Lecture 21: Introductory Linear Algebra

Solving Linear Systems with No Solutions or Infinitely Many Solutions

Example: Start with a linear system.
- Convert to an augmented matrix.
- Perform row operations to achieve row echelon form.
- Operations:
  - Row 2 minus double row 1
  - Row 3 minus row 1
- Resulting matrix reveals contradictions (e.g., row of zeros with a non-zero right side).
Conclusion:
- If a row reduces to 0 = non-zero, there is no solution.
- This contradiction is a consistent identifier for no solutions.

Challenge:
- Difficult to find and describe solutions.
- Need to express solutions without listing them.

Initial Setup:
- Given a linear system in reduced row echelon form.
- Identify leading entries in matrix columns.
Terminology:
- Leading Variables: Correspond to leading entries (e.g., v, x, y).
- Free Variables: No leading entry, free (e.g., w, z).
Solution Description:
- Solve leading variables in terms of free variables.
- Equations:
  - For v: v = 3 + 2w - 2z
  - For x: x = 7 + 3z
  - For y: y = 4 - z
- Free variables are unrestricted.
- Every solution vector can be expressed with these.

Row Reduction:
- Convert system to augmented matrix.
- Row operations lead to reduced row echelon form.
Solution Description with Row Reduction:
- Identify leading variables (w, y) and free variables (x, z).
- Describe solution set:
  - w = 2 + x - z
  - y = 1 + z
  - x = x and z = z (free variables)
- Conclusion:
  - Infinitely many solutions exist.
  - Number of free variables indicates dimensionality of solution set (e.g., two free variables imply a plane in 4D space).

Solving linear systems involves recognizing special cases like no solutions or infinitely many solutions.
Understanding and describing these cases are essential skills in linear algebra.