Evaluate Δ=∣0sin⁡α−cos⁡α−sin⁡α0sin⁡βcos⁡α−sin⁡β0∣\Delta=\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right| Solution Expanding along R1\mathrm{R}_{1}, we get Δ=0∣0sin⁡β−sin⁡β0∣−sin⁡α∣−sin⁡αsin⁡βcos⁡α0∣−cos⁡α∣−sin⁡α0cos⁡α−sin⁡β∣=0−sin⁡α(0−sin⁡βcos⁡α)−cos⁡α(sin⁡αsin⁡β−0)=sin⁡αsin⁡βcos⁡α−cos⁡αsin⁡αsin⁡β=0\begin{aligned} \Delta & =0\left|\begin{array}{cc} 0 & \sin \beta \\ -\sin \beta & 0 \end{array}\right|-\sin \alpha\left|\begin{array}{cc} -\sin \alpha & \sin \beta \\ \cos \alpha & 0 \end{array}\right|-\cos \alpha\left|\begin{array}{cc} -\sin \alpha & 0 \\ \cos \alpha & -\sin \beta \end{array}\right| \\ & =0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0) \\ & =\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0 \end{aligned}

Answer: The value of the determinant is 00 .

Step-by-step solution

Step 1: Expand along the first row

To evaluate the determinant of a 3×33 \times 3 matrix, we can expand along any row or column. Here, we choose to expand along the first row ( R1R_1 ). The formula for expansion along R1R_1 is a11C11+a12C12+a13C13a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} , where aija_{ij} are the elements and CijC_{ij} are their cofactors. The cofactors include a sign factor (−1)i+j(-1)^{i+j} .

Step 2: Calculate the 2x2 determinants

Now, we evaluate each of the 2×22 \times 2 determinants. The determinant of a 2×22 \times 2 matrix ∣abcd∣\left|\begin{array}{cc}a & b \\ c & d\end{array}\right| is given by ad - bc. We apply this formula to each of the three 2×22 \times 2 sub-determinants.

Step 3: Simplify the expression

We simplify the terms inside the parentheses. The first term becomes 0(0+sin⁡2β)=00(0 + \sin^2 \beta) = 0 . The second term simplifies to −sin⁡α(−sin⁡βcos⁡α)-\sin \alpha(-\sin \beta \cos \alpha) . The third term simplifies to −cos⁡α(sin⁡αsin⁡β)-\cos \alpha(\sin \alpha \sin \beta) .

Step 4: Combine and find the final value

After simplifying, we get two terms: sin⁡αsin⁡βcos⁡α\sin \alpha \sin \beta \cos \alpha and −cos⁡αsin⁡αsin⁡β-\cos \alpha \sin \alpha \sin \beta . These two terms are identical but with opposite signs.

Step 5: Final Result

Since the two terms are identical with opposite signs, they cancel each other out, resulting in a final determinant value of 00 .

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