JEE Main 2025 — 29 January, Morning Shift — Question 64

$|\mathrm{A}|=\frac{11}{2}$

$\mathrm{C}_{11}=\sum_{\mathrm{k}=1}^{2} \mathrm{a}_{1 \mathrm{k}} \cdot \mathrm{A}_{1 \mathrm{k}}=\mathrm{a}_{11} \mathrm{~A}_{11}+\mathrm{a}_{12} \mathrm{~A}_{12}=|\mathrm{A}|=\frac{11}{2}$

$\mathrm{C}_{12}=\sum_{\mathrm{k}=1}^{2} \mathrm{a}_{1 \mathrm{k}} \cdot \mathrm{A}_{2 \mathrm{k}}=0$

$\mathrm{C}_{21}=\sum_{\mathrm{k}=1}^{2} \mathrm{a}_{2 \mathrm{k}} \cdot \mathrm{A}_{1 \mathrm{k}}=0$

$\mathrm{C}_{22}=\sum_{\mathrm{k}=1}^{2} \mathrm{a}_{2 \mathrm{k}} \cdot \mathrm{A}_{2 \mathrm{k}}=|\mathrm{A}|=\frac{11}{2}$

$\mathrm{C}=\left[\begin{array}{cc}11 / 2 & 0 \\ 0 & 11 / 2\end{array}\right]$

$|C|=\frac{121}{4}$ $8|C|=242$