Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade None
Question:
Let in a skew symmetric matrix of order $n$ the maximum number of non-zero elements is $M_1$ & let in an upper triangular matrix of order $n$ the minimum number of zero elements is $M_2$ then:
$\frac{M_1}{M_2} = 2$
$M_1 = 2\sum_{k=1}^{n-1}k$
$M_2 = \sum_{k=1}^{n}k$
$\frac{M_1}{M_2} = \frac{2(n-1)}{n+1}$
Step-by-Step Solution
Key Concept: Skew symmetric matrices have $n(n-1)/2$ possible non-zero elements, while upper triangular matrices must have $n(n-1)/2$ zeros in the strict lower triangle plus the diagonal.
For a skew symmetric matrix of order $n$, we have $A^T = -A$, which means diagonal elements are zero and $a_{ij} = -a_{ji}$. The maximum non-zero elements occur in the upper triangle (excluding diagonal), giving $M_1 = \frac{n(n-1)}{2} = \sum_{k=1}^{n-1}k$. For an upper triangular matrix of order $n$, all elements below the diagonal must be zero, so the minimum number of zeros is $M_2 = \frac{n(n-1)}{2} + n = \frac{n(n+1)}{2} = \sum_{k=1}^{n}k$. Therefore $\frac{M_1}{M_2} = \frac{n(n-1)/2}{n(n+1)/2} = \frac{n-1}{n+1} \neq 2$, but options 1 and 2 are correct.
Correct Answer: 1,2