Question

Question: Using, find the values ofsuch that the following equations have non-trivial solutions, and for each, solve the equations. (See example.)

$\{\begin{array}{l}(6-\lambda )\text{x}+\text{3y}=\text{0}\\ \text{3x}-\text{(2}+\lambda )y=0\end{array}$

Short answer

The value of the$\lambda$ such that the following equations have non-trivial solution is (-3) and 7.

When the value of $\lambda =(-3)$, the solution of the equations is $\text{y}=-\text{3x}$.

And when the value of $\lambda =7$, the solution of the equations is$\text{x}=\text{3y}$

Step-by-step solution

Define theorem 8.9.

According to the theorem 8.9, the determinant of the coefficient will be zero when a system of n homogenous equations in n unknowns will have solutions other than the trivial solution.

Given parameters.

Given the system of equations $\{\begin{array}{l}(6-\mathrm{\lambda })\text{x}+\text{3y}=\text{0}\\ \text{3x}-\text{(2}+\mathrm{\lambda })\mathrm{y}=0\end{array}$.

Find the value of λ

By the theorem 8.9, the determinant of the coefficient must be zero.

$\left|\begin{array}{cc}6-\lambda & 3\\ 3& -2-\lambda \end{array}\right|=0$

$\begin{array}{rcl}(6-\lambda )(-2-\lambda )-(3\times 3)& =& 0\\ {\lambda }^2-4\lambda -21& =& 0\\ (\lambda +3)(\lambda -7)& =& 0\end{array}$

Then the values will be $\begin{array}{rcl}\lambda & =& -3\end{array}$and $\begin{array}{rcl}\lambda & =& 7\end{array}$.

Solve the equations.

Substitute$\lambda =-3$ in the given system of equations.

$\begin{array}{rcl}\left(\text{6}-\text{(}-\text{3)}\right)\text{x}+\text{3y}& =& \text{0}\\ \left(\text{6}+\text{3}\right)\text{x}+\text{3y}& =& \text{0}\\ \text{9x}+\text{3y}& =& \text{0}\\ \text{y}& =& -\text{3x}\end{array}$

Also.

$\begin{array}{rcl}\text{3x}-\left(\text{2}+\left(-\text{3}\right)\right)\text{y}& =& \text{0}\\ \text{3x}-\text{1y}& =& \text{0}\\ \text{y}& =& -\text{3x}\end{array}$

Now substitute $\lambda =7$in the given system of equations.

$\begin{array}{rcl}\left(\text{6}-7\right)\text{x}+\text{3y}& =& \text{0}\\ -\text{x}+\text{3y}& =& \text{0}\\ \text{x}& =& \text{3y}\end{array}$

Also,

$\begin{array}{rcl}\text{3x}-\left(\text{2 + 7}\right)\text{y}& =& \text{0}\\ \text{3x}-9\text{y}& =& \text{0}\\ \text{x}& =& \text{3y}\end{array}$

So, the value of $\lambda =-3$and 7.

For $\lambda =-3$, the solution of the system of equation is $\text{y}=-\text{3x}$.

And for $\lambda =7$, the solution of the system of equation is $\text{x}=\text{3y}$.