JEE Advanced 2024 Paper 1Question 1
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 $3\times3$ matrix $M=(a_{ij})_{3\times3}$, 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.
$$\begin{array}{ll}\textbf{List-I} & \textbf{List-II}\\ (P)\ \text{The number of matrices } M=(a_{ij})_{3\times3} \text{ with all entries in } T \text{ such that } R_i=C_j=0 \text{ for all } i,j \text{ is} & (1)\ 1\\ (Q)\ \text{The number of symmetric matrices } M=(a_{ij})_{3\times3} \text{ with all entries in } T \text{ such that } C_j=0 \text{ for all } j \text{ is} & (2)\ 12\\ (R)\ \text{Let } M=(a_{ij})_{3\times3} \text{ be a skew symmetric matrix such that } a_{ij}\in T \text{ for } i>j. \text{ Then the number of elements in the set } \left\{\begin{pmatrix}x\\y\\z\end{pmatrix}: x,y,z\in\mathbb{R},\ M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a_{12}\\0\\-a_{23}\end{pmatrix}\right\} \text{ is} & (3)\ \text{infinite}\\ (S)\ \text{Let } M=(a_{ij})_{3\times3} \text{ be a matrix with all entries in } T \text{ such that } R_i=0 \text{ for all } i. \text{ Then the absolute value of the determinant of } M \text{ is} & (4)\ 6\\ & (5)\ 0\end{array}$$
Matrices
Solution
$1+\alpha+\beta=0$, so a zero-sum row from $T$ is a permutation of $(1,\alpha,\beta)$. (P) Latin squares: $12$. (Q) symmetric Latin squares: $6$. (R) skew-symmetric $M$ is singular and the system is consistent, so infinitely many solutions. (S) $M(1,1,1)^T=0$ so $\det M=0$. Hence P-2, Q-4, R-3, S-5.