1. Tardigrade
  2. Question
  3. Mathematics
  4. If the system of equations x - k y - z = 0, kx - y - z = 0, x + y - z = 0 has a non-zero solution, then possible values of k are

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If the system of equations $x - k y - z = 0, kx - y - z = 0, x + y - z = 0$ has a non-zero solution, then possible values of $k$ are

IIT JEEIIT JEE 2000Determinants

Solution:

Since, the given system has non-zero solution.
$\therefore \begin {vmatrix}1 & -k & -1 \\k & -1 & a-1\\1 & 1 & -1 \end {vmatrix} =0$
Applying $C_1 \rightarrow C_1-C_2,C_2 \rightarrow C_2+C_3$
$\Rightarrow \begin {vmatrix}1+k & -k-1 & -1 \\1+k & -2 & -1 \\0 & 0 & -1 \end {vmatrix}=0$
$\Rightarrow 2(k+1)-(k+1)^2=0$
$\Rightarrow (k+1)(2-k-1)=0 \Rightarrow k= \pm 1$
NOTE There is a golden rule in determinant that $n$ one's $\Rightarrow (n - 1)$ zero's or $n$ (constant) $\Rightarrow (n - 1)$ zero's for all constant should be in a single row or a single column.