Matrices and Determinants
Matrix Inverses and Solving Linear Systems
The matrix that undoes multiplication, the 2 × 2 formula, the Gauss-Jordan method for larger cases, and solving Ax = b directly.
What this page covers
- State the defining property of an inverse
- Compute the inverse of a 2 × 2 matrix
- Find an inverse by Gauss-Jordan reduction
- Solve a linear system using the inverse
The idea
To solve 2x = 4 you multiply both sides by 12, or equivalently by 2-1. The source makes the analogy explicit before generalising it.
The same manoeuvre then solves a system:
Multiply on the left on both sides. Matrix multiplication does not commute, so A-1b and bA-1 are different, and the second may not even exist.
The 2×2 formula
For A = [[a, b], [c, d]] with |A| = ad - bc ≠ 0:
The recipe is short: swap the diagonal entries, negate the off-diagonal ones, and divide by the determinant.
Worked example — the source's case
For A = [[1, -3], [2, 1]]: the determinant is (1)(1) - (-3)(2) = 1 + 6 = 7.
| 1/7 | 3/7 |
| -2/7 | 1/7 |
AA-1 should be I. Top-left entry: (1)(17) + (-3)(-27) = 17 + 67 = 1 ✓. Top-right: (1)(37) + (-3)(17) = 0 ✓.
If ad - bc = 0 the formula divides by zero and no inverse exists. Such a matrix is called singular, and the corresponding system has either no solution or infinitely many — never exactly one. The source records the consequence as if |A| = 0, no solution.
Solving a system with the inverse
Worked example — the source's first
Solve x - 3y = 4, 2x + y = 1.
1 - 3(-1) = 4 ✓ and 2(1) + (-1) = 1 ✓. This agrees with the Gauss-Jordan solution of the same system.
Worked example — the source's second
Solve x - 3y = 2, 2x + y = -3. The same coefficient matrix, so the same inverse serves.
This is where the inverse method earns its place: the expensive step is inverting A, and once done it handles any right-hand side at negligible extra cost.
Worked example — a different matrix
Solve 4x + 5y = 18, 2x - 3y = -2.
Determinant: (4)(-3) - (5)(2) = -12 - 10 = -22.
4(2) + 5(2) = 18 ✓ and 2(2) - 3(2) = -2 ✓.
Inverting by Gauss-Jordan
For matrices larger than 2 × 2 there is no short formula worth using. The source gives a two-step procedure instead, and derives why it works.
Since row operations are left multiplications, reducing [A | I] until the left block becomes I applies A-1 to the whole augmented matrix — and so turns the right block into A-1.
- Form [A | I], the matrix beside an identity of the same size.
- Row-reduce until the left block is the identity.
- Read off the right block, which is now A-1.
- Check that AA-1 = I.
Worked example — the source's 3×3 case
Invert A = [[1, 0, 1], [2, 0, -1], [1, 2, 2]].
The source forms [A | I] and reduces. Its result:
| 1/3 | 1/3 | 0 |
| -5/6 | 1/6 | 1/2 |
| 2/3 | -1/3 | 0 |
The source then verifies by multiplying, computing AA-1 and A-1A and obtaining the identity both times — which is the right check, since both products must give I.
Row 1 of A times column 1 of A-1: (1)(13) + (0)(-56) + (1)(23) = 1 ✓.
When to use which method
| Situation | Method | Reason |
|---|---|---|
| One 2 × 2 system | Either | The formula is quick enough |
| One larger system | Gaussian elimination | Inverting costs more than solving |
| Same A, many different b | Inverse | Invert once, reuse indefinitely |
| |A| = 0 | Neither works | No inverse; use elimination to classify the system |
| Machine computation | Elimination with pivoting | Inverting is slower and numerically worse |
The inverse is more valuable as a concept than as a computational tool. It explains why a system has a unique solution exactly when the determinant is non-zero, which elimination shows but does not explain.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Swapping the off-diagonal entries | Swap a and d; negate b and c | The formula is specific |
| Forgetting to divide by the determinant | The whole matrix is scaled | Otherwise AA-1 ≠ I |
| Inverting a non-square matrix | Only square matrices have inverses | I must exist on both sides |
| Multiplying b on the right | Left-multiply: x = A-1b | Multiplication does not commute |
| Proceeding when |A| = 0 | No inverse exists | Compute the determinant first |
| Checking only AA-1 | Both products should give I | The source checks both |
Frequently asked questions
When does an inverse exist?
For a square matrix whose determinant is non-zero. The source records the consequence directly: if |A| = 0, no solution.
Why does the inverse method work?
Because Ax = b gives A-1Ax = A-1b, hence Ix = A-1b and so x = A-1b. It is the matrix version of dividing both sides by the coefficient.
Is the inverse method efficient?
Not for a single system — Gaussian elimination is faster. It pays off when the same coefficient matrix is used with many different right-hand sides.
How do I invert a larger matrix?
Form [A | I] and row-reduce until the left block is I. The right block is then A-1. The source sets this out as a two-step procedure.
Source. Handwritten teaching notes, Week 12, pages 4-6 and 9-10.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.