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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
The sum of distinct values of $\lambda$ for which the system of equations

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$

has non-zero solutions, is

  • ✓
    $3$
  • B
    $0$
  • C
    $6$
  • D
    $9$
✓

Answer

Correct option: A.
$3$
a
$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+(3 \lambda-3) z=0$

$\left|\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$R _{1} \rightarrow R _{1}- R _{2}$ and $R _{2} \rightarrow R _{2}- R _{3}$

$\left|\begin{array}{ccc}0 & 3-\lambda & \lambda-3 \\ \lambda-3 & \lambda-3 & -2(\lambda-3) \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$(\lambda-3)^{2}\left|\begin{array}{ccc}0 & -1 & 1 \\ 1 & 1 & -2 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$

$6 \lambda(\lambda-3)^{2}=0 \Rightarrow \lambda=0,3$

$\operatorname{Sum}=3$

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