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Systems of Linear Equations

Short Notes

A system of linear equations can be represented as Ax = b where A is the coefficient matrix, x is the variable vector, and b is the constant vector.

  • Consistent System: Has at least one solution
  • Inconsistent System: Has no solution
  • Homogeneous System: When b = 0 (always consistent, trivial solution exists)

Key Theories & Formulas

1. Existence and Uniqueness of Solutions

For a system Ax = b of n equations in n unknowns:

Condition Solution Type
rank(A) = rank(A b) = n
rank(A) = rank(A b) < n
rank(A) ≠ rank(A b)

2. Homogeneous Systems (Ax = 0)

  • Always consistent (x = 0 is always a solution)
  • Trivial solution only: if rank(A) = n (number of unknowns)
  • Non-trivial solutions exist: if rank(A) < n
  • Number of free variables = n - rank(A)
  • Key Result: If m < n (fewer equations than unknowns), non-trivial solutions always exist.

3. Row Echelon Form (REF) & Reduced REF (RREF)

Steps for Gaussian Elimination:

  1. Convert to upper triangular form (REF)
  2. Further reduce to RREF (leading 1s, zeros above and below pivots)

Example: $\(\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix} \xrightarrow{RREF} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\)$

4. Cramer's Rule

For a system with unique solution (det(A) ≠ 0): $\(x_i = \frac{\det(A_i)}{\det(A)}\)$ where \(A_i\) is A with column i replaced by b.

5. Matrix Inverse Method

If A is invertible: \(x = A^{-1}b\)

Example Problems

Problem 1: For what value of k does the system have infinite solutions? $\(\begin{align} x + 2y + 3z &= 4 \\ 2x + 3y + 4z &= 5 \\ 3x + 4y + 5z &= k \end{align}\)$

Solution:

  1. Form augmented matrix: \(\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 2 & 3 & 4 & | & 5 \\ 3 & 4 & 5 & | & k \end{pmatrix}\)
  2. \(R_2 \leftarrow R_2 - 2R_1\): \(\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 0 & -1 & -2 & | & -3 \\ 3 & 4 & 5 & | & k \end{pmatrix}\)
  3. \(R_3 \leftarrow R_3 - 3R_1\): \(\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 0 & -1 & -2 & | & -3 \\ 0 & -2 & -4 & | & k-12 \end{pmatrix}\)
  4. \(R_3 \leftarrow R_3 - 2R_2\): \(\begin{pmatrix} 1 & 2 & 3 & | & 4 \\ 0 & -1 & -2 & | & -3 \\ 0 & 0 & 0 & | & k-6 \end{pmatrix}\)
  5. For infinite solutions: \(k - 6 = 0 \Rightarrow \mathbf{k = 6}\)

Problem 2: Find the number of solutions for the homogeneous system with coefficient matrix: $\(A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}\)$

Solution:

  • det(A) = 0 (rows are in arithmetic progression)
  • rank(A) = 2 < 3 (number of unknowns)
  • Infinite solutions with 1 free variable

Hardest GATE Questions

Topic: Condition for Consistent System

Question (GATE 2016 Style): Consider the system: $\(\begin{align} x + y + z &= 1 \\ x + 2y + 3z &= 2 \\ x + 2y + (a^2-5)z &= a \end{align}\)$ For what values of 'a' does the system have (i) no solution, (ii) unique solution, (iii) infinite solutions?

Solution:

  1. Augmented matrix after row operations: \(\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 1 & 2 & | & 1 \\ 0 & 0 & a^2-8 & | & a-2 \end{pmatrix}\)

  2. Case 1: \(a^2 - 8 \neq 0\) (i.e., \(a \neq \pm 2\sqrt{2}\))

  3. rank(A) = rank(A|b) = 3 → Unique solution

  4. Case 2: \(a^2 - 8 = 0\) (i.e., \(a = \pm 2\sqrt{2}\))

  5. If \(a = 2\sqrt{2}\): \(a - 2 = 2\sqrt{2} - 2 \neq 0\) → No solution
  6. If \(a = -2\sqrt{2}\): \(a - 2 = -2\sqrt{2} - 2 \neq 0\) → No solution

Final Answer:

  • No solution: \(a = \pm 2\sqrt{2}\)
  • Unique solution: \(a \neq \pm 2\sqrt{2}\)
  • Infinite solutions: Never (for this system)

Why it's hard: Requires careful case analysis and understanding when rank conditions fail.

References