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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $f(x)=\begin{vmatrix}0 & x-a & x-b\\x-a & 0 & x-c\\x-b & x-c & 0\end{vmatrix},then$\[\begin{array}{1 1}(A)\quad f(a)=0 & (B)\quad f(b)=0\\(C)\quad f(0)=0 & (D)\quad f(1)=0\end{array}\]

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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 $\Delta=f(x)=\begin{vmatrix}0 & x-a & x-b\\x-a & 0 & x-c\\x-b & x-c & 0\end{vmatrix}$
Expanding along $R_1$ we get,
$\;\;\;=0-(x-a)[0-(x-b)(x-c)]+(x-b)[(x-a)(x-c)-0]$
$f(x)=(x-a)(x-b)(x-c)+(x-a)(x-b)(x-c)=2(x-a)(x-b)(x-c)$
Therefore f(a)=2(a-a)(a-b)(a-c)=0
Hence correct answer is A
answered Mar 26, 2013 by sreemathi.v
 

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