If $$A = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}$$, then A 2 is equal to:

Given,

$$A = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}$$

Substituting values we get :

$$A^2 = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} . \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.$$

Hence, option 3 is the correct option.