If `A=[(2,2,1),(1,3,1),(1,2,2)]` and the sum of eigen values of A is m anda product of eigen values of A is n, then m+n is equal to

Options

A. 10
B. 12
C. 14
D. 16

Answer

To solve the problem, we need to find the eigenvalues of the matrix \( A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix} \). The sum of the eigenvalues is denoted as \( m \) and the product of the eigenvalues as \( n \). We will then calculate \( m + n \).

### Step 1: Write the characteristic polynomial
To find the eigenvalues, we first need to compute the characteristic polynomial, which is given by the determinant of \( A - \lambda I \), where \( I \) is the identity matrix.

\[ A - \lambda I = \begin{pmatrix} 2 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 2 - \lambda \end{pmatrix} \]

### Step 2: Calculate the determinant
We need to calculate the determinant of the matrix \( A - \lambda I \):

\[ \text{det}(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 2 - \lambda \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix, we can expand this determinant:

\[ = (2 - \lambda) \begin{vmatrix} 3 - \lambda & 1 \\ 2 & 2 - \lambda \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ 1 & 2 - \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 3 - \lambda \\ 1 & 2 \end{vmatrix} \]

Calculating each of these \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} 3 - \lambda & 1 \\ 2 & 2 - \lambda \end{vmatrix} = (3 - \lambda)(2 - \lambda) - 2 = (6 - 5\lambda + \lambda^2 - 2) = \lambda^2 - 5\lambda + 4 \)

2. \( \begin{vmatrix} 1 & 1 \\ 1 & 2 - \lambda \end{vmatrix} = (1)(2 - \lambda) - (1)(1) = 2 - \lambda - 1 = 1 - \lambda \)

3. \( \begin{vmatrix} 1 & 3 - \lambda \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(3 - \lambda) = 2 - (3 - \lambda) = \lambda - 1 \)

Substituting these back into the determinant:

\[ \text{det}(A - \lambda I) = (2 - \lambda)(\lambda^2 - 5\lambda + 4) - 2(1 - \lambda) + (\lambda - 1) \]

### Step 3: Expand and simplify
Expanding the expression:

\[ = (2 - \lambda)(\lambda^2 - 5\lambda + 4) - 2 + 2\lambda + \lambda - 1 \]

After simplification, we will arrive at a cubic polynomial in \( \lambda \).

### Step 4: Find the eigenvalues
Once we have the cubic polynomial, we can find the eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \) using Vieta's formulas:

- The sum of the eigenvalues \( m = \lambda_1 + \lambda_2 + \lambda_3 \) is equal to the coefficient of \( \lambda^2 \) term (with a negative sign).
- The product of the eigenvalues \( n = \lambda_1 \cdot \lambda_2 \cdot \lambda_3 \) is equal to the constant term (with a sign depending on the degree of the polynomial).

### Step 5: Calculate \( m + n \)
After finding \( m \) and \( n \), we simply add them together to find \( m + n \).

### Conclusion
From the calculations, we find that:
- \( m = 7 \)
- \( n = 5 \)

Thus, \( m + n = 7 + 5 = 12 \).