ISRO CS 2025 Q.56: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. Write the columns of the matrix: $C_1=\left[\begin{array}{c}{\sin }^2\theta \\ {\cos }^2\theta \\ -3\end{array}\right]$, $C_2=\left[\begin{array}{c}1\\ 1\\ 4\end{array}\right]$, $C_3=\left[\begin{array}{c}{\cos }^2\theta \\ {\sin }^2\theta \\ 7\end{array}\right]$.

2. Add the first and third columns: $C_1+C_3=\left[\begin{array}{c}{\sin }^2\theta +{\cos }^2\theta \\ {\cos }^2\theta +{\sin }^2\theta \\ -3+7\end{array}\right]$.

3. Use ${\sin }^2\theta +{\cos }^2\theta =1$. Then $C_1+C_3=\left[\begin{array}{c}1\\ 1\\ 4\end{array}\right]=C_2$.

4. So $C_1-C_2+C_3=0$ is a non-trivial relation among the columns. The columns are linearly dependent, so the matrix is singular and $\det C=0$.

5. Check by direct expansion along the first row, with $s={\sin }^2\theta$ and $c={\cos }^2\theta$: $$\det C=s(7-4s)-1\cdot (7c+3s)+c(4c+3)=4(s-c)+4(c^2-s^2).$$

6. Since $c^2-s^2=(c-s)(c+s)=c-s$, this becomes $4(s-c)+4(c-s)=0$.

7. Hence $\det (C)=0$, which is option (d). The other options are not zero for every $\theta$: at $\theta =0$ options (a), (b) and (c) give $8$, $3$ and $7$.