- ✓$P < \frac{1}{10^{18}}$
- B$P=\frac{1}{10^{18}}$
- C$\frac{5^2}{10^{18}} \leq P \leq \frac{5^3}{10^{18}}$
- D$P \leq \frac{5^4}{10^{18}}$
We have, $A$ is $3 \times 3$ matrix consisting of integer entries from $\{-1000,-999, \ldots 999,1000\}$
Total entries $=2001$
Total number of outcomes $=(2001)^9$
Given, $A^2=-I, A=-A^{-1}$ (which is not possible)
$A$ is diagonal matrix.
$A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{array}\right]$
Total number of favourable outcomes
$=(2001)^3$
Required probability
$=\frac{(2001)^3}{(2001)^9}(2001)^{-6}=(2000+1)^{-6}$
$p=(2000)^{-6}\left(1+\frac{1}{2000}\right)^{-6}$
$p=\frac{1}{2^6 \times 10^{18}}\left(1+\frac{1}{2000}\right)^{-6}$
$P < \frac{1}{2^6 \times 10^{18}} \quad\left[\because 1+\frac{1}{2000}\right.$ is decreasing $]$
$P < \frac{1}{10^{18}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.