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Home  >>  CBSE XII  >>  Math  >>  Determinants
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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} $

Can you answer this question?
 
 

1 Answer

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  • 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.

 

answered Feb 28, 2013 by sreemathi.v
edited Mar 1, 2013 by balaji.thirumalai
 

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