Let $A = \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix}$. What is $f(A)$ if $f(x) = x^3 - 2x^2 - 5$?

Let $A$ be a matrix of order $2 \times 2$,whose entries are from the set $\{0, 1, 2, 3, 4, 5\}$. If the sum of all the entries of $A$ is a prime number $p$,where $2 < p < 8$,then the number of such matrices $A$ is:

If $A = \begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and $\alpha, \beta, \gamma$ are the roots of the characteristic equation $|A - xI| = 0$,then $\alpha^2 + \beta^2 + \gamma^2 = $

Let $R = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ be a non-zero $3 \times 3$ matrix,where $x \sin \theta = y \sin \left(\theta + \frac{2 \pi}{3}\right) = z \sin \left(\theta + \frac{4 \pi}{3}\right) \neq 0$,$\theta \in (0, 2 \pi)$. For a square matrix $M$,let $\text{trace}(M)$ denote the sum of all the diagonal entries of $M$. Then,among the statements:
$(I) \text{ Trace}(R) = 0$
$(II) \text{ If trace}(\text{adj}(\text{adj}(R))) = 0, \text{ then } R \text{ has exactly one non-zero entry.}$

How many $3 \times 3$ matrices $M$ with entries from $\{0, 1, 2\}$ are there,for which the sum of the diagonal entries of $M^T M$ is $5$?

If $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $\det(A^n - I) = 1 - \lambda^n$ for $n \in N$,then $\lambda$ is: