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)

$D = \begin{vmatrix}\lambda&3&2\\ 2\lambda&3&5\\ 4&\lambda&6\end{vmatrix} = \left(\lambda+8\right)\left(2-\lambda\right)$
for $\lambda = 2 ; D_{1} \ne 0$
Hence, no solution for $\lambda= 2$