Question

The system of equations $3x-y+4z=2;x+2y-z=3;4x-y+\lambda z=-1$ has no solution. The value of $\lambda$ is

• A. $1$
• B. $3$
• C. $5$
• D. $6$

Answer 1

Answer: (C)

We have

$3x-4+4z=2$

$x+2y-z=3$

$4x-y+\lambda z=-1$

Now,

$\Delta =\left[\begin{array}{ccc}3& -1& 4\\ 1& 2& -1\\ 4& -1& \lambda \end{array}\right]$

$=3\left(2\lambda -1\right)+\left(\lambda +4\right)+4\left(-1-8\right)$

$=7\lambda +1-36$

$=7\lambda -35$

${\Delta }_1=\left[\begin{array}{ccc}2& -1& 4\\ 3& 2& -1\\ -1& -1& \lambda \end{array}\right]$

$=2\left(2\lambda -1\right)+\left(3\lambda -1\right)+4\left(-3+2\right)$

$=7\lambda +5$

And,

${\Delta }_2=\left[\begin{array}{ccc}3& 2& 4\\ 1& 3& -1\\ 4& -1& \lambda \end{array}\right]$

$=7\lambda -63$

For $\lambda =5$ No solution

For $n$ solution $\Delta =0$

${\Delta }_1,{\Delta }_2,{\Delta }_3\ne 0$

Hence, the option $\left(C\right)$ is the correct answer.