Let $A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{bmatrix}$. Then,the roots of the equation $\operatorname{det}(A - \lambda I_{3}) = 0$ (where $I_{3}$ is the identity matrix of order $3$) are

Let $A$ be a $2 \times 2$ matrix with non-zero entries and let $A^2 = I$,where $I$ is the $2 \times 2$ identity matrix. Define $tr(A) = \text{sum of diagonal elements of } A$ and $|A| = \text{determinant of matrix } A$.
Statement $-1: tr(A) = 0$
Statement $-2: \det(A) = 1$

If $\alpha , \beta \neq 0$ and $f(n) = \alpha^n + \beta^n$ and $\begin{vmatrix} 3 & 1 + f(1) & 1 + f(2) \\ 1 + f(1) & 1 + f(2) & 1 + f(3) \\ 1 + f(2) & 1 + f(3) & 1 + f(4) \end{vmatrix} = K(1 - \alpha)^2 (1 - \beta)^2 (\alpha - \beta)^2$,then $K = \dots$

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:

Let $I$ be an identity matrix of order $2 \times 2$ and $P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$. Then the value of $n \in N$ for which $P^n = 5I - 8P$ is equal to ..... .

The value of $(1+\Delta)(1-\nabla)$ is