Gaussian Elimination & Linear System Solvability
“Augmented matrices, REF/RREF, Rouché-Capelli theorem, unique solutions, infinite solutions, and inconsistency.”
1.Linear Systems, Row Reduction & Solvability Regimes
§ 2.1.Consider the linear system of m equations in n unknowns: Ax = b.
Elementary Row Operations:
1. Swap two rows (R_i <-> R_j).
2. Multiply a row by non-zero scalar (R_i -> c * R_i).
3. Add a scalar multiple of one row to another (R_i -> R_i + c * R_j).
Row Echelon Form (REF) vs Reduced Row Echelon Form (RREF):
In RREF, every leading entry (pivot) is 1, and each pivot is the ONLY non-zero entry in its entire column.
The Three Solvability Regimes:
Let r = Rank(A) and r_aug = Rank([A | b]):
- r < r_aug: Inconsistent (0 = c contradiction; No solution).
- r = r_aug = n: Consistent & Unique (all columns have pivots).
- r = r_aug < n: Consistent & Infinite (n - r free parameters).
2. Worked Practice Problem: Parameterized System Consistency Analysis
Distinguish when the augmented column introduces an extra pivot versus when it falls into existing pivot structure.
Problem: Consider the system of linear equations in variables x, y, z:
x + y + z = 6
x + 2y + 3z = 10
x + 2y + k*z = mu
Determine the values of parameters k and mu for which the system has: (a) a unique solution, (b) infinitely many solutions, (c) no solution.
Step 1: Set up the augmented matrix [A | b]:
[ [1, 1, 1 | 6], [1, 2, 3 | 10], [1, 2, k | mu] ]
Step 2: Apply row operations to reach upper triangular form:
R2 -> R2 - R1: [ 0, 1, 2 | 4 ]
R3 -> R3 - R1: [ 0, 1, k - 1 | mu - 6 ]
R3 -> R3 - R2: [ 0, 0, k - 3 | mu - 10 ]
The reduced augmented matrix is:
[ [1, 1, 1 | 6], [0, 1, 2 | 4], [0, 0, k - 3 | mu - 10] ]
Step 3: Analyze the three cases:
Case (a) Unique Solution: Requires Rank(A) = 3 = n.
This requires k - 3 != 0 => k != 3 (mu can be any real value).
Case (b) Infinitely Many Solutions: Requires Rank(A) = Rank([A|b]) = 2 < 3.
This requires the entire last row to be zero:
k - 3 = 0 and mu - 10 = 0 => k = 3 and mu = 10.
Case (c) No Solution (Inconsistent): Requires Rank(A) = 2 != Rank([A|b]) = 3.
This requires k - 3 = 0 and mu - 10 != 0 => k = 3 and mu != 10.
Conclusion: Unique if k != 3; Infinite if k = 3, mu = 10; Inconsistent if k = 3, mu != 10.
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