UPSC MainsMathematics (Optional)Science and TechnologyPractice question

Solutions of Homogeneous Linear Equations with Parameter k

Determine all the values of k for the given system of equations to have a solution: 2x + 3ky + (3k + 4)z = 0 x + (k + 4)y + (4k + 2)z = 0 x + (k + 1)y + (3k + 4)z = 0

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How to approach

Identify the system as a homogeneous linear system of the form AX = 0, noting that homogeneous systems are unconditionally consistent. Set up the coefficient matrix and apply elementary row operations to compute its determinant. Analyze the discriminant of the resulting quadratic expression in k to determine whether non-trivial solutions exist or whether only the unique trivial solution exists for real k.

Model answer

437 words

Introduction

The given set of equations represents a linear homogeneous system of the form AX = 0 in three variables (x, y, z). A homogeneous linear system is inherently consistent because it unconditionally admits at least the trivial solution (x = y = z = 0). The nature of the solution set—whether it contains only the trivial solution or infinitely many non-trivial solutions—depends directly on the determinant and rank of the coefficient matrix A.

1. Matrix Formulation of the System

The system can be expressed in matrix form as AX = 0, where:

$$A = \begin{bmatrix} 2 & 3k & 3k+4 \\ 1 & k+4 & 4k+2 \\ 1 & k+1 & 3k+4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad 0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

2. Row Reduction and Determinant Calculation

To evaluate the determinant |A| efficiently, elementary row operations are performed:

  • Interchanging Rows: Swap R1 and R3 (R1 ↔ R3):
    $$A \sim \begin{bmatrix} 1 & k+1 & 3k+4 \\ 1 & k+4 & 4k+2 \\ 2 & 3k & 3k+4 \end{bmatrix}$$
  • Row Eliminations: Apply R2 → R2 - R1 and R3 → R3 - 2R1:
    $$A \sim \begin{bmatrix} 1 & k+1 & 3k+4 \\ 0 & 3 & k-2 \\ 0 & k-2 & -3k-4 \end{bmatrix}$$

Expanding the determinant along the first column:

$$|A| = 1 \cdot [3(-3k - 4) - (k - 2)^2]$$

$$|A| = -9k - 12 - (k^2 - 4k + 4) = -(k^2 + 5k + 16)$$

3. Analysis of Real Roots for k

For a homogeneous system, non-trivial solutions exist if and only if |A| = 0. Setting the quadratic factor to zero:

$$k^2 + 5k + 16 = 0$$

The discriminant (Δ) is given by:

$$\Delta = b^2 - 4ac = 5^2 - 4(1)(16) = 25 - 64 = -39$$

Because Δ < 0, the equation has no real roots. Consequently, |A| ≠ 0 for all real values of k.

4. Determination of Solution Set

  • Condition for Non-Trivial Solution: Requires |A| = 0, which is impossible for any real number k.
  • Condition for Trivial Solution: When |A| ≠ 0, Rank(A) = 3 = number of unknowns. Thus, the system possesses a unique trivial solution (x = 0, y = 0, z = 0).

Conclusion

Since |A| &ne; 0 for every real number k, the system never possesses non-trivial solutions. However, because every homogeneous system admits the trivial solution x = y = z = 0, the given system has a solution for all real values of k (k &isin; &reals;).

Key facts to remember

definition
Homogeneous Linear System

A system of linear equations in which all constant terms are zero (AX = 0). It is always consistent because the zero vector (trivial solution) satisfies every equation.

definition
Trivial vs Non-Trivial Solution

The trivial solution is the zero vector x = y = z = 0. A non-trivial solution is any non-zero vector satisfying the system, which exists if and only if the coefficient matrix is singular (|A| = 0).

Frequently asked questions

Why does a homogeneous system have a solution even when the determinant is non-zero?

When |A| is non-zero, the matrix is non-singular and has full rank, meaning the inverse matrix A⁻¹ exists. The unique solution is X = A⁻¹(0) = 0, which is the trivial solution.