Key Concept: Conditions for Non-Trivial Solutions of Homogeneous Linear Equations

A system of homogeneous linear equations is one where all constant terms are zero, typically written as $Ax = \mathbf{0}$. Here, $A$ is the coefficient matrix, $x$ is the column vector of variables (e.g., $x, y, z$), and $\mathbf{0}$ is the zero vector.

For such a system to possess non-zero solutions (also known as non-trivial solutions, meaning at least one variable is not zero), a crucial condition must be met: the determinant of the coefficient matrix must be equal to zero. If $\det(A) \neq 0$, the only solution is the trivial solution ($x=0, y=0, z=0$). If $\det(A) = 0$, the system has infinitely many non-trivial solutions.

Our goal is to find the values of $\lambda$ for which this condition, $\det(A) = 0$, holds true.

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Step-by-Step Solution

1. Formulate the Coefficient Matrix First, we extract the coefficients of $x, y, z$ from the given system of equations to form the coefficient matrix $A$: $A = \begin{pmatrix} \lambda - 1 & 3\lambda + 1 & 2\lambda \\ \lambda - 1 & 4\lambda - 2 & \lambda + 3 \\ 2 & 3\lambda + 1 & 3(\lambda - 1) \end{pmatrix}$ For non-zero solutions to exist, we set the determinant of this matrix to zero: $\left| \begin{matrix} \lambda - 1 & 3\lambda + 1 & 2\lambda \\ \lambda - 1 & 4\lambda - 2 & \lambda + 3 \\ 2 & 3\lambda + 1 & 3(\lambda - 1) \end{matrix} \right| = 0$

2. Simplify the Determinant using Row Operations To make the determinant expansion easier, we aim to introduce zeros into the matrix using elementary row operations. These operations do not change the value of the determinant.

• Operation 1: $R_2 \to R_2 - R_1$

  • Why this step? The first elements of $R_1$ and $R_2$ are both $(\lambda-1)$. Subtracting $R_1$ from $R_2$ will make the first element of the new $R_2$ zero, simplifying the determinant.
  • Applying the operation:
    • $R_2C_1: (\lambda-1) - (\lambda-1) = 0$
    • $R_2C_2: (4\lambda-2) - (3\lambda+1) = \lambda-3$
    • $R_2C_3: (\lambda+3) - (2\lambda) = 3-\lambda$

• Operation 2: $R_3 \to R_3 - R_1$

  • Why this step? This operation helps simplify the elements in the third row. Specifically, the second element $R_3C_2$ will become zero, as it's identical to $R_1C_2$.
  • Applying the operation:
    • $R_3C_1: 2 - (\lambda-1) = 3-\lambda$
    • $R_3C_2: (3\lambda+1) - (3\lambda+1) = 0$
    • $R_3C_3: 3(\lambda-1) - 2\lambda = 3\lambda-3-2\lambda = \lambda-3$

After these two row operations, the determinant becomes: $\left| \begin{matrix} \lambda - 1 & 3\lambda + 1 & 2\lambda \\ 0 & \lambda - 3 & 3 - \lambda \\ 3 - \lambda & 0 & \lambda - 3 \end{matrix} \right| = 0$ Notice the common factors $( \lambda - 3 )$ and $( 3 - \lambda )$ appearing in the second and third rows. Also, we have two zeros in strategic positions.

3. Further Simplification using a Column Operation To obtain even more zeros, ideally in a single row or column, we perform a column operation.

• Operation 3: $C_1 \to C_1 + C_3$
  • Why this step? Observe that the element $R_3C_1$ is $(3-\lambda)$ and $R_3C_3$ is $(\lambda-3)$. Adding these will result in zero. This will create a zero in $R_3C_1$, giving us two zeros in the third row, which is excellent for expansion.
  • Applying the operation:
    • $R_1C_1: (\lambda-1) + 2\lambda = 3\lambda-1$
    • $R_2C_1: 0 + (3-\lambda) = 3-\lambda$
    • $R_3C_1: (3-\lambda) + (\lambda-3) = 0$

Now, the determinant is transformed into: $\left| \begin{matrix} 3\lambda - 1 & 3\lambda + 1 & 2\lambda \\ 3 - \lambda & \lambda - 3 & 3 - \lambda \\ 0 & 0 & \lambda - 3 \end{matrix} \right| = 0$ We have successfully created two zeros in the third row.

