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 sum and product of the eigenvalues of the matrix \( A \) given as:

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

### Step 1: Find the Characteristic Polynomial
The eigenvalues of a matrix can be found by solving the characteristic polynomial, which is given by:

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

Where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue.

### Step 2: Set Up the Matrix \( A - \lambda I \)
The identity matrix \( I \) for a \( 3 \times 3 \) matrix is:

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Thus,

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

### Step 3: Calculate the Determinant
Now we need to calculate the determinant of \( A - \lambda I \):

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

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

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

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

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

### Step 4: Substitute Back into the Determinant
Now substituting these back into the determinant:

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

Expanding this:

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

### Step 5: Collect Like Terms
After expanding and simplifying, we will get a cubic polynomial in \( \lambda \):

\[ -\lambda^3 + 7\lambda^2 - 11\lambda + 5 = 0 \]

### Step 6: Find the Sum and Product of Eigenvalues
From Vieta's formulas, we know:

- The sum of the roots (eigenvalues) \( m = -\frac{b}{a} = 7 \)
- The product of the roots (eigenvalues) \( n = -\frac{d}{a} = 5 \)

### Step 7: Calculate \( m + n \)
Finally, we find:

\[ m + n = 7 + 5 = 12 \]

### Final Answer
Thus, the value of \( m + n \) is:

\[ \boxed{12} \]