Matrices And Determinants Ques 96
If the system of equations $x-k y-z=0, k x-y-z=0$, $x+y-z=0$ has a non-zero solution, then possible values of $k$ are
$(2000,2 \mathrm{M})$
(a) $-1,2$
(b) $1,2^{}$
(c) $0.1$
(d) $-1,1$
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Answer:
Correct Answer: 96.(d)
Solution:
Formula:
System of equations with 3 variables:
Since the given system has a non-zero solution.
$ \therefore \quad \begin{vmatrix} & 1 & -k & -1 \ & k & -1 & -1 \\ & 1 & 1 & -1 \end{vmatrix}=0 $
Applying $C_{1} \rightarrow C_{1}-C_{2}, C_{2} \rightarrow C_{2}+C_{3}$
$ \begin{vmatrix} & 1+k & -k-1 & -1 \\ & 1+k & -2 & -1 & =0 \\ & 0 & 0 & -1 \end{vmatrix} $
$ \Rightarrow 2(k+1)-(k+1)^2=0$
$ \Rightarrow (k+1)(1-k)=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.
BETA
