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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Matrices

The value of $\begin{vmatrix}a&a+b&a+2b\\a+2b&a&a+b\\a+b&a+2b&a\end{vmatrix}$ is equal to

$\begin{array}{1 1}(A)\;9a^2(a+b)&(B)\;9b^2(a+b)\\(C)\;a^2(a+b)&(D)\;b^2(a+b)\end{array}$

1 Answer

Given :
$\begin{vmatrix}a&a+b&a+2b\\a+2b&a&a+b\\a+b&a+2b&a\end{vmatrix}$
Applying $C_1=C_1+C_+C_3$ we get
$\begin{vmatrix}3a+3b&a+b&a+2b\\3a+3b&a&a+b\\3a+3b&a+2b&a\end{vmatrix}$
$3(a+b)\begin{vmatrix} 1&a+b&a+2b\\1&a&a+b\\1&a+2b&a\end{vmatrix}$
Applying $R_3=R_3-R_1,R_2-R_1=R_2$ we get
$3(a+b)\begin{vmatrix} 1&a+b&a+2b\\0&-b&a+b\\0&a+2b&a\end{vmatrix}$
$3(a+b)(2b^2+b^2)=(3a+3b)(2b^2+b^2)$
$\Rightarrow 6ab^2+3ab^2+6b^3+3b^3$
$\Rightarrow 9ab^2+9b^3$
$\Rightarrow 9b^2(a+b)$
Hence (B) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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