Let α and β be the distinct roots of the equation $x^2+x-\; <!--\; -\; -->1=0$. Consider the set T = {1, α, β}. For a 3 x 3 matrix $M=(a_{ij}{)}_{3\times \; <!--\; \times \; -->3})$, 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 $(a_{ij}{)}_{3\times <!--\; \times \; -->3}$ with all entries in T such that Ri = Cj = 0 for all i, j is	(1) 1
(Q) The number of symmetric matrices $(a_{ij}{)}_{3\times <!--\; \times \; -->3}$ with all entries in T such that Cj = 0 for all j, is	(2) 12
(R) Let $(a_{ij}{)}_{3\times <!--\; \times \; -->3}$ be a skew symmetric matrix such that $a_{ij}$ ∈ T for i > j. Then the number of elements in the set $\{\left(\begin{array}{l}x\\ y\\ z\end{array}\right):x,y\cdot <!--\; \cdot \; -->z\in <!--\; \in \; -->R,M\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}{a}_{12}\\ 0\\ -<!--\; -\; -->a_{23}\end{array}\right)\}$	(3) infinite
(S) Let $(a_{ij}{)}_{3\times <!--\; \times \; -->3}$ be a matrix with all entries in T such that Ri = 0 for all i. Then the absolute value of the determinant of M is	(4) 6
	(5) 0

The correct option is

Given Equation and Set T:
The quadratic equation is given by:
$x^2+x-1=0$
Since $\mathrm{��}$ and $\beta$ are the distinct roots of this equation, we have:
$\alpha +\beta =-1$
This implies:
$1+\alpha +\beta =0$
The roots of the equation are approximately $\alpha \approx 0.618$ and $\beta \approx -1.618$. The set is $T=\{1,\alpha ,\beta \}$. Since $1$, $\alpha$, and $\beta$ are distinct real numbers, the only way a sum of three elements from $T$ (with repetitions allowed) equals zero is if the three elements are exactly $1$, $\alpha$, and $\beta$ in some order.

Analysis of (P):
We want to find the number of matrices $M=(a_{ij}{)}_{3\times 3}$ with entries in $T$ such that $R_i=C_j=0$ for all $i,j$.
Since the sum of elements in each row and column is zero, each row and column must contain the elements $1$, $\alpha$, and $\beta$ exactly once. This is the definition of a Latin square of order 3.
- For the first row, there are $3!=6$ permutations.
- Once the first row is chosen (say, $A,B,C$), the remaining elements must satisfy the Latin square property. There are exactly 2 valid arrangements for the second and third rows:
Either:
Row 2: $B,C,A$ and Row 3: $C,A,B$
Or:
Row 2: $C,A,B$ and Row 3: $B,C,A$
Therefore, the total number of such matrices is:
$6\times 2=12$
Thus, (P) → (2).

Analysis of (Q):
We need to find the number of symmetric matrices $M=(a_{ij}{)}_{3\times 3}$ with all entries in $T$ such that $C_j=0$ for all $j$.
Since $M$ is symmetric, $R_i=C_i=0$ for all $i$. Thus, it must also form a Latin square.
Let the symmetric Latin square be:
$\left(\begin{array}{ccc}A& B& C\\ B& d& e\\ C& e& f\end{array}\right)$
For this matrix to be a Latin square:
- Row 2 requires $\{B,d,e\}=\{A,B,C\}\Rightarrow \{d,e\}=\{A,C\}$
- Row 3 requires $\{C,e,f\}=\{A,B,C\}\Rightarrow \{e,f\}=\{A,B\}$
Comparing the two, we must have $e=A$.
This uniquely determines $d=C$ and $f=B$.
Thus, for each of the $6$ permutations of the first row, there is exactly $1$ symmetric matrix:
$\left(\begin{array}{ccc}A& B& C\\ B& C& A\\ C& A& B\end{array}\right)$
So the total number of symmetric matrices is $6$.
Thus, (Q) → (4).

Analysis of (R):
Let $M=(a_{ij}{)}_{3\times 3}$ be a skew-symmetric matrix. Therefore, the diagonal entries are zero ($a_{11}=a_{22}=a_{33}=0$) and $a_{ji}=-a_{ij}$.
The matrix $M$ is:
$M=\left(\begin{array}{ccc}0& a_{12}& a_{13}\\ -a_{12}& 0& a_{23}\\ -a_{13}& -a_{23}& 0\end{array}\right)$
We are given the system:
$M\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}{a}_{12}\\ 0\\ -a_{23}\end{array}\right)$
This gives the following system of linear equations:
1) $a_{12}y+a_{13}z=a_{12}$
2) $-a_{12}x+a_{23}z=0\Rightarrow x=\frac{a_{23}}{a_{12}}z$
3) $-a_{13}x-a_{23}y=-a_{23}$
From (1), we get:
$y=1-\frac{a_{13}}{a_{12}}z$
Substituting $x$ and $y$ into (3):
$-a_{13}\left(\frac{a_{23}}{a_{12}}z\right)-a_{23}(1-\frac{a_{13}}{a_{12}}z)=-a_{23}$
$-\frac{a_{13}{a}_{23}}{a_{12}}$
$-a_{23}=-a_{23}$
This is an identity that is always satisfied, independent of the value of $z$. Therefore, for any real value $z\in ℝ$, there is a unique pair of values for $x$ and $y$. Hence, the system has an infinite number of solutions.
Thus, (R) → (3).

Analysis of (S):
We are given that $R_i=0$ for all $i$, which means:
$a_{i1}+a_{i2}+a_{i3}=0$
for $i=1,2,3$.
If we perform the column operation $C_1\to C_1+C_2+C_3$ on the matrix $M$, the first column of the resulting determinant becomes:
$\left(\begin{array}{c}{a}_{11}+a_{12}+a_{13}\\ a_{21}+a_{22}+a_{23}\\ a_{31}+a_{32}+a_{33}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$
Since an entire column consists of zeros, the determinant is $0$, and its absolute value is also $0$.
Thus, (S) → (5).