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

If $\omega$ is cube root of unity then $\Delta=\begin{vmatrix}x+1&\omega&\omega^2\\\omega&x+\omega^2&1\\\omega^2&1&x+\omega^2\end{vmatrix}=$

$\begin{array}{1 1}(A)\;x^3+1&(B)\;x^3=\omega\\(C)\;x^3+\omega^2&(D)\;x^3\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Applying $C_1=C_1+C_2+C_3$
Using $1+\omega+\omega^2=0$
$\Delta=\begin{vmatrix}x+1&\omega&\omega^2\\\omega&x+\omega^2&1\\\omega^2&1&x+\omega^2\end{vmatrix}$
$\Rightarrow x\begin{vmatrix}1&\omega&\omega^2\\1&x+\omega^2&1\\1&1&x+\omega^2\end{vmatrix}$
$\Rightarrow x\begin{vmatrix}1&\omega&\omega^2\\0&x+\omega^2-\omega&1-\omega^2\\0&1-\omega&x+\omega-\omega^2\end{vmatrix}$
$\Rightarrow x[(x+\omega-\omega^2)(x+\omega^2-\omega)-(1-\omega^2)(1-\omega)]$
$\Rightarrow x[x^2+1-\omega^2-\omega+1-1+\omega+\omega^2-1]$
$\Rightarrow x(x^2)$
$\Rightarrow x^3$
Hence (D) is the correct answer.
answered Apr 23, 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
...