Let M denote the set of all real matrices of order 3 × 3 and let S = {–3, –2, –1, 1, 2}. Let S1=A=aij∈M:A=AT and aij∈S,∀i,j S2=A=aij∈M:A=−AT and aij∈S,∀i,j S3=A=aij∈M:a11+a22+a33=0 and aij∈S,∀i,j If nS1∪S2∪S3=125α, then α equals.

a11 a12 a13a21 a22 a23a31 a32 a33No. of elements in S1:A=AT⇒53×53No. of elements in A=−AT⇒0since no. zero in 5 No. of elements in S3⇒a11+a22+a33=0⇒(1,2,−3)⇒31 or (1,1,−2)⇒3 or (−1,−1,2)⇒3⇒12×56nS1∩S3=12×53nS1∪S2∪S3=56(1+12)−12×53⇒53×13×53−12=125α α=1613

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Let M denote the set of all real matrices of order 3 × 3 and let S = {–3, –2, –1, 1, 2}. Let
$\begin{array}{l} \begin{aligned} S_{1} = \{A = [a_{ij}] \in M : A = A^{T} and a_{ij} \in S, \forall i, j\} \\ S_{2} = \{A = [a_{ij}] \in M : A = - A^{T} and a_{ij} \in S, \forall i, j\} \\ S_{3} = \{A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 and a_{ij} \in S, \forall i, j\} \end{aligned} \\ If n (S_{1} \cup S_{2} \cup S_{3}) = 125 α, then α equals. \end{array}$

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answer is 1613.

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Detailed Solution

$[\begin{array}{lll} a_{11} a_{12} a_{13} \\ a_{21} a_{22} a_{23} \\ a_{31} a_{32} a_{33} \end{array}]$

No. of elements in $S_{1} : A = A^{T} \Rightarrow 5^{3} \times 5^{3}$

No. of elements in $A = - A^{T} \Rightarrow 0$

since no. zero in 5 No. of elements in $S_{3} \Rightarrow$

$\begin{matrix} a_{11} + a_{22} + a_{33} = 0 \Rightarrow (1, 2, - 3) \Rightarrow 31 \\ or \\ (1, 1, - 2) \Rightarrow 3 \\ or \\ (- 1, - 1, 2) \Rightarrow 3 \end{matrix}\} \Rightarrow 12 \times 5^{6}$

$\begin{array}{l} n (S_{1} \cap S_{3}) = 12 \times 5^{3} \\ n (S_{1} \cup S_{2} \cup S_{3}) = 5^{6} (1 + 12) - 12 \times 5^{3} \\ \Rightarrow 5^{3} \times [13 \times 5^{3} - 12] = 125 α \\ α = 1613 \end{array}$