4. Expand the Determinant Expanding the determinant along the third row ($R_3$) is the most efficient method due to the presence of two zeros. The expansion formula for a $3 \times 3$ determinant along $R_3$ is: $\det(A) = a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33}$, where $C_{ij}$ is the cofactor. Since $a_{31}=0$ and $a_{32}=0$, only the term $a_{33}C_{33}$ contributes to the determinant. The element $a_{33}$ is $(\lambda-3)$. The cofactor $C_{33}$ is $(-1)^{3+3}$ times the determinant of the $2 \times 2$ submatrix obtained by removing the 3rd row and 3rd column: $C_{33} = (+1) \left| \begin{matrix} 3\lambda - 1 & 3\lambda + 1 \\ 3 - \lambda & \lambda - 3 \end{matrix} \right|$ So, the determinant equation becomes: $(\lambda - 3) \left[ (3\lambda - 1)(\lambda - 3) - (3\lambda + 1)(3 - \lambda) \right] = 0$

5. Solve the Equation for $\lambda$ Let's simplify the expression inside the square brackets. Notice that $(3-\lambda) = -(\lambda-3)$. Substitute this into the equation: $(\lambda - 3) \left[ (3\lambda - 1)(\lambda - 3) - (3\lambda + 1)(-(\lambda - 3)) \right] = 0$ $(\lambda - 3) \left[ (3\lambda - 1)(\lambda - 3) + (3\lambda + 1)(\lambda - 3) \right] = 0$ Now, we can factor out the common term $(\lambda-3)$ from the terms inside the square brackets: $(\lambda - 3) (\lambda - 3) \left[ (3\lambda - 1) + (3\lambda + 1) \right] = 0$ $(\lambda - 3)^2 \left[ 3\lambda - 1 + 3\lambda + 1 \right] = 0$ $(\lambda - 3)^2 (6\lambda) = 0$ This equation gives us the possible values for $\lambda$:

• From $6\lambda = 0$, we get $\lambda = 0$.
• From $(\lambda - 3)^2 = 0$, we get $\lambda - 3 = 0$, which means $\lambda = 3$.

Thus, the distinct values of $\lambda$ for which the system has non-zero solutions are $\lambda = 0$ and $\lambda = 3$.

6. Calculate the Sum of Distinct Values of $\lambda$ The distinct values of $\lambda$ found are $0$ and $3$. Their sum is $0 + 3 = 3$.

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Tips and Common Mistakes:

• Fundamental Condition: Always start by clearly stating that $\det(A) = 0$ is the necessary and sufficient condition for a homogeneous system to have non-trivial solutions.
• Arithmetic Precision: Double-check all calculations, especially when performing row/column operations and determinant expansion. A single sign error or arithmetic mistake can lead to incorrect values of $\lambda$.
• Strategic Operations: Choose row/column operations strategically to create zeros. Aim for a row or column with as many zeros as possible to simplify expansion.
• Factoring is Key: After expanding the determinant, look for common factors to simplify the polynomial equation in $\lambda$. This often reduces the complexity of solving for $\lambda$.
• Distinct Values: Pay attention to the word "distinct" in the question. If a value of $\lambda$ appears multiple times (like $\lambda=3$ here), it's counted only once when summing distinct values.

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Summary and Key Takeaway:

For a system of homogeneous linear equations $Ax = \mathbf{0}$, the existence of non-zero (non-trivial) solutions is directly linked to the singularity of the coefficient matrix $A$. This means $\det(A)$ must be zero. By systematically applying row and column operations to simplify the determinant, we can form a polynomial equation in $\lambda$. Solving this polynomial equation yields the values of $\lambda$ for which non-trivial solutions exist. In this problem, the distinct values of $\lambda$ are $0$ and $3$, and their sum is $3$.