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 the system of equations :$x-ky-z=0,kx-y-z=0,x+y-z=0$ has a non zero solutions then the possible value of k are

(A) -1,2

(B) 1,2

(C) 0,1

(D) -1,1

Can you answer this question?
 
 

1 Answer

0 votes
For the given homogeneous system to have non zero solution determinant of coefficient matrix should be zero.
(i.e) $\begin{vmatrix}1 &-k &-1\\k&-1&-1\\1 &1&-1\end{vmatrix}$
$\Rightarrow 1(1+1)+k(-k+1)-1(k+1)=0$
$\Rightarrow 2-k^2+k-k-1=0$
$k^2=1$
$k=\pm 1$
Hence (d) is the correct answer.
answered Nov 20, 2013 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
...