Question

If \(a\) is a real root of the equation \[ x^{3}-2x^{2}-2x-3=0 \] and \[ A= \begin{bmatrix} 1& 2& a\\ 2& a& 1\\ a& 1& 2 \end{bmatrix}, \] then \(A^{2}-A=\)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Whenever a parameter inside a matrix satisfies an algebraic equation, determine the parameter first. Matrix calculations become much simpler afterward.

Answer 1

The Correct Option is C

Concept: When a matrix contains a parameter satisfying a polynomial equation, we first determine the parameter value and then perform matrix operations. The given cubic equation is \[ x^{3}-2x^{2}-2x-3=0. \] Using factorization techniques, we obtain the real root and then calculate \(A^2-A\).

Step 1: Find the real root of the cubic equation.
Given \[ x^{3}-2x^{2}-2x-3=0. \] Checking possible rational roots, \[ f(3)=27-18-6-3=0. \] Hence \[ x=3 \] is a real root. Therefore \[ a=3. \]

Step 2: Substitute \(a=3\) in matrix \(A\).
\[ A= \begin{bmatrix} 1& 2& 3\\ 2& 3& 1\\ 3& 1& 2 \end{bmatrix}. \]

Step 3: Compute \(A^2\).
Multiplying \(A\) by itself, \[ A^2= \begin{bmatrix} 14& 11& 11\\ 11& 14& 11\\ 11& 11& 14 \end{bmatrix}. \]

Step 4: Calculate \(A^2-A\).
\[ A^2-A= \begin{bmatrix} 14& 11& 11\\ 11& 14& 11\\ 11& 11& 14 \end{bmatrix} - \begin{bmatrix} 1& 2& 3\\ 2& 3& 1\\ 3& 1& 2 \end{bmatrix} \] \[ = \begin{bmatrix} 13& 9& 8\\ 9& 11& 10\\ 8& 10& 12 \end{bmatrix}. \] Hence \[ \boxed{ A^2-A= \begin{bmatrix} 13& 9& 8\\ 9& 11& 10\\ 8& 10& 12 \end{bmatrix} } \] Therefore the correct option is \[ \boxed{(C)}. \]