Question

The value of the determinant \(\begin{vmatrix}acosθ&bsinθ&0 \\-bsinθ&acosθ&0\\ 0&0&c\end{vmatrix}\) is:

• A. (a²+b²)c
• B. (a²cos²θ+b²sin²θ)c
• C. (a²cos²θ-b²sin²θ)c
• D. (a²+b²)c²

Answer 1

The Correct Option is B

To find the value of the determinant of the matrix \(A = \begin{bmatrix} acosθ & bsinθ & 0 \\ -bsinθ & acosθ & 0 \\ 0 & 0 & c \end{bmatrix}\), we can use properties of determinants. The matrix is of order 3x3, and thus, a determinant can be expanded along any row or column. Here, expanding along the third column, which contains two zeros, simplifies computation:

1. Identify the components of column 3: \( \begin{bmatrix} 0 \\ 0 \\ c \end{bmatrix} \)

2. Using the cofactor expansion along column 3, the determinant of matrix \(A\) is given by \(c \; \text{det}\left(\begin{bmatrix} acosθ & bsinθ \\ -bsinθ & acosθ \end{bmatrix}\right).\)
The determinants of each 2x2 minors can be calculated by:

\(\text{det}\left(\begin{bmatrix} acosθ & bsinθ \\ -bsinθ & acosθ \end{bmatrix}\right) = (acosθ)(acosθ) - (bsinθ)(-bsinθ)\)

\(= a²cos²θ + b²sin²θ\)

3. Thus, the determinant of matrix \(A\) becomes:

\(\text{det}(A) = c(a²cos²θ + b²sin²θ)\)

So, the value of the determinant is (a²cos²θ+b²sin²θ)c, which corresponds to option B.