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Example of solving a 3-by-3 homogeneous matrix equation

Created over 8 years ago, updated 10 days ago

Example: Solving a 3-by-3 Homogeneous Matrix Equation

Solve Ax = 0 where:

A = [ 1  2  3]
    [ 2  5  3]
    [ 1  0  8]

Step 1: Form the augmented matrix [A | 0]

[ 1  2  3 | 0]
[ 2  5  3 | 0]
[ 1  0  8 | 0]

Step 2: R₂ → R₂ - 2R₁, R₃ → R₃ - R₁

[ 1  2  3 | 0]
[ 0  1 -3 | 0]
[ 0 -2  5 | 0]

Step 3: R₃ → R₃ + 2R₂

[ 1  2  3 | 0]
[ 0  1 -3 | 0]
[ 0  0 -1 | 0]

Step 4: R₃ → -R₃

[ 1  2  3 | 0]
[ 0  1 -3 | 0]
[ 0  0  1 | 0]

Step 5: Back-substitute:

  • z = 0
  • y - 3(0) = 0, so y = 0
  • x + 2(0) + 3(0) = 0, so x = 0

Solution: x = [0, 0, 0]ᵀ (the trivial solution)

The homogeneous system has only the trivial solution because A has 3 pivots (full rank). If A had fewer pivots, free variables would yield nontrivial solutions.