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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If A+B+C=0,then prove that $\begin{vmatrix}1& cosC &cos B\\cos C& 1 &cos A\\cos B & cos A &1\end{vmatrix}=0.$

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Toolbox:
  • If $A=\begin{vmatrix}a_{11} & a_{21} & a_{31}\\a_{12} & a_{22} & a_{32}\\a_{13} & a_{23} & a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_22\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
Let $\Delta=\begin{vmatrix}1 & cos C & cos B\\cos C & 1 & cos A\\cos B & cos A& 1\end{vmatrix}$
 
Expanding along $R_1$ we get,
 
$\Delta=1(1-cos^2A)-cos C(cos C-cos A.cos B)+cos B(cos C .cos A-cos B)$
 
$\quad=1-cos^2A-cos^2C+cos A.cos B.cos C+cos A.cos B.cos C-cos^2B.$
 
$\quad=1+2cos A.cos B.cos C-(cos^2A+cos^2B+cos^2C)$
 
But $cos^2A+cos^2B+cos^2C=1+2cos A cos B cosC.$
 
Substituting this we get,
 
$\Delta=1+2cos A cos B cos C-1-2cos A cos B cos C=0.$
 
Hence $\Delta=0.$

 

answered Mar 15, 2013 by sreemathi.v
edited Mar 15, 2013 by sreemathi.v
 

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