Linear Algebra · step 1 of 13

Matrix arithmetic & inverses

Matrix multiplication combines rows with columns: entry (i,j) of AB is the dot product of row i of A with column j of B, so the inner dimensions must match. Multiplication is associative but NOT commutative — AB and BA are usually different and may not even both exist. A square matrix is invertible exactly when its determinant is non-zero, and (AB)⁻¹ = B⁻¹A⁻¹ with the order reversed.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Check the dimensionsAn m×n times an n×p gives an m×p. If the inner numbers differ the product is undefined.
Compute entry by entryEntry (i,j) is row i of A dotted with column j of B. Work along one row at a time.
Invert a 2×2 with the formulaFor [[a,b],[c,d]] the inverse is (1/(ad−bc))·[[d,−b],[−c,a]] — swap the diagonal, negate the off-diagonal, divide by the determinant.
Invert larger matrices by row reductionAugment with the identity, row reduce to [I | A⁻¹]. If the left side cannot reach I, the matrix is singular.

Worked example

Find the inverse of A = [[3, 1], [5, 2]] and verify it.

det A = 3·2 − 1·5 = 6 − 5 = 1, non-zero so the inverse exists.
Apply the formula: swap the diagonal to get 2 and 3, negate the off-diagonal to get −1 and −5.
A⁻¹ = (1/1)·[[2, −1], [−5, 3]].
Check: [[3,1],[5,2]]·[[2,−1],[−5,3]] = [[6−5, −3+3],[10−10, −5+6]] = [[1,0],[0,1]] ✓.

Answer. A⁻¹ = [[2, −1], [−5, 3]].

Where marks get dropped

These are the specific errors that cost credit on matrix arithmetic & inverses questions — QED's rubric penalises each of them separately.

Assuming AB = BA. Matrix multiplication is non-commutative, and most exam counterexamples are 2×2.
Inverting entrywise. The inverse is not the matrix of reciprocals — that operation has no meaning here.
Getting (AB)⁻¹ wrong. It is B⁻¹A⁻¹, not A⁻¹B⁻¹; the order reverses just as it does for transposes.

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Matrix arithmetic & inverses — frequently asked questions

When is a matrix invertible?

Exactly when det ≠ 0, equivalently when the rows are linearly independent, the rank is full, and 0 is not an eigenvalue. All these conditions are equivalent.

Why does the order reverse in (AB)⁻¹?

Because (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I. Undoing a composition means undoing the last operation first.

Is AB = 0 possible with A, B ≠ 0?

Yes — matrices have zero divisors. [[1,0],[0,0]]·[[0,0],[0,1]] = 0 despite neither factor being zero.

The rest of Linear Algebra

Matrices, systems, determinants, eigenvalues. Each subtopic below has its own method, worked example and mark-losing traps.

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