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Three equations, three unknowns, and a question that wants only z. Cramer's rule writes z straight away as one determinant divided by another. Gabriel Cramer published it in 1750, in a book about curves. This post covers who he was, what his appendix actually says, a worked example of the kind of problem he had in mind, and the mistakes that cost marks.
Who was Gabriel Cramer?
His MacTutor biography gives his birth in Geneva on 31 July 1704 and his death at Bagnols-sur-Cèze, France, on 4 January 1752. He earned a doctorate in 1722, aged 18, with a thesis on sound. In 1724 Geneva's Academy split a chair so that Cramer and Giovanni Calandrini could share the teaching of mathematics. Cramer taught his courses in French rather than Latin, so that people without Latin could follow them.
From 1727 to 1729 he travelled, working with Johann Bernoulli in Basel and meeting Halley, de Moivre and Stirling in England. He later edited the collected works of Johann Bernoulli (1742) and Jacob Bernoulli (1744), work that shows how much the leading mathematicians trusted him.
Why did a book on curves need a rule for equations?
His major book, Introduction à l'analyse des lignes courbes algébriques (1750), classifies algebraic curves. MacTutor explains the step that led to the rule: an equation of degree n has ½n² + (3/2)n adjustable constants, so a curve of degree n can be made to pass through that many given points. For n = 2 that is 2 + 3 = 5, so a conic can be passed through five points.
Each point gives one equation that is linear in the unknown constants. Five points give five linear equations in five unknowns. In a footnote, Cramer says he believes he has found a convenient general rule for any number of equations and unknowns of the first degree, and sends the reader to the appendix.
What did the appendix say?
On pages 657 to 659, Cramer writes each unknown as a fraction. The common denominator has as many terms as there are arrangements of n things, so 1 × 2 × 3 = 6 terms for three unknowns. Each term multiplies one coefficient from each equation, and the equation numbers run through the orders 123, 132, 213, 231, 312 and 321.
For the sign, he counts what he calls a dérangement: a number followed, anywhere later in the order, by a smaller one. An even count, or none, gives +; an odd count gives −. His own example: 312 has two, so it takes +, while 321 has three, so it takes −. The numerator for an unknown comes from the denominator by replacing that unknown's coefficients with the constant terms. He also notes that when the common denominator is zero, the problem may become indeterminate or impossible.
He was not the first. The MacTutor history of matrices and determinants notes that Colin Maclaurin's Treatise of Algebra, published in 1748, proved the rule for two and three unknowns. Cramer stated it for n unknowns, without proof. Our post on Arthur Cayley picks up the story a century later.
Worked example: a parabola through three points
Here is a smaller version of Cramer's problem. Find y = ax² + bx + c through (1, 2), (2, 3) and (3, 6). Substituting gives:
a + b + c = 2
4a + 2b + c = 3
9a + 3b + c = 6
D, from rows (1, 1, 1), (4, 2, 1), (9, 3, 1): 1(2 − 3) − 1(4 − 9) + 1(12 − 18) = −1 + 5 − 6 = −2.
Da, with the constants (2, 3, 6) in the first column: 2(2 − 3) − 1(3 − 6) + 1(9 − 12) = −2 + 3 − 3 = −2. So a = −2/−2 = 1.
Db, constants in the second column: 1(3 − 6) − 2(4 − 9) + 1(24 − 27) = −3 + 10 − 3 = 4. So b = 4/−2 = −2.
Dc, constants in the third column: 1(12 − 9) − 1(24 − 27) + 2(12 − 18) = 3 + 3 − 12 = −6. So c = −6/−2 = 3.
The curve is y = x² − 2x + 3. Check: x = 1 gives 2, x = 2 gives 3, x = 3 gives 6.
What mistakes do students make?
Replacing a row. The constants replace the column of the unknown you want.
Middle-term sign. In expanding along the first row, the second term carries a minus sign.
Over-reading D = 0. With three unknowns, D = 0 and all three numerator determinants zero does not mean infinitely many solutions. x + y + z = 1, x + y + z = 2, x + y + z = 3 has every determinant zero and no solution. Our post on consistency of linear equations shows the full test.
Using it on big systems. The number of terms grows as n!, so five unknowns need 120. MacTutor records that Laplace in 1772 called the methods of Cramer and Bezout impractical.
Where is it in the syllabus?
The JEE (Main) 2026 syllabus lists the "solution of simultaneous linear equations in two or three variables using matrices", and the JEE (Advanced) 2026 brochure lists solutions of such equations. The current NCERT Class 12 determinants chapter teaches the matrix method and does not name Cramer's rule. If a board question asks for the matrix method, use it and keep Cramer's rule for checking. Check your exam year's documents.
If sign slips keep creeping into your expansions, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can compare each cofactor with your own.
What can a student take from Cramer's way of working?
Tools come from problems. He needed curves through points, and the rule followed.
Spot the pattern, then state it generally. He went from small cases to n unknowns with one sign rule.
Teach in the reader's language. Choosing French over Latin opened his lectures to more people.
Next step: use Cramer's rule to find the parabola through (0, 1), (1, 0) and (2, 3), then check by substitution.
