Matching Properties of Matrices with Entries in Root Set

To solve the given matching problem, we analyze each part individually.

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Preliminary Information:

The given quadratic equation is: $x^2 + x - 1 = 0$

Since $\alpha$ and $\beta$ are the distinct roots of this equation, by Vieta's formulas, we have: $\alpha + \beta = -1 \implies 1 + \alpha + \beta = 0$

The set of entries is $T = \{1, \alpha, \beta\}$. Since $\alpha$ and $\beta$ are irrational numbers ($\frac{-1 \pm \sqrt{5}}{2}$), the equation $a + b + c = 0$ for $a, b, c \in T$ holds if and only if $\{a, b, c\} = \{1, \alpha, \beta\}$ as a multiset. That is, a sum of three elements chosen from $T$ is $0$ if and only if the three elements are distinct.

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Part (P):

We are looking for the number of $3 \times 3$ matrices $M = (a_{ij})$ with all entries in $T$ such that $R_i = 0$ and $C_j = 0$ for all $i, j = 1, 2, 3$.

1. Since $R_i = 0$ for each row $i$, every row must contain all three elements $\{1, \alpha, \beta\}$ exactly once.
2. Similarly, since $C_j = 0$ for each column $j$, every column must contain all three elements $\{1, \alpha, \beta\}$ exactly once.
3. Such a matrix is a Latin Square of order $3$ formed by the three symbols $\{1, \alpha, \beta\}$.
4. The number of ways to choose the first row is $3! = 6$.
5. For any chosen first row, say $(1, \alpha, \beta)$, there are exactly $2$ ways to complete the remaining two rows to satisfy the Latin square property:
   • Row 2: $(\alpha, \beta, 1)$, Row 3: $(\beta, 1, \alpha)$
   • Row 2: $(\beta, 1, \alpha)$, Row 3: $(\alpha, \beta, 1)$

Thus, the total number of such matrices is: $6 \times 2 = 12$

Hence, $(\text{P}) \rightarrow (2)$.

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Part (Q):

We are looking for the number of symmetric matrices $M = (a_{ij})_{3 \times 3}$ with entries in $T$ such that $C_j = 0$ for all $j = 1, 2, 3$.

1. Since $M$ is symmetric ($M = M^T$), $R_i = C_i = 0$ for all $i$. Thus, $M$ is a symmetric Latin square of order $3$.
2. In a symmetric Latin square of odd order, each symbol must appear on the main diagonal $a_{11}, a_{22}, a_{33}$ exactly once.
3. The number of ways to arrange the three distinct elements $\{1, \alpha, \beta\}$ on the main diagonal is $3! = 6$.
4. For each fixed choice of the main diagonal, say $a_{11} = 1, a_{22} = \alpha, a_{33} = \beta$, the off-diagonal entries are uniquely determined:
   • $a_{12} = a_{21} = \beta$
   • $a_{13} = a_{31} = \alpha$
   • $a_{23} = a_{32} = 1$

Thus, for each diagonal arrangement, there is exactly $1$ valid symmetric matrix. The total number of symmetric matrices is: $6 \times 1 = 6$

Hence, $(\text{Q}) \rightarrow (4)$.

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Part (R):

Let $M$ be a $3 \times 3$ skew-symmetric matrix. By definition of a skew-symmetric matrix, $a_{ii} = 0$ for all $i$, and $a_{ji} = -a_{ij}$. So $M$ has the form: $M = \begin{pmatrix} 0 & -a_{21} & -a_{31} \\ a_{21} & 0 & -a_{32} \\ a_{31} & a_{32} & 0 \end{pmatrix}$ where $a_{21}, a_{31}, a_{32} \in T = \{1, \alpha, \beta\}$.

Note that:

1. $\det(M) = 0$ because $M$ is an odd-order skew-symmetric matrix.
2. The vector $V = \begin{pmatrix} a_{32} \\ -a_{31} \\ a_{21} \end{pmatrix}$ satisfies $M V = \mathbf{0}$, and $V \neq \mathbf{0}$ since $a_{ij} \in T \implies a_{ij} \neq 0$. Thus, $\text{rank}(M) = 2$.
3. We are given the system of linear equations $M X = B$, where $B = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix} = \begin{pmatrix} -a_{21} \\ 0 \\ a_{32} \end{pmatrix}$.
4. The inner product of $B$ with the nullspace vector $V$ is: $B^T V = (-a_{21})(a_{32}) + (0)(-a_{31}) + (a_{32})(a_{21}) = 0$

Since $B$ is orthogonal to the nullspace of $M^T = -M$, $B$ lies in the column space (range) of $M$. Therefore, the non-homogeneous system $M X = B$ is consistent.

Because $\det(M) = 0$ and the system is consistent, it possesses infinitely many solutions.

Hence, $(\text{R}) \rightarrow (3)$.

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Part (S):

We are given a matrix $M$ with all entries in $T$ such that $R_i = 0$ for $i = 1, 2, 3$.

1. The condition $R_i = a_{i1} + a_{i2} + a_{i3} = 0$ means that multiplying $M$ by the non-zero vector $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ yields: $M \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} R_1 \\ R_2 \\ R_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
2. Since $M \mathbf{v} = \mathbf{0}$ for $\mathbf{v} \neq \mathbf{0}$, $M$ is singular, which means: $\det(M) = 0 \implies |\det(M)| = 0$

Hence, $(\text{S}) \rightarrow (5)$.

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Conclusion:

• $(\text{P}) \rightarrow (2)$
• $(\text{Q}) \rightarrow (4)$
• $(\text{R}) \rightarrow (3)$
• $(\text{S}) \rightarrow (5)$

This corresponds to Option C.