- 2073 Magh · 8 marks
Gauss-Jordan inverse of matrix
Short answer
Gauss-Jordan method on [A|I] yields inverse after row operations. A⁻¹ = [0.5000 0.5000 -0.5000; 0.5000 5.5000 -3.5000; -0.5000 -3.5000 2.5000].
Compute the inverse of the following matrix using the Gauss-Jordan Method. [3 1 2; 1 2 3; 2 3 5]
Gauss-Jordan method for the inverse: write the augmented matrix [A | I] and apply row operations until the left half becomes the identity matrix I. The right half is then A⁻¹.
Step 1: Check that the inverse exists
det A = 3(2·5 − 3·3) − 1(1·5 − 3·2) + 2(1·3 − 2·2) = 3(1) − 1(−1) + 2(−1) = 2 ≠ 0, so A⁻¹ exists.
Step 2: Row operations
Augmented matrix [A | I]:
3.0000 1.0000 2.0000 | 1.0000 0.0000 0.0000
1.0000 2.0000 3.0000 | 0.0000 1.0000 0.0000
2.0000 3.0000 5.0000 | 0.0000 0.0000 1.0000
R1 ← R1 / 3.0000:
1.0000 0.3333 0.6667 | 0.3333 0.0000 0.0000
1.0000 2.0000 3.0000 | 0.0000 1.0000 0.0000
2.0000 3.0000 5.0000 | 0.0000 0.0000 1.0000
R2 ← R2 − (1.0000) R1; R3 ← R3 − (2.0000) R1:
1.0000 0.3333 0.6667 | 0.3333 0.0000 0.0000
0.0000 1.6667 2.3333 | -0.3333 1.0000 0.0000
0.0000 2.3333 3.6667 | -0.6667 0.0000 1.0000
R2 ← R2 / 1.6667:
1.0000 0.3333 0.6667 | 0.3333 0.0000 0.0000
0.0000 1.0000 1.4000 | -0.2000 0.6000 0.0000
0.0000 2.3333 3.6667 | -0.6667 0.0000 1.0000
R1 ← R1 − (0.3333) R2; R3 ← R3 − (2.3333) R2:
1.0000 0.0000 0.2000 | 0.4000 -0.2000 0.0000
0.0000 1.0000 1.4000 | -0.2000 0.6000 0.0000
0.0000 0.0000 0.4000 | -0.2000 -1.4000 1.0000
R3 ← R3 / 0.4000:
1.0000 0.0000 0.2000 | 0.4000 -0.2000 0.0000
0.0000 1.0000 1.4000 | -0.2000 0.6000 0.0000
0.0000 0.0000 1.0000 | -0.5000 -3.5000 2.5000
R1 ← R1 − (0.2000) R3; R2 ← R2 − (1.4000) R3:
1.0000 0.0000 0.0000 | 0.5000 0.5000 -0.5000
0.0000 1.0000 0.0000 | 0.5000 5.5000 -3.5000
0.0000 0.0000 1.0000 | -0.5000 -3.5000 2.5000
Step 3: Result
The left half is I, so
A⁻¹ =
0.5000 0.5000 -0.5000
0.5000 5.5000 -3.5000
-0.5000 -3.5000 2.5000
In fractions: A⁻¹ = ½ × [1 1 −1; 1 11 −7; −1 −7 5].
Check: row 1 of A times column 1 of A⁻¹ = 3(0.5) + 1(0.5) + 2(−0.5) = 1, and row 1 of A times column 2 of A⁻¹ = 3(0.5) + 1(5.5) + 2(−3.5) = 0, as required for A·A⁻¹ = I.
Answer: A⁻¹ = [0.5000 0.5000 -0.5000; 0.5000 5.5000 -3.5000; -0.5000 -3.5000 2.5000]
✓ Answer checked by Claude Opus 5.5 (high effort)
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