Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
Let $\{\Delta_1, \Delta_2, \Delta_3, \ldots \Delta_k\}$ be the set of third order determinants that can be made with the distinct non-zero real numbers $a_1, a_2, \ldots a_9$. Then:
$k = 9!$
$\sum_{i=1}^{k} \Delta_i = 0$
At least one $\Delta_i = 0$
None of these
Step-by-Step Solution
Key Concept: The set of all 3×3 determinants formed by permutations of 9 distinct non-zero reals has cardinality 9! since each determinant corresponds to a unique arrangement of the 9 numbers in the 9 positions. The sum equals zero because for every permutation σ producing determinant Δ, the permutation that swaps two rows (an odd permutation) produces -Δ, and these pair up completely due to the symmetric nature of all possible permutations.
The number of third-order determinants using nine different numbers in nine positions equals the number of permutations of nine distinct objects, which is $9!$. For each determinant formed, interchanging two consecutive rows or columns produces a related determinant with opposite sign. The sum of pairs of determinants obtained by such interchanges equals zero, as each determinant and its negation cancel.
Correct Answer: 1,2