lambda के उन भिन्न मानों का योग, जिनके लिए सम | vedclass

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Question

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

$3$

Vedclass · Answer

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+(3 \lambda-3) z=0$

$\left|\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \ \lambda-1 & 4 \lambda-2 & \lambda+3 \ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}ight|=0$

$R _{1} ightarrow R _{1}- R _{2}$ and $R _{2} ightarrow R _{2}- R _{3}$

$\left|\begin{array}{ccc}0 & 3-\lambda & \lambda-3 \ \lambda-3 & \lambda-3 & -2(\lambda-3) \ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}ight|=0$

$(\lambda-3)^{2}\left|\begin{array}{ccc}0 & -1 & 1 \ 1 & 1 & -2 \ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}ight|=0$

$(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$

$6 \lambda(\lambda-3)^{2}=0 \Rightarrow \lambda=0,3$

$\operatorname{Sum}=3$

Similar Questions

एक ऐसा क्रमित युग्म $(\alpha, \beta)$ जिसके लिये रैखिक समीकरण निकाय $(1+\alpha) x +\beta y + z =2$, $\alpha x +(1+\beta) y + z =3$, $\alpha x +\beta y +2 z =2$ का एकमात्र एक हल है

सारणिक $\left| {\,\begin{array}{*{20}{c}}2&8&4\\{ - 5}&6&{ - 10}\\1&7&2\end{array}\,} \right|$ का मान है

यदि $\alpha+\beta+\gamma=2 \pi$ है, तो समीकरण निकाय

$x+(\cos \gamma) y+(\cos \beta) z=0$

$(\cos \gamma) x+y+(\cos \alpha) z=0$

$(\cos \beta) x+(\cos \alpha) y+z=0$

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $

$x$ के मान ज्ञात कीजिए यदि

$\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{ll}x & 3 \\ 2 x & 5\end{array}\right|$