Question

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

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

Answer 1

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

$x+y+z=6,x+2y+3z=0,x+2y+\lambda z=\mu$ has no solution
$\left[\begin{array}{ccccc}1& 1& 1& .& 6\\ 1& 2& 3& .& 10\\ 1& 2& \lambda & .& \mu \end{array}\right]$
$\left[\begin{array}{ccccc}1& 1& 1& .& 6\\ 1& 2& 3& .& 10\\ 0& 0& 3-\lambda & .& 10-\mu \end{array}\right]$
Now $\left|A\right|=0$ if $\lambda =3$
for $\lambda =3$ either infinite solutions exist or no solution exist for no solution $\mu \ne 10$ if $\mu =0$ infinite solutions exist