ICAR AIEEA PG 2024 - Statistical Sciences

Linear dependence in vector spaces:
The vectors in $S$ are 2-dimensional vectors belonging to $\mathbb{R}^2$.
The dimension of $\mathbb{R}^2$ is 2 ($\dim(\mathbb{R}^2) = 2$).
By the fundamental dimension theorem of linear algebra, any set of more than $n$ vectors in an $n$-dimensional vector space must be linearly dependent.
Here, the set $S$ contains 4 vectors in $\mathbb{R}^2$. Since $4 \gt 2$, $S$ is necessarily linearly dependent.
Notice also that removing 1 vector leaves 3 vectors, which still exceeds 2, so the subsets in Options 3 and 4 are also linearly dependent.

Inequality analysis for real numbers:
Given $0 \lt a \lt b$:
Multiplying the inequality $a \lt b$ by the strictly positive real number $a$ preserves the inequality:

This proves Option 4 is correct.

Testing other options:

• Since $a \lt b$, taking reciprocals reverses the inequality $\frac{1}{a} \gt \frac{1}{b}$ (false).
• Multiplying $a \lt b$ by $b \gt 0$ gives $ab \lt b^2$ (false).
• $a^2 \lt ab$ means $ab \gt a^2$ (false).

Given $c \gt 1$:
Since $c \gt 1 \gt 0$, $c$ is strictly positive.
Multiplying both sides of the inequality $1 \lt c$ by $c$ yields:

Fundamental theorem of skew-symmetric matrices:
By definition of skew-symmetric matrix: $A^T = -A$.
Taking the determinant of both sides:

Since $\det(A^T) = \det(A)$ and for an $n \times n$ matrix $\det(-A) = (-1)^n \det(A)$:

Here, the order $n = 5$ is an odd integer: