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Using properties of determinants, prove that: $\begin{vmatrix} sin\alpha&cos\alpha&cos(\alpha+\delta)\\ sin\beta&cos\beta&cos(\beta+\delta)\\ sin\gamma&cos\gamma&cos(\gamma+\delta)) \end{vmatrix}= 0.$

1 Answer

  • If each element of a row(or a column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
  • By this property we can take out any common factor from any one row or any column .
  • Elementary transformations can be done by
  • 1. Interchanging any two rows or columns.
  • 2. Mutiplication of the elements of any row or column by a non-zero number
  • 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
$\Delta=\begin{vmatrix} sin\alpha&cos\alpha&cos(\alpha+\delta)\\ sin\beta&cos\beta&cos(\beta+\delta)\\ sin\gamma&cos\gamma&cos(\gamma+\delta)) \end{vmatrix}$
Multiply $C_1$ by $\sin\delta$ and $C_2$ by $\cos\delta$
$\Delta=\frac{1}{sin\delta cos\delta}\begin{vmatrix} sin\alpha sin\delta&cos\alpha cos\delta&cos(\alpha+\delta)\\ sin\beta sin\delta&cos\beta cos\delta&cos(\beta+\delta)\\ sin\gamma sin\delta&cos\gamma cos\delta&cos(\gamma+\delta)) \end{vmatrix}$
We know cos(A+B)=cos Acos B-sin Asin B.
$\Delta=\frac{1}{sin\delta cos\delta}\begin{vmatrix} sin\alpha sin\delta&cos\alpha cos\delta&cos\alpha cos\delta-sin\alpha sin\delta\\ sin\beta sin\delta&cos\beta cos\delta&cos\beta cos\delta-sin\beta sin\delta\\ sin\gamma sin\delta&cos\gamma cos\delta&cos\gamma cos\delta-sin\beta sin\delta \end{vmatrix}$
Apply $C_1\rightarrow C_1+C_3$
$\Delta=\frac{1}{sin\delta cos\delta}\begin{vmatrix} cos\alpha cos\delta&cos\alpha cos\delta&cos\alpha cos\delta-sin\alpha sin\delta\\ cos\beta cos\delta&cos\beta cos\delta&cos\beta cos\delta-sin\beta sin\delta\\ cos\gamma cos\delta&cos\gamma cos\delta&cos\gamma cos\delta-sin\beta sin\delta \end{vmatrix}$
Since two columns $C_1$ and $C_2$ the value of the determinant is 0.
Therefore $\Delta=0.$
Hence proved.


answered Mar 3, 2013 by sreemathi.v
edited Mar 5, 2013 by vijayalakshmi_ramakrishnans

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