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 soiution 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

Answer: (A)

$D=\left|\begin{array}{ccc}\lambda & 3& 2\\ 2\lambda & 3& 5\\ 4& \lambda & 6\end{array}\right|=\left(\lambda +8\right)\left(2-\lambda \right)$
for $\lambda =2;D_1\ne 0$
Hence, no solution for $\lambda =2$