Matrices and Determinants 4 Question 7
7. The set of all values of $\lambda$ for which the system of linear equations $x-2 y-2 z=\lambda x, x+2 y+z=\lambda y$ and $-x-y=\lambda z$
has a non-trivial solution
(2019 Main, 12 Jan II)
(a) contains exactly two elements.
(b) contains more than two elements.
(c) is a singleton.
(d) is an empty set.
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Answer:
Correct Answer: 7. (c)
Solution:
- The given system of linear equations is
$ \begin{aligned} x-2 y-2 z & =\lambda x \\ x+2 y+z & =\lambda y \\ -x-y-\lambda z & =0, \end{aligned} $
which can be rewritten as
$ \begin{aligned} (1-\lambda) x-2 y-2 z & =0 \\ x+(2-\lambda) y+z & =0 \\ x+y+\lambda z & =0 \end{aligned} $
Now, for non-trivial solution, we should have
$ \begin{aligned} & \left|\begin{array}{ccc} 1-\lambda & -2 & -2 \\ 1 & 2-\lambda & 1 \\ 1 & 1 & \lambda \end{array}\right|=0 \\ & {\left[\because \text { If } a_{1} x+b_{1} y+c_{1} z=0 ; a_{2} x+b_{2} y+c_{2} z=0\right.} \\ & \left.a_{3} x+b_{3} y+c_{3} z=0\right] \\ & \text { has a non-trivial solution, then }\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=0 \\ & \Rightarrow \quad(1-\lambda)[(2-\lambda) \lambda-1]+2[\lambda-1]-2[1-2+\lambda]=0 \\ & \Rightarrow \quad(\lambda-1)\left[\lambda^{2}-2 \lambda+1+2-2\right]=0 \\ & \Rightarrow \quad(\lambda-1)^{3}=0 \\ & \Rightarrow \quad \lambda=1 \end{aligned} $
