The number of real values of [ ] for which the system of linear equations 2x + 4y - [ ]z =…

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 :

Accepted Answer

The system of equations can be written in the matrix form as

$$\left[ {\begin{pmatrix} 2 & 4 & { - \lambda }  \
   4 & \lambda  & 2  \
   \lambda  & 2 & 2  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} x  \
   y  \
   z  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 0  \
   0  \
   0  \

\end{pmatrix} } \right]$$

The system has infinite solutions; thus, we get

$$\left| {\begin{pmatrix} 2 & 4 & { - \lambda }  \
   4 & \lambda  & 2  \
   \lambda  & 2 & 2  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow 0 = 2(2\lambda  - 4) - 4(8 - 2\lambda ) - \lambda (8 - {\lambda ^2})$$

$$ \Rightarrow 4\lambda  - 8 - 32 + 8\lambda  - 8\lambda  + {\lambda ^3} = 0$$

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

We can solve this by graphical method:

$y = {x^3} + 4x - 40$

For x = 0, y = -40: If we take y = -40, then we have

$- 40 = {x^3} + 4x - 40$

$$ \Rightarrow {x^3} + 4x = 0$$

$$ \Rightarrow x({x^2} + 4) = 0$$

$$ \Rightarrow x = 0,{x^2} + 4 = 0$$

$$ \Rightarrow x =  \pm \,2i$$

The given equation of line intersects x only at one point; therefore, the real value of $$\lambda$$ is only one.

Solutions · 1