Question

The characteristic polynomial of 3 × 3 matrix A is |A - λl| = λ³ + 3λ² + 4λ + 3. Let x = trace(A) and y = |A|, the determinant of A. Then  
1. x = 3, y = -3
2. x = y = -3
3. \(\dfrac{x}{y} = \dfrac{4}{3}\)
4. \(\dfrac{x}{y} = \dfrac{3}{4}\)

Answer 1

Correct Answer - Option 2 : x = y = -3

Concept: 

• The characteristics polynomial of an n × n matrix A is a polynomial whose roots are the eigenvalues of matrix A.
• It is defined as a determinant (A - λI) where I is the identity matrix.
• The coefficient of the polynomial is a determinant and trace of the matrix.
• For 3 × 3 matrix A, the characteristics polynomial can be found using the formula, 

|A - λl| = - λ³ - tr(A) λ² - (1/2) [ tr(A)² - tr(A²)λ ] - det (A)    -------(i)

where tr(A)tr(A) is the trace of AA and det(A)det(A) is the determinant of AA.

Calculation:

Given:

 |A - λl| = λ³ + 3λ² + 4λ + 3 

On comparing with equation, (i) we get,

tr(A) = -3, det(A) = -3        --------(ii)

But in the question it is given that 

x = trace(A) and y = |A|          --------(iii)

from euation (ii) and (iii) we get, x = y = -3