If $$A = \left(\begin{array}{c}1 & 0\\ -1 & 1\end{array}\right)$$, then $$A^{50}$$ is

Given $$A = \left(\begin{array}{c}1 & 0\\ -1 & 1\end{array}\right)$$, then $$A^{50}$$

$$\ A^2$$= $$\begin{bmatrix}1 & 0\\ -1 &1 \end{bmatrix}.\begin{bmatrix}1 & 0\\ -1 &1 \end{bmatrix}$$

=$$\begin{bmatrix}(1)(1)+0(-1) & 1(0)+0(1)\\ (1)(-1)+(1)(-1) &(-1)(0)+(1)(1) \end{bmatrix}$$

=$$\begin{bmatrix}1 & 0\\ -2 &1 \end{bmatrix}$$.

Now, $$\ A^3$$= $$\begin{bmatrix}1 & 0\\ -2 &1 \end{bmatrix}.\begin{bmatrix}1 & 0\\ -1 &1 \end{bmatrix}$$

=$$\begin{bmatrix}(1)(1)+0(-1) & 1(0)+0(1)\\ (1)(-2)+(1)(-1) &(-2)(0)+(1)(1) \end{bmatrix}$$

=$$\begin{bmatrix}1 & 0\\ -3 &1 \end{bmatrix}$$.

So, in general $$\ A^n$$= $$\begin{bmatrix}1 & 0\\ -n &1 \end{bmatrix}$$.

Therefore, $$\ A^{50}$$= $$\begin{bmatrix}1 & 0\\ -50 &1 \end{bmatrix}$$, which is given in Option C