If A= \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\), f(x) = x² - 2x - I , then the of f(A) is ?

Concept: 

Let , f(x) = a₀ xⁿ + a₁ x^n-1 + a₂ x^n-2 + ... + a_n-1 x + a_n  be a polynomial and let matrix A be a square matrix of order n . Then, 

f(A) = a₀ Aⁿ + a₁ A^n-1 + a₂ A^n-2 + ... + a_n-1 A + a_n I_n  is called matrix polynomial . 

Calculation: 

Given , A = \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) and  f(x) = x2 - 2x - I . 

Matrix A is the square matrix of order 2 , then its satisfying the given polynomial 

⇒ f(A) = A² - 2A - I 

⇒ f(A) = A A - 2A - I 

⇒ f(A) = \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\)- 2 \(\rm \begin{bmatrix} 2 &1 \\ 1& 2 \end{bmatrix}\) -  \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\) 

⇒ f(A) = \(\rm \begin{bmatrix} 5 &4 \\ 4& 5 \end{bmatrix}\) - \(\rm \begin{bmatrix} 4 &2 \\ 2& 4 \end{bmatrix}\) - \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\)

⇒ f(A) = \(\rm \begin{bmatrix} 1 &2 \\ 2& 1 \end{bmatrix}\) - \(\rm \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}\) 

⇒ f(A) = \(\rm \begin{bmatrix} 0 &2 \\ 2& 0 \end{bmatrix}\) 

The correct option is 1.