JEE MAIN - Mathematics (2025 - 28th January Morning Shift - No. 21)
Let M denote the set of all real matrices of order $3 \times 3$ and let $\mathrm{S}=\{-3,-2,-1,1,2\}$. Let
$$\begin{aligned} & \mathrm{S}_1=\left\{\mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: \mathrm{A}=\mathrm{A}^{\mathrm{T}} \text { and } a_{\mathrm{ij}} \in \mathrm{~S}, \forall \mathrm{i}, \mathrm{j}\right\}, \\ & \mathrm{S}_2=\left\{\mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: \mathrm{A}=-\mathrm{A}^{\mathrm{T}} \text { and } a_{\mathrm{ij}} \in \mathrm{~S}, \forall \mathrm{i}, \mathrm{j}\right\}, \\ & \mathrm{S}_3=\left\{\mathrm{A}=\left[a_{\mathrm{ij}}\right] \in \mathrm{M}: a_{11}+a_{22}+a_{33}=0 \text { and } a_{\mathrm{ij}} \in \mathrm{~S}, \forall \mathrm{i}, \mathrm{j}\right\} . \end{aligned}$$
If $n\left(S_1 \cup S_2 \cup S_3\right)=125 \alpha$, then $\alpha$ equls __________.
Explanation
$$\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$$
No. of elements in $\mathrm{S}_1: \mathrm{A}=\mathrm{A}^{\mathrm{T}} \Rightarrow 5^3 \times 5^3$
No. of elements in $S_2: A=-A^T \Rightarrow 0$
since no zero in $\mathrm{S}_2$
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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