Question

If the system of linear equations $x+y+z=6,x+2y+3z=14and2x+5y+\lambda z=\mu ,(\lambda ,\mu ϵR)$ has no solution, then

• A. $\lambda \ne 8$
• B. $\lambda =8,\mu \ne 36$
• C. $\lambda =8,\mu =36$
• D. $Noneofthese$

Answer 1

The correct option is B $\lambda =8,\mu \ne 36$

$\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& 2& 3& 14\\ 2& 5& \lambda & \mu \end{array}\right]R_2\to R_2-R_1;R_3\to R_3-2R_1\bigsqcup \left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 8\\ 0& 3& \lambda -2& \mu -12\end{array}\right]R_3\to R_3-3R_2\bigsqcup \left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 8\\ 0& 0& \lambda -8& \mu -36\end{array}\right]\lambda -8=0\&\mu -36\ne 0$