Let the system of equations x+2 y+3 z=5, 2 x+ | vedclass

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Question

$ x+2 y+3 z=5 $

$2 x+3 y+z=9 $

$ 4 x+3 y+\lambda z=\mu$

for infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$

$\Delta=\left|\begin{

Vedclass · Accepted Answer

$17$

Vedclass · Answer

$ x+2 y+3 z=5 $

$2 x+3 y+z=9 $

$ 4 x+3 y+\lambda z=\mu$

for infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$

$\Delta=\left|\begin{array}{lll}1 & 2 & 3 \ 2 & 3 & 1 \ 4 & 3 & \lambda\end{array}ight|=0 \Rightarrow \lambda=-13$

$\Delta_1=\left|\begin{array}{ccc}5 & 2 & 3 \ 9 & 3 & 1 \ \mu & 3 & -13\end{array}ight|=0 \Rightarrow \mu=15$

$\Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \ 2 & 9 & 1 \ 4 & 15 & -13\end{array}ight|=0$

$\Delta_3=\left|\begin{array}{ccc}1 & 2 & 5 \ 2 & 3 & 9 \ 4 & 3 & 15\end{array}ight|=0$

for $\lambda=-13, \mu=15$ system of equation has infinite solution hence $\lambda+2 \mu=17$

Let the system of equations $x+2 y+3 z=5$, $2 x+3 y+z=9,4 x+3 y+\lambda z=\mu$ have infinite number of solutions. Then $\lambda+2 \mu$ is equal to :

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