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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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?

Similar Questions

Number of values of $m$ for which the lines $x + y - 1 = 0$, $(m - 1) x + (m^2 - 7) y - 5 = 0 \,\,\&\,\, (m - 2) x + (2m - 5) y = 0$ are concurrent, are

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\b&c&{b + c}\\{a + b}&{b + c}&0\end{array}\,} \right| = 0$; then $a,b,c$ are in

The system of equations $kx + y + z =1$ $x + ky + z = k$ and $x + y + zk = k ^{2}$ has no solution if $k$ is equal to

The system of equations $x + y + z = 6$, $x + 2y + 3z = 10,x + 2y + \lambda z = \mu $, has no solution for

If the system of linear equations $2 \mathrm{x}+2 \mathrm{ay}+\mathrm{az}=0$ ; $2 x+3 b y+b z=0$ ; $2 \mathrm{x}+4 \mathrm{cy}+\mathrm{cz}=0$ ; where $a, b, c \in R$ are non-zero and distinct; has a non-zero solution, then