Question

The characteristic polynomial of a third order matrix \( A \) is given by \[ f(x) = x^3 - 6x^2 + 11x - 6. \] If one of the eigenvalues of \( A \) is 1, then the sum of the other two eigenvalues is:

Hint:

The sum of all eigenvalues of a matrix is equal to the trace (sum of diagonal elements) of the matrix. Eigenvalues are the roots of the characteristic polynomial.

Answer 1

Step 1: Use the property of characteristic polynomials. 
The characteristic polynomial of a matrix gives the eigenvalues as the roots of the polynomial. Given: \[ f(x) = x^3 - 6x^2 + 11x - 6. \] This polynomial can be factored. 
Step 2: Use the given eigenvalue to factor the polynomial. 
Since one of the eigenvalues is 1, then \( (x - 1) \) is a factor of the polynomial. Using polynomial division or factorization: \[ f(x) = (x - 1)(x^2 - 5x + 6). \] Step 3: Factor the quadratic term. \[ x^2 - 5x + 6 = (x - 2)(x - 3). \] So the complete factorization is: \[ f(x) = (x - 1)(x - 2)(x - 3). \] Step 4: Identify eigenvalues. 
The eigenvalues of matrix \( A \) are \( 1, 2, 3 \). Since one eigenvalue is given as 1, the other two are 2 and 3. 
Step 5: Find the sum of the other two eigenvalues. \[ 2 + 3 = 5. \]