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 $M=\left(a_{i j}\right)_{3 \times 3}$, define $R_i=a_{i 1}+a_{i 2}+a_{i 3}$ and $C_j=a_{1 j}+a_{2 j}+a_{3 j}$ 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})_{3x3} $ 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})_{3x3} $ with all entries in $ T $ such that $ C_j = 0 $ for all $ j $, is	(2) 12
(R) Let $ M = (a_{ij})_{3x3} $ be a skew symmetric matrix such that $ a_{ij} \in T $ for $ i > j $. Then the number of elements in the set $ \left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in \mathbb{R}, M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ a_{13} \end{pmatrix} \right\} $ is	(3) infinite
(S) Let $ M = (a_{ij})_{3x3} $ 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

The correct option is

Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations

$$ \begin{aligned} & x+2 y+z=7 \\\\ & x+\alpha z=11 \\\\ & 2 x-3 y+\beta z=\gamma \end{aligned} $$

Match each entry in List-I to the correct entries in List-II.

List - I	List - II
(P) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has	(1) a unique solution
(Q) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has	(2) no solution
(R) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has	(3) infinitely many solutions
(S) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has	(4) $x=11, y=-2$ and $z=0$ as a solution
	(5) $x=-15, y=4$ and $z=0$ as a solution

The correct option is:

If $M=\left(\begin{array}{rr}\frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}\end{array}\right)$, then which of the

following matrices is equal to $M^{2022} ?$

Let $$p, q, r$$ be nonzero real numbers that are, respectively, the $$10^{\text {th }}, 100^{\text {th }}$$ and $$1000^{\text {th }}$$ terms of a harmonic progression. Consider the system of linear equations

$$$ \begin{gathered} x+y+z=1 \\ 10 x+100 y+1000 z=0 \\ q r x+p r y+p q z=0 \end{gathered} $$$

List-I	List-II
(I) If $$\frac{q}{r}=10$$, then the system of linear equations has	(P) $$x=0, \quad y=\frac{10}{9}, z=-\frac{1}{9}$$ as a solution
(II) If $$\frac{p}{r} \neq 100$$, then the system of linear equations has	(Q) $$x=\frac{10}{9}, y=-\frac{1}{9}, z=0$$ as a solution
(III) If $$\frac{p}{q} \neq 10$$, then the system of linear equations has	(R) infinitely many solutions
(IV) If $$\frac{p}{q}=10$$, then the system of linear equations has	(S) no solution
	(T) at least one solution

The correct option is: