Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:

If $A = \begin{bmatrix} 1 + a^2 + a^4 & 1 + ab + a^2b^2 & 1 + ac + a^2c^2 \\ 1 + ab + a^2b^2 & 1 + b^2 + b^4 & 1 + bc + b^2c^2 \\ 1 + ac + a^2c^2 & 1 + bc + b^2c^2 & 1 + c^2 + c^4 \end{bmatrix}$ and $\det(A) = \det(4I)$,where $I$ is a $3 \times 3$ identity matrix,then $(a - b)^3 + (b - c)^3 + (c - a)^3$ can be equal to -

Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=(a_{ij})$,define $R_i=a_{i1}+a_{i2}+a_{i3}$ and $C_j=a_{1j}+a_{2j}+a_{3j}$ for $i=1,2,3$ and $j=1,2,3$. Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P)$ The number of matrices $M=(a_{ij})_{3 \times 3}$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is	$(1)$ $1$
$(Q)$ The number of symmetric matrices $M=(a_{ij})_{3 \times 3}$ with all entries in $T$ such that $C_j=0$ for all $j$ is	$(2)$ $2$
$(R)$ Let $M=(a_{ij})_{3 \times 3}$ be a skew-symmetric matrix such that $a_{ij} \in T$ for $i>j$. Then the number of elements in the set $\{\begin{bmatrix} x \\ y \\ z \end{bmatrix}: x, y, z \in \mathbb{R}, M\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a_{12} \\ 0 \\ -a_{23} \end{bmatrix}\}$ is	$(3)$ $\text{Infinite}$
$(S)$ Let $M=(a_{ij})_{3 \times 3}$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is	$(4)$ $6$
	$(5)$ $0$

If $\Delta=\left|\begin{array}{lll}1 & 5 & 6 \\ 0 & 1 & 7 \\ 0 & 0 & 1\end{array}\right|$ and $\Delta^{\prime}=\left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$, then

Let $a = \min \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{x \to 0} \frac{\sin x \cos x}{e^x - e^{-x}}$. Then the value of $\sum_{r=0}^n a^r b^{n-r}$ is

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