Target Audience: Anyone who can compute a determinant and wants to know why the rules are what they are.
Prerequisites: Vector Spaces and Linear Maps — row reduction, rank and linear independence are assumed throughout, and the phrase “the rows are dependent” does a great deal of work on this page.
Most courses introduce the determinant as a formula and then list its properties. This page runs the other way round, because that is how the proofs on this site are organised: three properties come first, they determine a single function, and every formula afterwards is proved by checking those three properties and invoking uniqueness.
The consequence is worth stating up front. You never have to show that two formulas for the determinant agree. You show each one satisfies the axioms, and uniqueness does the rest. Sections 6 and 7 both work this way, and the explorer in Section 10 computes all three definitions on the same matrix so you can watch them land on the same number.
Everything numerical on this page was verified in exact rational arithmetic before it was written: the three definitions were checked against each other on 248 matrices, and the identities in Sections 8 to 10 on every one of them.
The determinant attaches a single number to a square matrix, and that number answers one question: does this matrix collapse space?
Think of the rows of \(A\) as vectors. They span a parallelogram in two dimensions, a parallelepiped in three, and an \(n\)-dimensional box in general. Then \(|\det A|\) is the volume of that box, and the sign of \(\det A\) records whether \(A\) preserves orientation or reverses it.
This single reading explains most of the algebra that follows. Multiplicativity is scale factors multiplying. A zero determinant is a flattened box. Scaling one row scaling the determinant is one edge of the box getting longer.
The collapse question is the useful one. If the rows are linearly dependent they all lie inside some lower-dimensional slab, the box built on them is flat, and its volume is \(0\). So these are two descriptions of the same situation:
Everything else the determinant is used for — invertibility, Cramer’s rule, eigenvalues in the next tutorial — runs through that equivalence.
When \(\det A = 0\) for an \(n \times n\) matrix, all of the following hold at once: \(A\) has no inverse; \(\operatorname{rank} A \lt n\); the rows are dependent; the columns are dependent; and \(A\mathbf{x} = \mathbf{0}\) has a solution other than \(\mathbf{0}\). Recognising that these are one fact rather than five is most of what makes the determinant useful.
Write \(A\) as a stack of rows \(r_1, \ldots, r_n\). The determinant is defined as the function of those rows satisfying three conditions.
Hold every row fixed except one. As a function of that single remaining row, \(\det\) is linear: \[ \det(\ldots, c\,r_p, \ldots) = c \det(\ldots, r_p, \ldots), \qquad \det(\ldots, u + v, \ldots) = \det(\ldots, u, \ldots) + \det(\ldots, v, \ldots). \]
If two rows of \(A\) are equal, then \(\det A = 0\).
\(\det I = 1\).
Two questions immediately arise, and the site answers both on one page.
Existence. A function satisfying D1, D2 and D3 can be built — by the cofactor recursion of Section 6.
Uniqueness. There is only one. Any function satisfying the three axioms takes the same value on every square matrix.
The Determinant: Existence and Uniqueness proves both.
Without D3 the function that is identically \(0\) satisfies D1 and D2 perfectly well, and so does \(7\det\). Normalisation is what reduces a whole family of functions to one.
It is tempting to treat existence as the substantial theorem and uniqueness as tidying-up. The opposite is true in practice. Every later result on this page — cofactor expansion along any row, the Leibniz formula, the product rule — is proved the same way: show the candidate satisfies D1, D2 and D3, then conclude it is the determinant. Not one of them requires comparing two formulas term by term.
D1 does not say \(\det(A + B) = \det A + \det B\), and that equation is false. Take \(A = B = I\) in the \(2 \times 2\) case: the left side is \(\det(2I) = 4\) and the right side is \(1 + 1 = 2\). For the same reason \(\det(cA) = c^{n}\det A\), not \(c \det A\) — one factor of \(c\) comes out of each of the \(n\) rows.
For a matrix with rows \((a, b)\) and \((c, d)\) the three axioms force
Those two terms are not arbitrary. Each is a product of one entry from each row and each column — there are exactly two such choices — and the signs are the alternating property showing through. That observation, generalised, is the whole content of Section 7.
Rows \((3, 8)\) and \((4, 6)\): \(\det = 3 \cdot 6 - 8 \cdot 4 = 18 - 32 = -14\). The box built on those two rows has area \(14\), and the negative sign says the pair is oriented the opposite way round from the standard basis.
For three rows the count is \(3! = 6\) terms, and writing them out by hand is already tedious enough that the methods in the next three sections earn their keep. The matrix
has \(\det A = -85\). It is used as the running example for the rest of the page, and the explorer computes it three separate ways.
The rule taught for \(3 \times 3\) matrices — add the three downward diagonals, subtract the three upward ones — is correct, and it is correct only there. It produces \(6\) terms, and \(4 \times 4\) determinants need \(24\). Extending the pattern to larger matrices is a genuine and common error.
