Question

Investigate for what values of $\lambda ,\mu$ the simultaneous equation $x+y+z=6;x+2y+3z=10$ & $x+2y+\lambda z=\mu$ have No solution

• A. $\lambda =3,\mu \ne 10$
• B. $\lambda =10,\mu \ne 3$
• C. $\lambda \ne 3,\mu =10$
• D. $\lambda =3,\mu \ne 10,3$

Answer 1

The correct option is A $\lambda =3,\mu \ne 10$

Given,

$x+y+z=6$

$x+2y+3z=10$

$x+2y+\lambda z=\mu$

We have,

$a_{11}=1,a_{12}=1,a_{13}=1,b_{11}=1,b_{12}=2,b_{13}=3,c_{11}=1,c_{12}=2,c_{13}=\lambda ,d_{11}=6,d_{12}=10,d_{13}=\mu$

By cramer's rule we have,

$\Delta =\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ b_{11}& b_{12}& b_{13}\\ c_{11}& c_{12}& c_{13}\end{array}\right|$

${\Delta }_1=\left|\begin{array}{ccc}{d}_1& a_{12}& a_{13}\\ d_2& b_{12}& b_{13}\\ d_3& c_{12}& c_{13}\end{array}\right|$

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 2& \lambda \end{array}\right|=1(2\lambda -6)-1(\lambda -3)+1(2-2)=2\lambda -6-\lambda +3+0=\lambda -3{\Delta }_1=\left|\begin{array}{ccc}6& 1& 1\\ 10& 2& 3\\ \mu & 2& \lambda \end{array}\right|=6(2\lambda -6)-1(10\lambda -3\mu )+1(20-2\mu )=12\lambda -36-10\lambda +3\mu +20-2\mu =2\lambda +\mu -16$

For no solution $\Delta =0,{\Delta }_1\ne 0\Rightarrow \lambda =3\Rightarrow \mu \ne 10$