To determine the values of $\lambda$ for which the system of equations has a non-trivial solution, we must solve the following system of linear equations:

$3x + y - \lambda z = 0$ $4x - 2y - 3z = 0$ $2\lambda x + 4y + \lambda z = 0$

This is a system of linear equations in $x$, $y$, and $z$, and for there to be a non-trivial solution (other than the zero solution), the determinant of the coefficient matrix must be zero.

Step 1: Write the system in matrix form

We can write the system as:

3 & 1 & -\lambda \\ 4 & -2 & -3 \\ 2\lambda & 4 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0$$ ### Step 2: Find the determinant of the coefficient matrix The determinant of the matrix is: $$\text{det}\left(\begin{bmatrix} 3 & 1 & -\lambda \\ 4 & -2 & -3 \\ 2\lambda & 4 & \lambda \end{bmatrix}\right)$$ We compute this determinant using cofactor expansion along the first row. $$\text{det} = 3 \begin{vmatrix} -2 & -3 \\ 4 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 4 & -3 \\ 2\lambda & \lambda \end{vmatrix} + (-\lambda) \begin{vmatrix} 4 & -2 \\ 2\lambda & 4 \end{vmatrix}$$ ### Step 3: Calculate the 2x2 determinants 1. For the first 2x2 determinant: $$\begin{vmatrix} -2 & -3 \\ 4 & \lambda \end{vmatrix} = (-2)(\lambda) - (-3)(4) = -2\lambda + 12 = 12 - 2\lambda$$ 2. For the second 2x2 determinant: $$\begin{vmatrix} 4 & -3 \\ 2\lambda & \lambda \end{vmatrix} = (4)(\lambda) - (-3)(2\lambda) = 4\lambda + 6\lambda = 10\lambda$$ 3. For the third 2x2 determinant: $$\begin{vmatrix} 4 & -2 \\ 2\lambda & 4 \end{vmatrix} = (4)(4) - (-2)(2\lambda) = 16 + 4\lambda = 16 + 4\lambda$$ ### Step 4: Substitute these into the determinant formula $$\text{det} = 3(12 - 2\lambda) - 1(10\lambda) - \lambda(16 + 4\lambda)$$ Simplify: $$\text{det} = 36 - 6\lambda - 10\lambda - \lambda(16 + 4\lambda)$$ $$= 36 - 16\lambda - \lambda(16 + 4\lambda)$$ $$= 36 - 16\lambda - 16\lambda - 4\lambda^2$$ $$= 36 - 32\lambda - 4\lambda^2$$ ### Step 5: Set the determinant to zero for a non-trivial solution For a non-trivial solution to exist, the determinant must be zero: $$36 - 32\lambda - 4\lambda^2 = 0$$ ### Step 6: Solve the quadratic equation Rearrange the equation: $$-4\lambda^2 - 32\lambda + 36 = 0$$ Divide the entire equation by -4: $$\lambda^2 + 8\lambda - 9 = 0$$ Now solve for $$\lambda$$ using the quadratic formula: $$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $$a = 1$$, $$b = 8$$, and $$c = -9$$: $$\lambda = \frac{-8 \pm \sqrt{8^2 - 4(1)(-9)}}{2(1)}$$ $$\lambda = \frac{-8 \pm \sqrt{64 + 36}}{2}$$ $$\lambda = \frac{-8 \pm \sqrt{100}}{2}$$ $$\lambda = \frac{-8 \pm 10}{2}$$ So, we get two possible values for $$\lambda$$: $$\lambda = \frac{-8 + 10}{2} = 1 \quad \text{or} \quad \lambda = \frac{-8 - 10}{2} = -9$$ ### Step 7: Solve the system for each value of $$\lambda$$ Now, we solve the system for $$\lambda = 1$$ and $$\lambda = -9$$. #### Case 1: $$\lambda = 1$$ The system becomes: $$3x + y - z = 0$$ $$4x - 2y - 3z = 0$$ $$2x + 4y + z = 0$$ We can solve this system using Gaussian elimination or substitution. #### Case 2: $$\lambda = -9$$ The system becomes: $$3x + y + 9z = 0$$ $$4x - 2y - 3z = 0$$ $$-18x + 4y - 9z = 0$$ Similarly, solve this system using Gaussian elimination or substitution. Would you like to go through these cases step-by-step in detail? Let me know!