Question

The system of linear equations
$\lambda x+2y+2z=5$
$2\lambda x+3y+5z=8$
$4x+\lambda y+6z=10$ has:

• A. no solution when $\lambda =2$
• B. infinitely many solutions when $\lambda =2$
• C. no solution when $\lambda =8$
• D. a unique solution when $\lambda =-8$

Answer 1

The correct option is A no solution when $\lambda =2$

Using Cramer's Rule
$D=\left[\begin{array}{ccc}\lambda & 2& 2\\ 2\lambda & 3& 5\\ 4& \lambda & 6\end{array}\right]\Rightarrow D=18\lambda -5{\lambda }^2-24\lambda +40+4{\lambda }^2-24$
$\Rightarrow D=-{\lambda }^2-6\lambda +16$
Now, $D=0$
$\Rightarrow {\lambda }^2+6\lambda -16=0$
$\Rightarrow \lambda =-8\text{or}2$
For $\lambda =2$
$D_1=\left[\begin{array}{ccc}5& 2& 2\\ 8& 3& 5\\ 10& 2& 6\end{array}\right]=40+4-28\ne 0$
$\therefore$ Given system of equations have no solution for $\lambda =2$