IB Maths AI HL Matrices Paper 1 & 2 Determinant & inverse ~8 min read

Determinants & Inverses

The determinant is a single number that tells you whether a square matrix has an inverse. When it does, the inverse undoes matrix multiplication — turning matrix equations like AX = B into one-line solutions.

📘 What you need to know

Only square matrices have determinants and possible inverses.
For a 2×2 matrix A = ((a, b), (c, d)): det A = ad − bc. In the formula booklet.
A is invertible iff det A ≠ 0. If det A = 0, A is singular and has no inverse.
For invertible 2×2: A⁻¹ = 1det A ((d, −b), (−c, a)). In the formula booklet.
AA⁻¹ = A⁻¹A = I.
Matrix equations: AX = B ⇒ X = A⁻¹B (pre-multiply); XA = B ⇒ X = BA⁻¹ (post-multiply).

Determinants of square matrices

The determinant of a 2×2 matrix is a cross-multiplication: the product of the main diagonal minus the product of the anti-diagonal. For A = ((a, b), (c, d)), det A = ad − bc. You only need to compute 2×2 determinants by hand; for 3×3 and larger, use your GDC.

Useful properties: det(I) = 1, det(O) = 0, det(AB) = det(A) det(B), and det(kA) = kⁿ det(A) for an n × n matrix (so k² for 2×2).

Inverses and the 2×2 formula

The inverse A⁻¹ is the matrix that satisfies AA⁻¹ = I. It exists exactly when det A ≠ 0. For 2×2 matrices the formula has a memorable pattern: swap the diagonal, negate the off-diagonal, divide by the determinant.

The 2×2 inverse always follows the same recipe — swap the diagonal entries, negate the other two, and divide everything by the determinant.

Determinant and inverse (2×2) A = ( ( a, b ), ( c, d ) ) ⇒ det A = ad − bc A⁻¹ = 1det A ( ( d, −b ), ( −c, a ) ), det A ≠ 0

Solving matrix equations with inverses

An inverse cancels the matrix it inverts, just like dividing by x cancels multiplication by x. But because matrix multiplication is non-commutative, which side you multiply on matters. Pre-multiplying isolates X when A is on the left; post-multiplying when A is on the right.

Pre- vs post-multiply: from AX = B, multiply both sides on the left by A⁻¹ to get X = A⁻¹B. From XA = B, multiply on the right by A⁻¹ to get X = BA⁻¹. Always keep the inverse on the same side as the matrix you’re cancelling.

🧭 Recipe — solving a matrix equation by inversion

Identify the form: is it AX = B (left-multiplied) or XA = B (right-multiplied)?
Compute det A; if it is zero, the equation cannot be solved by inversion.
Find A⁻¹ using the formula (2×2 by hand, larger via GDC).
Apply the inverse on the matching side: pre-multiply for AX = B, post-multiply for XA = B.
Compute the product A⁻¹B or BA⁻¹ to get X explicitly.

Worked examples

WE 1

Determinant of a 2×2 matrix

Find det A for A = ((5, 2), (−3, 4)).

apply ad − bc det A = (5)(4) − (2)(−3) = 20 − (−6) = 20 + 6 det A = 26 watch the double negative — subtracting a negative becomes adding.

WE 2

Find a missing entry from a given determinant

Given B = ((2, −3), (4, k)) and det B = 10, find k.

set up the determinant equation det B = (2)(k) − (−3)(4) = 2k + 12 equate to 10 and solve 2k + 12 = 10 ⇒ 2k = −2 k = −1 check: det((2, −3),(4, −1)) = −2 + 12 = 10 ✓

WE 3

Properties: scalar and product determinants

Let C be a 2×2 matrix with det C = 7. Find (a) det(3C) and (b) det(C²).

(a) det(kA) = k² det(A) for a 2×2 det(3C) = 3² · 7 = 9 · 7 = 63 (b) det(AB) = det(A) det(B); take B = C det(C²) = det(C) · det(C) = 7 · 7 = 49 (a) 63 · (b) 49 both follow from determinant property rules — no need to know C explicitly.

WE 4

Inverse of a 2×2 matrix

Find the inverse of M = ((3, 5), (2, 4)).

compute the determinant first det M = (3)(4) − (5)(2) = 12 − 10 = 2 swap diagonal, negate off-diagonal, divide by det M⁻¹ = (1/2) · ((4, −5), (−2, 3)) = ((2, −5/2), (−1, 3/2)) M⁻¹ = ((2, −5/2), (−1, 3/2)) check: M · M⁻¹ should give the 2×2 identity.

WE 5

For what value is the matrix singular?

Find the value of a for which N = ((a, 4), (3, 6)) is singular.

singular means det = 0 det N = 6a − 12 = 0 solve for a 6a = 12 ⇒ a = 2 a = 2 at a = 2 the matrix has no inverse — rows or columns are proportional.

WE 6

Full question: solve PX = Q for X

Let P = ((1, 2), (3, 7)) and Q = ((2, 1), (5, 8)). (a) Find P⁻¹. (b) Solve PX = Q for X.

(a) det P = (1)(7) − (2)(3) = 1 P⁻¹ = (1/1) · ((7, −2), (−3, 1)) = ((7, −2), (−3, 1)) (b) PX = Q ⇒ X = P⁻¹ Q (pre-multiply) X₁₁ = (7)(2) + (−2)(5) = 14 − 10 = 4 X₁₂ = (7)(1) + (−2)(8) = 7 − 16 = −9 X₂₁ = (−3)(2) + (1)(5) = −1 X₂₂ = (−3)(1) + (1)(8) = 5 (a) P⁻¹ = ((7, −2), (−3, 1)) · (b) X = ((4, −9), (−1, 5)) check: PX should equal Q — quick multiply confirms.

💡 Top tips

Always compute det first — if it is zero, stop; there is no inverse.
The 2×2 inverse pattern: swap diagonal, negate off-diagonal, divide by det.
For 3×3 (or bigger) determinants and inverses, use your GDC — you are not expected to do them by hand.
Use det(AB) = det(A) det(B) and det(kA) = kⁿ det(A) to handle products and scalars without expanding.
Match the side of the inverse: pre-multiply for AX = B; post-multiply for XA = B.

⚠ Common mistakes

Trying to invert a non-square matrix or one with det = 0 — impossible.
Forgetting to divide by det A in the 2×2 inverse formula — the entries alone aren’t enough.
Wrong signs: it’s swap-then-negate, not just negate every entry.
Pre-multiplying when you should post-multiply (or vice versa) — X = A⁻¹B only solves AX = B.
Treating det(kA) as k det(A) — for an n × n matrix it is kⁿ det(A).

Next up: Solving Systems of Linear Equations with Matrices — using the inverse trick from this note to crack 2×2 and 3×3 systems of equations in one matrix step.

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