Let $ M $ denote the set of all real matrices of order 3 x 3 and let $ S = \{-3, -2, -1, 1, 2\} $. Let $ S_1 = \{A = [a_{ij}] \in M : A = A^T \text{ and } a_{ij} \in S, \forall i, j\} $, $ S_2 = \{A = [a_{ij}] \in M : A = -A^T \text{ and } a_{ij} \in S, \forall i, j\} $, $ S_3 = \{A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j\} $. If $n(S_1 \cup S_2 \cup S_3) = 125$, then $ \alpha $ equals:

Question

Let $ M $ denote the set of all real matrices of order 3 x 3 and let $ S = \{-3, -2, -1, 1, 2\} $. Let $ S_1 = \{A = [a_{ij}] \in M : A = A^T \text{ and } a_{ij} \in S, \forall i, j\} $, $ S_2 = \{A = [a_{ij}] \in M : A = -A^T \text{ and } a_{ij} \in S, \forall i, j\} $,  $ S_3 = \{A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j\} $. If $n(S_1 \cup S_2 \cup S_3) = 125$, then $ \alpha $ equals:

cdquestions Admin · Accepted Answer

Step 1: Analyze each set.  $ S_1 $ includes symmetric matrices, so elements above the diagonal determine the matrix. With 5 choices for each, and 6 such positions: $$ |S_1| = 5^6 $$ $ S_2 $ includes skew-symmetric matrices, where non-diagonal elements are independent, and diagonal elements must be 0 (which are not in $ S $), invalidating $ S_2 $. Thus: $$ |S_2| = 0 $$ $ S_3 $ must balance the trace to be zero. Choosing two elements freely allows the third to be determined: $$ |S_3| = 5^2 \times (\text{number of valid third elements}) $$ Step 2: Calculate the union of sets. Using the inclusion-exclusion principle, find $ n(S_1 \cup S_2 \cup S_3) $: $$ n(S_1 \cup S_2 \cup S_3) = |S_1| + |S_2| + |S_3| - (\text{intersections}) = 125 $$

Show Hint

Correct Answer: 125

Solution and Explanation

Top Questions on Linear Algebra

Questions Asked in JEE Main exam