Let M denote the set of all real matrices of order 3 × 3 and let S=\ -3,-2,-1,1,2\ . Let aligned & S_ 1 =\ A=[a_ ij ] ∈ M: A=A^ T and a_ ij ∈ S, i, j\ , & S _ 2 =\ A=[a_ ij ] ∈ M: A=-A^ T and a_ ij ∈ S, i, j\ , & S_ 3

Correct answer: 0 [ array lll a_ 11 & a_ 12 & a_ 13 a_ 21 & a_ 22 & a_ 23 a_ 31 & a_ 32 & a_ 33 array ] No. of elements in S_1: A=A ^ T 5^3 × 5^3 No. of elements in A=-A^T 0 since no. zero in 5 No. of elements in S_3 . array c a_ 11 +a_ 22 +a_ 33 =0 (1,2,-3) 31 or (1,1,-2) 3 or (-1,-1,2) 3 array \ 12 × 5^6 aligned & n(S_1 S_3)=12 × 5^3 & n(S_1 S_2 S_3)=5^6(1+12)-12 × 5^3 & 5^3 ×[13 × 5^3-12]=125 α & α=1613 aligned

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Matrices

Let M denote the set of all real matrices of order 3 × 3 and let...

Let M denote the set of all real matrices of order

3 \times 3

and let

\mathrm{S}=\{-3,-2,-1,1,2\}

. Let

\begin{aligned}& \mathrm{S}_{1}=\left\{ \mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: \mathrm{A}=\mathrm{A}^{\mathrm{T}} \text{\text{ and } }a_{\mathrm{ij}} \in \mathrm{ S}, \forall \mathrm{i}, \mathrm{j}\right\}, \\ & \mathrm{S} }_{2}=\left\{\mathrm{A}=\left[a_{\mathrm{ij} }\right] \in \mathrm{M}: \mathrm{A}=-\mathrm{A}^{\mathrm{T}} \text{ and }a_{\mathrm{ij}} \in \mathrm{ S}, \forall \mathrm{i}, \mathrm{j}\right\}, \\ & \mathrm{S}_{3}=\left\{ \mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: a_{11}+a_{22}+a_{33}=0 \text{ and }a_{\mathrm{ij}} \in \mathrm{ S}, \forall \mathrm{i}, \mathrm{j}\right\} . \end{aligned}

If n\left(\mathrm{ S}_1

\cup_2

\mathrm{US}_3\right)= 125 \alpha,

then \alpha

equals _______$

Correct answer

Step-by-step solution

\left[

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

\right]

No. of elements in

\mathrm{S}_1:

\mathrm{A}=\mathrm{A}

^{\mathrm{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 \mathrm{S}_3 \Rightarrow \left.

\begin{array}{c}a_{11}+a_{22}+a_{33}=0 \Rightarrow(1,2,-3) \Rightarrow 31 \\ \text{ or } \\ (1,1,-2) \Rightarrow 3 \\ \text{ or } \\ (-1,-1,2) \Rightarrow 3\end{array}

\right\} \Rightarrow 12 \times 5^6

\begin{aligned} & \mathrm{n}\left(\mathrm{S}_1 \cap \mathrm{ S}_3\right)=12 \times 5^3 \\ & \mathrm{n}\left(\mathrm{S}_1 \cup \mathrm{ S}_2 \cup \mathrm{ S}_3\right)=5^6(1+12)-12 \times 5^3 \\ & \quad \Rightarrow 5^3 \times\left[13 \times 5^3-12\right]=125 \alpha \\ & \quad \alpha=1613\end{aligned}

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