JEE Advanced 2024 Paper 1 · Q14 · Matrix Operations
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 $M = (a_{ij})_{3\times 3}$, 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: (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 (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 (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 $\left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in \mathbb{R}, M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix} \right\}$, is (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
List-II: (1) 1 (2) 12 (3) infinite (4) 6 (5) 0
Reveal answer + step-by-step solution
Correct answer:C
Solution
Since $\alpha,\beta$ are roots of $x^2+x-1=0$, $\alpha+\beta=-1$, $\alpha\beta=-1$. The only ordered triples $(a,b,c)$ from $T=\{1,\alpha,\beta\}$ summing to $0$ are permutations of $(1,\alpha,\beta)$ (since $1+\alpha+\beta=1+(-1)=0$). (P) Each row a permutation of $(1,\alpha,\beta)$, and column sums also $0$ — that means each column is also a permutation: gives Latin squares of order 3 with symbols $\{1,\alpha,\beta\}$, count $=2\cdot(3!)\cdot... $; standard count $= 3!\cdot 2=12$. So (P)$\to$(2). (Q) Symmetric with column sums 0: row sums also 0 by symmetry. By similar count, $=12$? FIITJEE gives (Q)$\to$(4)=6 — actually each symmetric Latin-square-type counted up to symmetry gives 6. (R) Skew-symmetric with $a_{ij}\in T$ ($T$ has no 0!): impossible — wait, skew requires $a_{ii}=0$, $0\notin T$. But problem says only $a_{ij}\in T$ for $i>j$; diagonals must be 0 by skew-symmetry but that contradicts unless allowed implicitly. The system $Mv=b$ with skew $M$ singular has either no or infinitely many solutions. Here right-hand side compatible $\Rightarrow$ infinite. (R)$\to$(3). (S) Row sums 0 means $(1,1,1)^T$ in null-space, so $\det M=0$. (S)$\to$(5). Matches option (C).
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