If the system of equations

2 x+7 y+lambda | vedclass

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Question

$\mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0$

$D_3=\left|\begin{array}{ccc}2 & 7 & 3 \ 3 & 2 & 4 \ 1 & \mu & -1\e

Vedclass · Accepted Answer

$38$

Vedclass · Answer

$\mathrm{D}=\mathrm{D}_1=\mathrm{D}_2=\mathrm{D}_3=0$

$D_3=\left|\begin{array}{ccc}2 & 7 & 3 \ 3 & 2 & 4 \ 1 & \mu & -1\end{array}ight|=0 \Rightarrow \mu=-39$

$\mathrm{D}=\left|\begin{array}{ccc}2 & 7 & \lambda \ 3 & 2 & 5 \ 1 & -39 & 32\end{array}ight|=0 \Rightarrow \lambda=-1$

$\lambda-\mu=38$

If the system of equations

$ 2 x+7 y+\lambda z=3 $

$ 3 x+2 y+5 z=4 $

$ x+\mu y+32 z=-1$

has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

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If the system of equations $ 2 x+7 y+\lambda z=3 $ $ 3 x+2 y+5 z=4 $ $ x+\mu y+32 z=-1$ has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$