Consider the system of linear equation x+y+z= | vedclass

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Question

$ x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda $ $ { }^2 z=\mu^2+15$

$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 &

Vedclass · Accepted Answer

The system is inconsistent if $\lambda=\frac{1}{2}$ and $\mu 
eq 1$

Vedclass · Answer

$ x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda $ $ { }^2 z=\mu^2+15$

$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & 2 \lambda \ 1 & 3 & 4 \lambda^2\end{array}ight|=(2 \lambda-1)^2$

For unique solution $\Delta 
eq 0,2 \lambda-1 
eq 0,\left(\lambda 
eq \frac{1}{2}ight)$

Let $\Delta=0, \lambda=\frac{1}{2}$

$\Delta_{\mathrm{y}}=0, \Delta_{\mathrm{x}}=\Delta_{\mathrm{z}}=\left|\begin{array}{ccc}4 \mu & 1 & 1 \ 10 \mu & 2 & 1 \ \mu^2+15 & 3 & 1\end{array}ight|$

$=(\mu-15)(\mu-1)$

For infinite solution $\lambda=\frac{1}{2}, \mu=1$ or 15

Consider the system of linear equation $x+y+z=$ $4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15$, where $\lambda, \mu \in R$. Which one of the following statements is $NOT$ correct?

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