​$$\text{If } A = \begin{pmatrix}3 & -2 \\2 & -1\end{pmatrix}, \text{ then } A^{20} \text{ equals:}$$​​

Correct option is B

​​$$A = \begin{pmatrix}3 & -2 \\2 & -1\end{pmatrix}\\\implies \det(A - xI) = x^2 - 2x + 1 = (x - 1)^2 \\\text{Characteristic equation of } A \text{ is:} \\C(x) = x^2 - 2x + 1 = (x - 1)^2 = 0 \\\text{By Cayley-Hamilton theorem:} \\C(A) = (A - I)^2 = 0 \\\text{Now let } x^{20} = (x - 1)^2 q(x) + ax + b \quad \cdots \quad (1) \\\implies 20x^{19} = (x - 1)^2 q'(x) + 2(x - 1)q(x) + a \quad \cdots \quad (2) \\\text{(By differentiating both sides of (1))}\\\text{By putting } x = 1 \text{ on both sides of equations (1) and (2), we get:} \\1 = a + b \quad \text{and} \quad 20 = a \\\implies b = -19 \\\implies A^{20} = (A - I)^2 q(A) + aA + bI \\\implies A^{20} = 20A - 19I \\\implies A^{20} = \begin{pmatrix}41 & -40 \\40 & -39\end{pmatrix}$$​​