Matrix Determinant and Inverse (2×2 and 3×3) by Hand

By Aniket Bhardwaj · 2 September 2026 · Maths & Physics

Determinants and inverses are where Class 12 algebra gets mechanical — and where marks disappear to sign errors rather than to misunderstanding. This article does the whole job by hand: a full 3×3 determinant by cofactor expansion (cross-checked a second way), a 2×2 inverse multiplied back to the identity to prove it is right, the same matrix used to solve a pair of equations, and a 3×3 inverse by the adjugate method.

The formulas

2×2: A = [ a b ; c d ] → det A = ad − bc and A⁻¹ = (1/det A) [ d −b ; −c a ]

n×n cofactor expansion along row i: det A = Σ_j a_ij C_ij, where C_ij = (−1)^i+j M_ij

Inverse, any size: A⁻¹ = adj(A) / det A, where adj(A) is the transpose of the matrix of cofactors. Exists only if det A ≠ 0.

What the pieces mean

Minor M_ij — the determinant of what is left after you delete row i and column j.
Cofactor C_ij — the minor with a sign attached. The signs follow a chessboard pattern:
+ − + / − + − / + − +
Adjugate adj(A) — the cofactor matrix, transposed. The transpose step is the one people forget.
det A — a single number. Geometrically it is the factor by which the matrix scales area (2×2) or volume (3×3). If det A = 0 the transformation flattens space, nothing can be un-flattened, and no inverse exists.

Worked example 1 — a full 3×3 determinant by cofactor expansion

Question: Find det A for
A = [ 2 −1 3 ; 0 5 2 ; 1 −1 4 ]

Expand along row 1. The entries are a₁₁ = 2, a₁₂ = −1, a₁₃ = 3, and the row-1 signs are + − +.

Minor M₁₁ — delete row 1 and column 1, leaving [ 5 2 ; −1 4 ]:
M₁₁ = (5)(4) − (2)(−1) = 20 + 2 = 22 → C₁₁ = +22

Minor M₁₂ — delete row 1 and column 2, leaving [ 0 2 ; 1 4 ]:
M₁₂ = (0)(4) − (2)(1) = 0 − 2 = −2 → C₁₂ = −(−2) = +2

Minor M₁₃ — delete row 1 and column 3, leaving [ 0 5 ; 1 −1 ]:
M₁₃ = (0)(−1) − (5)(1) = 0 − 5 = −5 → C₁₃ = +(−5) = −5

Combine:
det A = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃
det A = (2)(22) + (−1)(2) + (3)(−5)
det A = 44 − 2 − 15 = 27

Cross-check — expand along column 1 instead. Column 1 contains a zero, so one whole term vanishes. The column-1 signs are + − +.
a₁₁ = 2, C₁₁ = +22 → 44
a₂₁ = 0 → contributes 0 (no minor needed)
a₃₁ = 1, M₃₁ = det[ −1 3 ; 5 2 ] = (−1)(2) − (3)(5) = −2 − 15 = −17, sign +, so C₃₁ = −17 → 1 × (−17) = −17
det A = 44 + 0 − 17 = 27 ✔ — the two expansions agree.

Exam tip built into that check: always expand along the row or column with the most zeros. Here column 1 needed only two minors instead of three, and it doubles as a free verification of the row-1 answer.

Worked example 2 — a 2×2 inverse, verified against the identity

Question: Find A⁻¹ for A = [ 4 7 ; 2 6 ] and prove your answer is correct.

Step 1 — determinant: det A = (4)(6) − (7)(2) = 24 − 14 = 10. Non-zero, so the inverse exists.

Step 2 — swap the diagonal, negate the off-diagonal:
adj(A) = [ 6 −7 ; −2 4 ]

Step 3 — divide by the determinant:
A⁻¹ = (1/10) [ 6 −7 ; −2 4 ] = [ 0.6 −0.7 ; −0.2 0.4 ]

Step 4 — multiply back and check you get the identity. Every entry of A · A⁻¹ is a row of A dotted with a column of A⁻¹:

Row 1 × Col 1: (4)(0.6) + (7)(−0.2) = 2.4 − 1.4 = 1
Row 1 × Col 2: (4)(−0.7) + (7)(0.4) = −2.8 + 2.8 = 0
Row 2 × Col 1: (2)(0.6) + (6)(−0.2) = 1.2 − 1.2 = 0
Row 2 × Col 2: (2)(−0.7) + (6)(0.4) = −1.4 + 2.4 = 1

A · A⁻¹ = [ 1 0 ; 0 1 ] = I ✔ The inverse is correct.

