Key Concept for Non-Trivial Solutions

For a system of homogeneous linear equations (where all equations are equal to zero), a non-trivial solution (i.e., a solution other than $x=0, y=0, z=0$) exists if and only if the determinant of the coefficient matrix is equal to zero. This is a fundamental property in linear algebra.

Given the system of linear equations:

1. $4x + \lambda y + 2z = 0$
2. $2x - y + z = 0$
3. $\mu x + 2y + 3z = 0$

Since all equations are homogeneous (equal to 0), for a non-trivial solution to exist, the determinant of the coefficient matrix must be zero.

Setting up the Coefficient Matrix

First, we construct the coefficient matrix $A$ from the coefficients of $x, y,$ and $z$ in each equation: $A = \begin{pmatrix} 4 & \lambda & 2 \\ 2 & -1 & 1 \\ \mu & 2 & 3 \end{pmatrix}$

Calculating the Determinant

Next, we set the determinant of this matrix to zero: $\det(A) = \left| \begin{matrix} 4 & \lambda & 2 \\ 2 & -1 & 1 \\ \mu & 2 & 3 \end{matrix} \right| = 0$

Now, we expand the determinant. We will expand along the first row for clarity, using the formula for a $3 \times 3$ determinant: $\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Applying this to our matrix: $4 \times ((-1)(3) - (1)(2)) - \lambda \times ((2)(3) - (1)(\mu)) + 2 \times ((2)(2) - (-1)(\mu)) = 0$

Let's simplify each term:

• First term: $4 \times (-3 - 2) = 4 \times (-5) = -20$
• Second term: $-\lambda \times (6 - \mu)$
• Third term: $2 \times (4 + \mu)$

Substitute these back into the determinant equation: $-20 - \lambda (6 - \mu) + 2 (4 + \mu) = 0$ $-20 - 6\lambda + \lambda\mu + 8 + 2\mu = 0$ Combine the constant terms: $-12 - 6\lambda + \lambda\mu + 2\mu = 0$

Solving the Determinant Equation

Our goal is to factor this equation to find the conditions on $\lambda$ and $\mu$. Notice that we have terms involving $(6-\mu)$ and terms that can be manipulated to form $(6-\mu)$. Rearrange the terms: $-12 + 2\mu - 6\lambda + \lambda\mu = 0$ Factor $-2$ from the first two terms and $\lambda$ from the last two terms: $-2(6 - \mu) - \lambda(6 - \mu) = 0$ Now, we can clearly see a common factor of $(6 - \mu)$: $(6 - \mu) (-2 - \lambda) = 0$ This can also be written as: $-(6 - \mu) (\lambda + 2) = 0$ Since the product is zero, we can remove the negative sign: $(6 - \mu) (\lambda + 2) = 0$

For this product to be zero, at least one of the factors must be zero. This leads to two possibilities:

1. $6 - \mu = 0 \implies \mu = 6$ In this case, if $\mu = 6$, the first factor is zero, making the entire product zero, regardless of the value of $\lambda$. So, if $\mu = 6$, then $\lambda$ can be any real number ($\lambda \in \mathbb{R}$).

2. $\lambda + 2 = 0 \implies \lambda = -2$ In this case, if $\lambda = -2$, the second factor is zero, making the entire product zero, regardless of the value of $\mu$. So, if $\lambda = -2$, then $\mu$ can be any real number ($\mu \in \mathbb{R}$).

Analyzing the Solutions and Options

The condition for the system to have a non-trivial solution is that either $\mu = 6$ (with $\lambda \in \mathbb{R}$) or $\lambda = -2$ (with $\mu \in \mathbb{R}$). We need to find which of the given options matches one of these conditions.

Let's examine the options:

• (A) $\mu = 6, \lambda \in \mathbb{R}$: This option perfectly matches our first derived condition. If $\mu=6$, the determinant is indeed zero for any real value of $\lambda$, ensuring non-trivial solutions.
• (B) $\lambda = 3, \mu \in \mathbb{R}$: This does not match. If $\lambda=3$, for the determinant to be zero, we must have $\mu=6$. It cannot be $\mu \in \mathbb{R}$.
• (C) $\mu = -6, \lambda \in \mathbb{R}$: This does not match our conditions.
• (D) $\lambda = 2, \mu \in \mathbb{R}$: This does not match. If $\lambda=2$, for the determinant to be zero, we must have $\mu=6$. It cannot be $\mu \in \mathbb{R}$.

Therefore, option (A) is the correct statement.

Tips for Success & Common Mistakes

• Homogeneous vs. Non-homogeneous Systems: Remember that the condition $\det(A)=0$ is specific to homogeneous systems ($Ax=0$) for the existence of non-trivial solutions. For non-homogeneous systems ($Ax=B$ where $B \neq 0$), $\det(A)=0$ implies either no solution or infinitely many solutions, but not necessarily a non-trivial solution in the context of $(0,0,0)$ being a trivial solution.
• Careful Determinant Calculation: Errors in arithmetic or signs during determinant expansion are common. Always double-check your calculations, especially with negative numbers.
• Algebraic Manipulation and Factorization: When simplifying the determinant equation, look for opportunities to factor terms. Recognizing patterns like $ax+by+cx+dy$ where factors can be grouped is key. In this case, $-12 + 2\mu$ could be seen as $-2(6 - \mu)$, which helped in factoring.
• Understanding "OR": The result $(A)(B)=0$ means $A=0$ or $B=0$. This is crucial for interpreting the conditions on $\lambda$ and $\mu$. If one factor is zero, the other variable is free to take any real value.

Summary and Key Takeaway

To determine when a system of homogeneous linear equations has a non-trivial solution, we must calculate the determinant of its coefficient matrix and set it equal to zero. Solving this determinant equation provides the specific conditions on the parameters. In this problem, the condition simplifies to $(\mu-6)(\lambda+2)=0$, implying that either $\mu=6$ (with $\lambda \in \mathbb{R}$) or $\lambda=-2$ (with $\mu \in \mathbb{R}$) must hold for non-trivial solutions to exist. Comparing these findings with the given options, we find that $\mu=6, \lambda \in \mathbb{R}$ is the correct choice.