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  • If the system of equations <img alt="\begin{aligned} & x+2 y+3 z=3 \\ & 4 x+3 y-4 z=4 \\ & 8 x+4 y-\lambda z=9+\mu \end{aligned}" src="https://entrancecorner.oncodecogs.com/gi

If the system of equations

\begin{aligned} & x+2 y+3 z=3 \\ & 4 x+3 y-4 z=4 \\ & 8 x+4 y-\lambda z=9+\mu \end{aligned}

has infinitely many solutions, then the ordered pair (\lambda ,\mu ) is equal to :

Option: 1

\left(-\frac{72}{5}, \frac{21}{5}\right)


Option: 2

\left(-\frac{72}{5} \cdot-\frac{21}{5}\right)


Option: 3

\left(\frac{72}{5},-\frac{21}{5}\right)


Option: 4

\left(\frac{72}{5}, \frac{21}{5}\right)


Answers (1)

Planes are not parallel

\begin{aligned} & \therefore(x+2 y+3 z-3)+a(4 x+3 y-4 z-4) \\ & =8 x+4 y-\lambda z-9-\mu=0 \\ & \frac{1+4 a}{8}=\frac{2+3 a}{4}=\frac{3-4 a}{-\lambda}=\frac{-3-4 a}{-9-\mu} \end{aligned}

(i) 1+4 \mathrm{a}=4+6 \mathrm{a}

$$ \mathrm{a}=\frac{-3}{2}

\text { (ii) } \frac{2-\frac{9}{2}}{4}=\frac{3+6}{-\lambda}

\begin{aligned} & -\lambda=\frac{36}{-5} \times 2 \\ & \lambda=\frac{72}{5} \end{aligned}

\begin{aligned} & \text { (iii) } \frac{-5}{8}=\frac{-3-4 a}{-9-\mu} \\ & \frac{5}{8}=\frac{3-6}{-9-\mu} \\ & -9-\mu=\frac{-24}{5} \end{aligned}

\begin{aligned} \mu & =\frac{-45+24}{5} \\ \mu & =\frac{-21}{5} \end{aligned}

Posted by

Kshitij

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