Matrices And Determinants Ques 94
Let $\lambda$ be a real number for which the system of linear equations
$x+y+z=6,4 x+\lambda y-\lambda z=\lambda-2$ and
$3 x+2 y-4 z=-5$
has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation
(2019 Main, 10 April II)
(a) $\lambda^{2}-3 \lambda-4=0$
(b) $\lambda^{2}+3 \lambda-4=0$
(c) $\lambda^{2}-\lambda-6=0$
(d) $\lambda^{2}+\lambda-6=0$
Show Answer
Answer:
Correct Answer: 94.(c)
Solution:
Formula:
System of equations with 3 variables:
- Given a system of linear equations
$ \begin{aligned} x+y+z & =6 \\ 4 x+\lambda y-\lambda z & =\lambda-2 \end{aligned} $
and $\quad 3 x+2 y-4 z=-5$
has infinitely many solutions, then $\Delta=0$
$\Rightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4\end{array}\right|=0$
$\Rightarrow 1(-4 \lambda+2 \lambda)-1(-16+3 \lambda)+1(8-3 \lambda)=0$
$\Rightarrow-8 \lambda+24=0 \quad \Rightarrow \quad \lambda=3$
From, the option $\lambda=3$, satisfies the quadratic equation $\lambda^{2}-\lambda-6=0$
BETA
