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 solution is:

Answer 1

Here, we will use the infinitely many solutions, the determinant $D$ is 0, ${D_x}$ is 0, ${D_y}$ is 0 and ${D_z}$ is 0. Then we will use the given system of linear equations to find the determinant $D$, then differentiate the given linear equations with respect to $x$, differentiate the given linear equations with respect to $y$ and differentiate the given linear equations with respect to $z$. Then we will use the Rolle's theorem and intermediate theorem to find the number of real values of lambda.

Complete step-by-step answer:
We are given the system of linear equations $2x + 4y - \lambda z = 0$, $4x + \lambda y + 2z = 0$, $\lambda x + 2y + 2z = 0$ has infinitely many solutions.
We know that the infinitely many solutions, the determinant $D$ is 0, ${D_x}$ is 0, ${D_y}$ is 0 and ${D_z}$ is 0.
Using the given system of linear equations to find the determinant $D$, we get

\Rightarrow D = \left| {\begin{array}{*{20}{c}} 2&4&{ - \lambda } \\\ 4&\lambda &2 \\\ \lambda &2&2 \end{array}} \right| \\\ \Rightarrow D = 2\left( {2\lambda - 4} \right) - 4\left( {8 - 2\lambda } \right) - \lambda \left( {8 - {\lambda ^2}} \right) \\\ \Rightarrow D = 4\lambda - 8 - 32 + 8\lambda - 8\lambda + {\lambda ^3} \\\ \Rightarrow D = {\lambda ^3} + 4\lambda - 40 \\\

Substituting the value of $D$ in the above equation, we get

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

Differentiating the given linear equations with respect to $x$, we get
$\Rightarrow 2 + 4y - \lambda z = 0$
$\Rightarrow 4 + \lambda y + 2z = 0$
$\Rightarrow \lambda + 2y + 2z = 0$
Using the above system of linear equations to find the determinant ${D_x}$, we get

\Rightarrow {D_x} = \left| {\begin{array}{*{20}{c}} 0&4&{ - \lambda } \\\ 0&\lambda &2 \\\ 0&2&2 \end{array}} \right| \\\ \Rightarrow {D_x} = 0 \\\

Differentiating the given linear equations with respect to $y$, we get
$\Rightarrow 2x + 4 - \lambda z = 0$
$\Rightarrow 4x + \lambda + 2z = 0$
$\Rightarrow \lambda x + 2 + 2z = 0$
Using the above system of linear equations to find the determinant ${D_y}$, we get

\Rightarrow {D_y} = \left| {\begin{array}{*{20}{c}} 2&0&{ - \lambda } \\\ 4&0&2 \\\ \lambda &0&2 \end{array}} \right| \\\ \Rightarrow {D_y} = 0 \\\

Differentiating the given linear equations with respect to $z$, we get
$\Rightarrow 2x + 4y - \lambda = 0$
$\Rightarrow 4x + \lambda y + 2 = 0$
$\Rightarrow \lambda x + 2y + 2 = 0$
Using the above system of linear equations to find the determinant ${D_y}$, we get

\Rightarrow {D_z} = \left| {\begin{array}{*{20}{c}} 2&4&0 \\\ 3&\lambda &0 \\\ \lambda &2&0 \end{array}} \right| \\\ \Rightarrow {D_z} = 0 \\\

Since by intermediate value property, the equation ${\lambda ^3} + 4\lambda - 40$ has one solution.
Differentiating the equation ${\lambda ^3} + 4\lambda - 40$ by $\lambda$, we get

$$\dfrac{d}{{d\lambda }}\left( {{\lambda ^3} + 4\lambda - 40} \right) = 3{\lambda ^2} + 4 \\\ > 0 \\\$$

Therefore by Rolle's theorem, the equation cannot have $\beta$, $y$ such that $\lambda \left( \beta \right) = 0$ and $\lambda \left( y \right) = 0$.
Hence, the number of real values of lambda is one.

Note: In solving these types of questions, the key concept is to use the intermediate theorem and rolle's theorem appropriately to easily get the answer. Students need to differentiate the given linear equation with respect to the distinct variables to use the fact that when the infinitely many solutions, the determinant $D$ is 0, ${D_x}$ is 0, ${D_y}$ is 0 and ${D_z}$ is 0.