Verified Solution • 2025–26 Session moderate

QJEEM25Mae94c: JEE Main 2025 Mathematics - Class 12 JEEM25Mae94c Solution: Let be a matrix such that for all and . Let the random variable …

Let A=[aij]A = [a_{ij}] be a 2×22 \times 2 matrix such that aij∈{0,1}a_{ij} \in \{0, 1\} for all ii and jj . Let the random variable XX denote the possible values of the determinant of the matrix AA . Then, the variance of XX is:

Multiple Choice Options

  • A 58\frac{5}{8}
  • B 14\frac{1}{4}
  • C 34\frac{3}{4}
  • D 38\frac{3}{8}

Step-by-Step Verified Solution

amp;∣A∣=∣a11amp;a12a21amp;a22∣amp;=a11a22−a21a12amp;={−1,0,1}\begin{aligned}& |A|=\left|\begin{array}{ll}a_{11} & a_{12} \\a_{21} & a_{22}\end{array}\right| \\& =a_{11} a_{22}-a_{21} a_{12} \\& =\{-1,0,1\}\end{aligned}

xamp;Piamp;PiXiamp;P1Xi2−1amp;316amp;−316amp;3160amp;1016amp;0amp;01amp;316amp;316amp;316amp;amp;∑PiXi=0amp;∑PiXi2=38\begin{array}{c|c|c|c}\mathrm{x} & \mathrm{P}_{\mathrm{i}} & \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}} & \mathrm{P}_1 \mathrm{X}_{\mathrm{i}}{ }^2 \\-1 & \frac{3}{16} & -\frac{3}{16} & \frac{3}{16} \\0 & \frac{10}{16} & 0 & 0 \\1 & \frac{3}{16} & \frac{3}{16} & \frac{3}{16} \\\hline & & \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=0 & \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}{ }^2=\frac{3}{8}\end{array}

amp;∴var⁡(x)=∑PiXi2−(∑PiXi)2amp;=38−0=38\begin{aligned}& \therefore \operatorname{var}(\mathrm{x})=\sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}^2-\left(\sum \mathrm{P}_{\mathrm{i}} X_{\mathrm{i}}\right)^2 \\& =\frac{3}{8}-0=\frac{3}{8}\end{aligned}

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