Matrix $A$ satisfies $A^2 = 2A - I$,where $I$ is the identity matrix. Then for $n \ge 2$,$A^n$ is equal to $(n \in N)$:

If $A$ and $B$ are $3 \times 3$ order matrices and $|A|=5$,$|B|=3$,then $|3AB|=$ . . . . . . .

Let $\alpha \in(0, \infty)$ and $A=\begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(2A-A^{T}) \cdot \operatorname{adj}(A-2A^{T}))=2^8$,then $(\operatorname{det}(A))^2$ is equal to:

Match the items of List-$I$ with those of List-$II$. The correct match is:

Let $M$ and $N$ be two invertible square matrices over $\mathbb{R}$ of order $2$ such that $N$ is diagonal. Then $M N M^{-1}$ is diagonal . . . . . .

Let $a = \text{Minimum} \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2}$. The value of $\sum_{r = 0}^n a^r \cdot b^{n - r}$ is