MTH2021 · Linear Algebra with Applications
Matrix Algebra, Inverses & Elementary Matrices
Weeks 1–2 build the algebra of matrices — addition, the product (AB)ᵢⱼ = Σₖ aᵢₖbₖⱼ, transpose, trace — and then the inverse, computed by the inversion algorithm that row-reduces [A | I] to [I | A⁻¹]. Elementary matrices tie the story together and lead to the Equivalent-Statements Theorem, the recurring list of conditions equivalent to invertibility. This material is examined in Quizzes 1–2 and Test 1 (Weeks 1–2), and the equivalent-statements list is a staple of the proof-heavy final.
What this chapter covers
- 01Matrix operations and laws: (A+B), kA, product (AB)ᵢⱼ = Σₖ aᵢₖbₖⱼ; multiplication is associative but NOT commutative
- 02Transpose (Aᵀ)ⱼᵢ = (A)ᵢⱼ, symmetric matrices Aᵀ = A, trace tr(A) = Σ aᵢᵢ with tr(AB) = tr(BA)
- 03Ax as a linear combination of the columns of A (Ax = x₁a₁ + … + xₙaₙ)
- 04The inverse: A invertible ⇔ ∃B, AB = BA = I; uniqueness; (AB)⁻¹ = B⁻¹A⁻¹, (Aᵀ)⁻¹ = (A⁻¹)ᵀ
- 05Elementary matrices E = ρ(I); ρ(A) = EA; every elementary matrix is invertible
- 06The inversion algorithm: row-reduce [A | I] → [I | A⁻¹]; a zero row on the left ⇒ singular
- 07Equivalent-Statements Theorem: A invertible ⇔ Ax = 0 only trivially ⇔ RREF(A) = I ⇔ A is a product of elementary matrices
- 08A invertible ⇔ Ax = b has exactly one solution for every b
Inverting a 3×3 matrix by the [A | I] algorithm
- +1Form [A | I] = [[2,1,1|1,0,0],[1,2,1|0,1,0],[1,1,2|0,0,1]]. Swap R1 ↔ R2 to get a leading 1: [[1,2,1|0,1,0],[2,1,1|1,0,0],[1,1,2|0,0,1]].
- +1Clear column 1: R2 → R2 − 2R1 and R3 → R3 − R1 give [[1,2,1|0,1,0],[0,−3,−1|1,−2,0],[0,−1,1|0,−1,1]].
- +1Get a clean pivot in column 2: scale R3 → −R3 to (0,1,−1 | 0,1,−1) and use it to clear column 2. R1 → R1 − 2R3 gives (1,0,3 | 0,−1,2); R2 → R2 + 3R3 gives (0,0,−4 | 1,1,−3).
- +1Pivot in column 3: R2 → −(1/4)R2 gives (0,0,1|−1/4,−1/4,3/4). Clear column 3: R1 → R1 − 3R2 gives (1,0,0|3/4,−1/4,−1/4); R3 → R3 + R2 gives (0,1,0|−1/4,3/4,−1/4).
- +1Read off the right block: A⁻¹ = (1/4)·[[3,−1,−1],[−1,3,−1],[−1,−1,3]]. Check one entry: row 1 of A times column 1 of A⁻¹ = (1/4)(2·3 + 1·(−1) + 1·(−1)) = (1/4)(6−1−1) = 1 ✓.
Key terms
- Matrix product
- (AB)ᵢⱼ = Σₖ (A)ᵢₖ(B)ₖⱼ, defined when A is m×r and B is r×n. Associative, distributes over addition, but generally AB ≠ BA.
- Transpose
- The matrix Aᵀ with (Aᵀ)ⱼᵢ = (A)ᵢⱼ; rows become columns. Key law (AB)ᵀ = BᵀAᵀ. A is symmetric when Aᵀ = A.
- Trace
- tr(A) = Σᵢ (A)ᵢᵢ, the sum of the diagonal entries. Linear, and tr(AB) = tr(BA), though tr(AB) ≠ tr(A)tr(B) in general.
- Inverse matrix
- For square A, the unique B with AB = BA = I, written A⁻¹; A is then invertible (nonsingular). (AB)⁻¹ = B⁻¹A⁻¹ and (Aᵀ)⁻¹ = (A⁻¹)ᵀ.
- Elementary matrix
- The matrix E obtained by performing a single elementary row operation on I. Then EA is A with that same row operation applied; every elementary matrix is invertible with an elementary inverse.
- Equivalent-Statements Theorem
- For an n×n matrix A the following are equivalent: A is invertible; Ax = 0 has only the trivial solution; RREF(A) = Iₙ; A is a product of elementary matrices; Ax = b has a unique solution for every b.
Matrix Algebra, Inverses & Elementary Matrices FAQ
How does the inversion algorithm actually find A⁻¹?
Row-reducing [A | I] to RREF applies a sequence of elementary matrices E_k…E₁ that turns A into I. So E_k…E₁·A = I, which means E_k…E₁ = A⁻¹. The same operations acting on the I block on the right therefore assemble A⁻¹ directly. If the left block cannot reach I (a zero row appears), A is singular.
Why is matrix multiplication not commutative?
Because the product mixes rows of the first matrix with columns of the second in an order-dependent way; even the shapes may only fit one way round. A concrete 2×2 counterexample shows AB ≠ BA, so you must preserve order in every identity — for instance (AB)⁻¹ = B⁻¹A⁻¹ and (AB)ᵀ = BᵀAᵀ reverse the factors.
What is the point of the Equivalent-Statements Theorem?
It bundles many different-looking conditions — invertibility, only-trivial null space, RREF = I, unique solutions for every b, later det ≠ 0 and full rank — into one interchangeable list. In a proof you can enter the list through whichever condition you are handed and exit through whichever you need, which is exactly what the handwritten exam questions reward.
Do I ever need a one-sided inverse?
For a square matrix a one-sided inverse is automatically a full inverse: if AB = I (or BA = I) for square A, B, then B = A⁻¹. This is why you only have to verify one product in practice. For non-square matrices the situation differs, which motivates the pseudoinverse in the SVD chapter.
How is this assessed?
Matrix algebra and inversion are drilled in Quizzes 1–2 and appear in Test 1 (Weeks 1–2, closed book). In the final's handwritten section expect a short proof using the inverse laws or the equivalent-statements list rather than only a numerical inversion, so rehearse the arguments, not just the row reduction.
Exam move
Practise the [A | I] → [I | A⁻¹] algorithm until the bookkeeping is automatic, and always spot-check one entry of A·A⁻¹ = I. Memorise the order-reversing laws (AB)⁻¹ = B⁻¹A⁻¹ and (AB)ᵀ = BᵀAᵀ and the Equivalent-Statements list, because the exam tests them as short proofs, not just computations. Week 1–2 material is examined in Quizzes 1–2 and Test 1 (closed book); keep it warm since the equivalent-statements list recurs in the determinant, rank and eigenvalue chapters. When your inversion arithmetic drifts, ask Sia to re-check each row operation.
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