Question

The number of real values of $\lambda$ for which the system of linear equations. $2x+4y-\lambda z=0$, $4x+\lambda y+2z=0,\lambda x+2y+2z=0$ has infinitely many solutions, is

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is A $1$

For infinitely many solutions, $D=0,D_x=0,D_y=0$ and $D_z=0$

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

$=2(2\lambda -4)-4(8-2\lambda )-\lambda (8-{\lambda }^2)$

$={\lambda }^3+4\lambda -40$

If $D=0$, then

${\lambda }^3+4\lambda -40=0$

Again, $D_x=0$

$D_x=\left[\begin{array}{ccc}0& 4& -\lambda \\ 0& \lambda & 2\\ 0& 2& 2\end{array}\right]=0$

Similarly,

$D_y=0$ and $D_z=0$

The equation, ${\lambda }^3+4\lambda -40$ has only one solution

Since $\lambda (-\infty )=-\infty$ and $\lambda \left(\infty \right)=\infty$

Here $\alpha$ is only one solution, so $\lambda \left(a\right)=0$

Using intermediate value property

Now, differentiating ${\lambda }^3+4\lambda -40$ w.r.t $\lambda$ we get

$3{\lambda }^2+4>0$

The equation can not have y.m, so

$\lambda \left(m\right)=0$ and $\lambda \left(y\right)=0$

Thus, the number of real value of $\lambda$ is $1$.