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# Evaluate $\begin{vmatrix} cos\alpha \; cos\beta & cos\alpha \; sin\beta & -sin\alpha \\ -sin\beta & cos\beta & 0 \\ sin \alpha \; cos\beta & sin\alpha \; sin\beta & cos\alpha \end{vmatrix}$

$\begin{array}{1 1} 0 \\ -1 \\ 1 \\ 2 \end{array}$

Toolbox:
• The value of a determinant can be obtained by expanding along any its rows or columns and multiply with its corresponding cofactors.
Let $\bigtriangleup= \begin{vmatrix} cos\alpha \; cos\beta & cos\alpha \; sin\beta & -sin\alpha \\ -sin\beta & cos\beta & 0 \\ sin \alpha \; cos\beta & sin\alpha \; sin\beta & cos\alpha \end{vmatrix}$

Now expanding along $R_1$ we get,

$cos\alpha cos\beta(cos\beta cos\alpha-0\times sin\alpha sin\beta)-cos\alpha sin\beta[-sin\beta cos\alpha-0]-sin \alpha(-sin\beta sin\alpha sin\beta-cos\beta sin\alpha cos \beta)$

On simplifying we get,

$=cos\alpha cos\beta(cos\beta cos\alpha)+cos\alpha sin\beta(sin\beta cos\alpha)+sin^2\alpha.sin^2\beta+sin^2\alpha cos^2\beta$

$=cos^2\alpha cos^2\beta+cos^2\alpha sin^2\beta+sin^2\alpha sin^2\beta+sin^2\alpha cos^2\beta$

Taking the common factors,

$=sin^2\alpha(sin^2\beta+cos^2\beta)+cos^2\alpha(sin^2\beta+cos^2\beta)$

$=(sin^2\beta+cos^2\beta)(sin^2\alpha+cos^2\alpha)$

But $sin^2\beta+cos^2\beta=sin^2\alpha+cos^2\alpha$=1.

Therefore $\mid\bigtriangleup\mid$=1.

Hence the value of the determinant is 1.

edited Mar 1, 2013