That last step takes about thirty seconds and it is the only proof that matters. If the product is not exactly I, you have a sign or an arithmetic error and you still have time to find it.

Worked example 3 — using the inverse to solve equations

Question: Solve 4x + 7y = 22 and 2x + 6y = 16.

In matrix form AX = B with the same A as above and B = [ 22 ; 16 ], the solution is X = A⁻¹B:

x = (0.6)(22) + (−0.7)(16) = 13.2 − 11.2 = 2
y = (−0.2)(22) + (0.4)(16) = −4.4 + 6.4 = 2

Substitute back: 4(2) + 7(2) = 8 + 14 = 22 ✔ and 2(2) + 6(2) = 4 + 12 = 16 ✔

Worked example 4 — the 3×3 inverse by the adjugate method

Question: Find A⁻¹ for the matrix of example 1, whose determinant we already know is 27.

All nine cofactors (chessboard signs + − + / − + − / + − +):
C₁₁ = +[(5)(4) − (2)(−1)] = +22    C₁₂ = −[(0)(4) − (2)(1)] = +2    C₁₃ = +[(0)(−1) − (5)(1)] = −5
C₂₁ = −[(−1)(4) − (3)(−1)] = −(−4 + 3) = +1    C₂₂ = +[(2)(4) − (3)(1)] = +5    C₂₃ = −[(2)(−1) − (−1)(1)] = −(−2 + 1) = +1
C₃₁ = +[(−1)(2) − (3)(5)] = −17    C₃₂ = −[(2)(2) − (3)(0)] = −4    C₃₃ = +[(2)(5) − (−1)(0)] = +10

Transpose to get the adjugate (rows of cofactors become columns):
adj(A) = [ 22 1 −17 ; 2 5 −4 ; −5 1 10 ]

Divide by det A = 27: A⁻¹ = (1/27) × adj(A)

Verification, row 1 of A against each column of adj(A):
(2)(22) + (−1)(2) + (3)(−5) = 44 − 2 − 15 = 27 → 27/27 = 1
(2)(1) + (−1)(5) + (3)(1) = 2 − 5 + 3 = 0 → 0
(2)(−17) + (−1)(−4) + (3)(10) = −34 + 4 + 30 = 0 → 0

Repeating for rows 2 and 3 gives (0, 27, 0) and (0, 0, 27), which divided by 27 are (0, 1, 0) and (0, 0, 1). The product is the 3×3 identity ✔

Common mistakes

Forgetting to transpose the cofactor matrix. adj(A) is the transpose. Skip it and only the diagonal of your check will come out right — which is exactly why you must test all nine products, not one.
Losing the (−1)^i+j sign. C₁₂ = −M₁₂. In example 1 the minor was −2 and the cofactor was +2; dropping the sign turns det A = 27 into 23.
Dividing by the determinant twice — once inside the adjugate and again outside. Divide exactly once, at the end.
Trying to invert a singular matrix. For B = [ 3 6 ; 1 2 ], det B = (3)(2) − (6)(1) = 6 − 6 = 0, so B⁻¹ does not exist and the corresponding equations have either no solution or infinitely many. Always compute the determinant before starting the inverse.
Assuming AB = BA. Matrix multiplication is not commutative, though A · A⁻¹ = A⁻¹ · A = I is a genuine exception.

Where this appears in exams

Exam	Typical question
CBSE/ICSE Class 12	Determinants and minors, adjugate and inverse, solving three equations by the matrix method
JEE Main & Advanced	Properties of determinants, consistency of a system, adj(A) identities such as \|adj A\| = \|A\|ⁿ⁻¹
Class 11–12 Physics	Solving simultaneous circuit equations from Kirchhoff's laws
IIT-JAM / GATE Chemistry	Secular determinants in Hückel MO theory, symmetry and character tables, least-squares fitting

Check the determinant before you spend ten minutes on an inverse. The Matrix calculator handles 2×2 and 3×3 determinants, inverses and products, so you can confirm your cofactors first and your adjugate second — and see immediately if det A = 0.

Open the Matrix Calculator →

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