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Description:Cleaned up 3x3 matrix equation example
# Example of solving a 3-by-3 matrix equation ## Example: Solving a 3-by-3 Matrix Equation Solve `Ax = b` where: ``` A = [1 2-1] b = [8] [-3 -14] [2] [-11] [-25 -1] [ 7] [1 1 2] [-35] ``` **Step 1:** Form the augmented matrix `[A | b]` ``` [ 1 2-1 |8] [-3 -14] [ 2|-11] [-25 -1 | 7] [ 1 1 2 |-35] ``` **Step 2:**R₂ → R₂ + 3R₁, R₃ → R₃ + 2R₁R2 = R2 - 2*R1, R3 = R3 - R1 ``` [ 1 2-1 |84] [ 05 -1| 13-3 | -1] [ 05 0 | 13-1 1 | 1] ``` **Step 3:**R₃ → R₃ - R₂R3 = R3 + R2 ``` [ 1 2-1 |84] [ 05 -1| 13-3 | -1] [ 0 01-2 | 0] ``` **Step 4:** FromR₃:R3: -2z = 0, so z = 0.⏎ FromR₂: 5y = 13R2: y - 3(0) = -1, so y =13/5. Wait -- let me redo with integer-friendly steps. ⏎ Using RREF directly: ``` [-1. From R1: x + 2(-1| 8] [-3) + 0 = 4, so x = 6. ⏎ **Solution:** x = [6, -12 |-11] [-2, 0]^ ⏎ **Verification:** - Row 1: 1(6) + 2| -3] ``` R₂ → R₂ + 3R₁: [0, 5,(-1| 13] R₃ → R₃ + 2R₁: [) + 1(0,) = 6 - 2 = 4 ✓ - Row 2: 2(6) + 5, 0 | 13] R₃ → R₃ - R₂: [0, 0,(-1|) + (-1)(0] z = 0 From R₂: 5y - 0 = 13, y = 13/) = 12 - 5⏎ From R₁: x= 7 ✓ - Row 3: 1(6) + 1(-1) + 2(13/5) -0= 8, x = 8 - 26/) = 6 - 1 = 5= 14/5 ⏎ **Solution:** x = 14/5, y = 13/5, z = 0✓ # Parents * Matrix equations
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