InterviewSolution
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InterviewSolution
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Evaluate : Δ = \(\begin{vmatrix}0& sin\,\alpha& -cos\,\alpha \\[0.3em]-sin\,\alpha& 0 & sin\,\beta \\[0.3em]cos\,\alpha & -sin\,\beta &0\end{vmatrix}\) |
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Answer» Δ = \(\begin{vmatrix}0& sin\,\alpha& -cos\,\alpha \\[0.3em]-sin\,\alpha& 0 & sin\,\beta \\[0.3em]cos\,\alpha & -sin\,\beta &0\end{vmatrix}\) Expanding along the first row, |A| = 0\(\begin{vmatrix}0& sin\,\beta \\[0.3em]-sin\,\beta& 0 \\[0.3em]\end{vmatrix}\) - sin α \(\begin{vmatrix}-sin\,\alpha& sin\,\beta \\[0.3em]cos\,\alpha& 0 \\[0.3em]\end{vmatrix}\) - cos α \(\begin{vmatrix}-sin\,\alpha& 0 \\[0.3em]cos\,\alpha& -sin\,\beta \\[0.3em]\end{vmatrix}\) ⇒ |A| = 0(0 – sinβ(–sinβ) ) –sinα(–sinα× 0 – sinβ cosα ) – cosα((–sinα)(–sinβ) – 0× cosα ) |A| = 0 + sinα sinβ cosα – cosα sinα sinβ |A| = 0 |
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