Topic: Engineering Mathematics
Let $C=\begin{pmatrix}\sin^2\theta & 1 & \cos^2\theta\\\cos^2\theta & 1 & \sin^2\theta\\-3 & 4 & 7\end{pmatrix}$
Then det (C) will be
- A. $7\cos^2\theta+3\sin^2\theta+1$
- B. $4\sin^2\theta+3\cos^2\theta$
- C. $7-4\sin^2\theta$
- D. 0
Answer: D. 0
Explanation: To evaluate the determinant of matrix C, apply the column operation: Replace Column 1 with (Column 1 + Column 3).
The new Column 1 becomes:
Row 1: $\sin^2\theta + \cos^2\theta = 1$
Row 2: $\cos^2\theta + \sin^2\theta = 1$
Row 3: $-3 + 7 = 4$
The matrix is now $\begin{pmatrix}1 & 1 & \cos^2\theta \\\ 1 & 1 & \sin^2\theta \\\ 4 & 4 & 7\end{pmatrix}$.
Because Column 1 and Column 2 are identical, the determinant of the matrix is 0.