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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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If $A,B,C$ are the angles of a $\Delta$le then the value of the determinant $\small\begin{vmatrix}-1+\cos B&\cos C+\cos B&\cos B\\\cos C+\cos A&-1+\cos A&\cos A\\-1+\cos B&-1+\cos B&-1\end{vmatrix}$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;-1\qquad(d)\;2$

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1 Answer

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We have
$\begin{vmatrix}-1+\cos B&\cos C+\cos B&\cos B\\\cos C+\cos A&-1+\cos A&\cos A\\-1+\cos B&-1+\cos B&-1\end{vmatrix}$
Apply $C_1\Rightarrow C_1-C_3$ and $C_2\rightarrow C_2-C_3$
$\begin{vmatrix}-1&\cos C&\cos B\\cos C&-1&\cos A\\\cos B&\cos A&-1\end{vmatrix}$
Multiply and divide by a in column 1
Apply $C_1\rightarrow C_1+bC_2+cC_3$
$\large\frac{1}{a}$$\begin{vmatrix}-a&\cos C&\cos B\\a\cos C&-1&\cos A\\acos B&\cos A&-1\end{vmatrix}$
$\Rightarrow \large\frac{1}{a}$$\begin{vmatrix}0&\cos C&\cos B\\0&-1&\cos A\\0&\cos A&-1\end{vmatrix}$
$\Rightarrow 0$
Hence (a) is the correct answer.
answered Nov 25, 2013 by sreemathi.v
edited Mar 20, 2014 by sharmaaparna1
 

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