1(v)
If $$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$, then A2 is equal to:
- (a)$$\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$
- (b)$$\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$$
- (c)$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
- (d)$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
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Given,
$$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
Substituting values we get :
$$A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} . \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\\[1em] = \begin{bmatrix} 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix}\\[1em] = \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix}\\[1em] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$
Hence, option 3 is the correct option.