Question ID: #453
Let M denote the set of all real matrices of order $3 \times 3$ 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 $n(S_1 \cup S_2 \cup S_3) = 125\alpha$, then $\alpha$ equals.
Solution:
Set $S = \{-3, -2, -1, 1, 2\}$. Size $n(S) = 5$.
1. Calculate $n(S_2)$ (Skew-Symmetric):
For skew-symmetric matrices, diagonal elements must be 0.
Since $0 \notin S$, no such matrix exists.
$$ n(S_2) = 0 $$
2. Calculate $n(S_1)$ (Symmetric):
A $3 \times 3$ symmetric matrix is determined by its diagonal (3 elements) and upper triangular elements (3 elements).
Total positions to choose = $3 + 3 = 6$.
$$ n(S_1) = 5^6 $$
3. Calculate $n(S_3)$ (Trace = 0):
Condition: $a_{11} + a_{22} + a_{33} = 0$.
We find triplets from $S$ that sum to 0:
(i) $\{1, 1, -2\}$: Permutations = $\frac{3!}{2!} = 3$.
(ii) $\{1, 2, -3\}$: Permutations = $3! = 6$.
(iii) $\{-1, -1, 2\}$: Permutations = $\frac{3!}{2!} = 3$.
Total ways for diagonal = $3 + 6 + 3 = 12$.
The non-diagonal elements (6 positions) can be any value from $S$.
$$ n(S_3) = 12 \times 5^6 $$
4. Calculate $n(S_1 \cap S_3)$ (Symmetric & Trace=0):
Diagonal ways = 12 (same as above).
Upper triangular elements (3 positions) can be chosen in $5^3$ ways.
Lower triangular elements are fixed by symmetry.
$$ n(S_1 \cap S_3) = 12 \times 5^3 $$
5. Total Union:
$$ n(S_1 \cup S_2 \cup S_3) = n(S_1) + n(S_3) – n(S_1 \cap S_3) \quad (\text{Since } S_2 = \emptyset) $$
$$ = 5^6 + 12 \times 5^6 – 12 \times 5^3 $$
$$ = 5^6(1 + 12) – 12 \times 5^3 $$
$$ = 13 \times 5^6 – 12 \times 5^3 $$
$$ = 5^3 (13 \times 125 – 12) $$
$$ = 125 (1625 – 12) $$
$$ = 125 (1613) $$
Given count = $125\alpha$.
$$ \alpha = 1613 $$
Ans. 1613
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