \\ 3 & 5 & 7 \end{bmatrix} \) are as follows:

### Eigenvalues:
1. \(\lambda_1 = 2\)
2. \(\lambda_2 = 3\)

### Eigenvectors:
1. For \(\lambda_1 = 2\), the eigenvector is:
   \[
   \mathbf{v_1} = \begin{bmatrix} -1 \\ -\frac{2}{5} \\ 1 \end{bmatrix}
   \]

2. For \(\lambda_2 = 3\), the eigenvector is:
   \[
   \mathbf{v_2} = \begin{bmatrix} -\frac{1}{2} \\ -\frac{1}{2} \\ 1 \end{bmatrix}
   \]

Would you like more details or have any further questions?

### Follow-up Questions:
1. How would you normalize the eigenvectors?
2. Can you explain the geometric significance of eigenvalues and eigenvectors?
3. What happens if a matrix has repeated eigenvalues?
4. How are eigenvectors used in diagonalizing a matrix?
5. Can the eigenvectors be complex for real matrices?

### Tip:
Eigenvectors corresponding to distinct eigenvalues are linearly independent, which is useful for diagonalizing matrices.

Learn how to find eigenvalues and eigenvectors of the matrix A = [3 10 5; -2 -3 -4; 3 5 7]. Step-by-step solution including characteristic equation, determinant calculation, and eigenvector computation. Suitable for undergraduate level students studying linear algebra.

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