logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
0 votes

The value of $\begin{vmatrix}1&a&a^2-bc\\1&b& b^2-ca\\1&c&c^2-ab\end{vmatrix}$ is

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;(a-b)(b-c)(c-a)&(D)\;\text{None of these}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
$\Delta=\begin{vmatrix}1&a&a^2\\1&b& b^2\\1&c&c^2\end{vmatrix}-\begin{vmatrix}1&a&bc\\1&b& ca\\1&c&ab\end{vmatrix}$
$\Rightarrow \begin{vmatrix}1&a&a^2\\1&b& b^2\\1&c&c^2\end{vmatrix}-\large\frac{1}{abc}$$\begin{vmatrix}a&a^2&abc\\b&b^2& abc\\c&c^2&abc\end{vmatrix}$
$\Rightarrow \begin{vmatrix}1&a&a^2\\1&b& b^2\\1&c&c^2\end{vmatrix}-\large\frac{abc}{abc}$$\begin{vmatrix}a&a^2&1\\b&b^2& 1\\c&c^2&1\end{vmatrix}$
$\Rightarrow \begin{vmatrix}1&a&a^2\\1&b& b^2\\1&c&c^2\end{vmatrix}-\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$
$\Rightarrow 0$
Hence (A) is the correct answer.
answered Apr 22, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...