The three elementary row operations each have a known effect on the determinant. This short table is the bridge between the axioms and any practical computation.
| Operation | Effect on \(\det\) | Where it comes from |
|---|---|---|
| Swap two rows | Negated | D1 and D2 together |
| Scale a row by \(c \neq 0\) | Multiplied by \(c\) | D1 directly |
| Add a multiple of one row to another | No change | D1 to split, then D2 to kill the second piece |
Adding \(c\,r_q\) to \(r_p\) gives, by linearity in row \(p\),
\[ \det(\ldots, r_p + c\,r_q, \ldots) = \det(\ldots, r_p, \ldots) + c \det(\ldots, r_q, \ldots) \]
where the second determinant has \(r_q\) in two row positions and is therefore \(0\) by D2. Two axioms, used once each, and the operation that does all the work in elimination turns out to be free.
The swap rule deserves a note, since it is a consequence rather than an assumption. Applying D2 to a matrix whose rows \(p\) and \(q\) are both \(r_p + r_q\), then expanding by D1, leaves \(\det(\ldots, r_p, \ldots, r_q, \ldots) + \det(\ldots, r_q, \ldots, r_p, \ldots) = 0\) — which says precisely that swapping negates.
A matrix with a zero row has determinant \(0\): scaling that row by \(2\) changes nothing, but by D1 it must double the determinant, so the determinant is its own double. And a triangular matrix has determinant equal to the product of its diagonal entries, since elimination has no work left to do.
Put the dictionary together with the triangular corollary and a method falls out. Reduce the matrix to triangular form using only the operation that costs nothing, multiply the diagonal, and account for any swaps.
Starting from \(A\) above, two eliminations suffice and no swap is needed:
\(R_3 \to R_3 - \tfrac{5}{2}R_1\), then \(R_3 \to R_3 - \tfrac{9}{8}R_2\), giving the triangular matrix with diagonal \(2,\; 4,\; -\tfrac{85}{8}\).
\[ \det A = 2 \cdot 4 \cdot \left(-\tfrac{85}{8}\right) = -85. \]
Notice the fractions. Row reduction on an integer matrix produces rational entries even when the answer is an integer, which is why the explorer on this page carries exact fractions rather than decimals: a determinant test written as “is this value \(0\)?” against accumulated floating-point error is one of the classic numerical traps.
Reduction needs roughly \(n^{3}/3\) arithmetic operations. Keep that number in mind for Section 7, where the alternative costs \(n!\).
Two words that are easy to conflate, and must not be.
The minor \(M_{ij}\) is the determinant of the matrix obtained by deleting row \(i\) and column \(j\) — a determinant one size smaller.
The cofactor is that minor with a sign: \(C_{ij} = (-1)^{i + j} M_{ij}\).
The sign \((-1)^{i + j}\) is a checkerboard with \(+\) in the top-left corner. Expansions and the adjugate are always stated in cofactors, never minors, which is why keeping the two words apart matters.
That every row and every column gives the same answer is a theorem, proved on Cofactor Expansion and the Adjugate Formula. The recursion along the first row is also the construction used to prove existence in Section 2, so this formula is the one definition that is true by building rather than by argument.
For the running example \(A\), expanding along row \(1\):
\[ \det A = 2 \cdot (-10) \;+\; (-1) \cdot 5 \;+\; 3 \cdot (-20) = -20 - 5 - 60 = -85. \]
The three cofactors are \(C_{11} = -10\), \(C_{12} = 5\) and \(C_{13} = -20\). Note \(C_{12}\) has the opposite sign to its minor \(M_{12} = -5\); that is the checkerboard at work.
Each zero entry removes an entire minor from the sum. On a matrix with no zeros at all, cofactor expansion is the expensive method, and reduction is the one to reach for.
The \(ad - bc\) observation in Section 3 generalises exactly. Every term of a determinant takes one entry from each row and each column, which is the same thing as choosing a permutation.
The sum runs over all \(n!\) permutations \(\sigma\) of \(\{1, \ldots, n\}\), and \(\operatorname{sgn}(\sigma)\) is \(+1\) for even permutations and \(-1\) for odd ones.
An inversion is a pair of positions whose entries are out of their natural order. Then \(\operatorname{sgn}(\sigma) = (-1)^{k}\), where \(k\) is the number of inversions.
Equivalently, write \(\sigma\) as a product of transpositions and count them. That decomposition is not unique, so the equivalence needs proof.
Since a permutation can be written as a product of transpositions in many ways, the parity of that count has to be shown independent of the route. It is: every single transposition changes the inversion count by an odd amount, so the parity of the inversion count is an invariant of the permutation itself. The Leibniz Formula for the Determinant proves this first, then verifies the formula satisfies all three axioms.
For the running example the six permutations of \(\{1, 2, 3\}\) contribute \(-16,\; -4,\; 0,\; -5,\; 0,\; -60\), and those total \(-85\) — the same value the other two methods gave.
Two terms vanish because they pick up the entry \(a_{21} = 0\). That is the only economy available here, and it is the reason the formula is impractical: nothing prunes the sum in general.
\(n!\) terms means a \(10 \times 10\) determinant would need \(3{,}628{,}800\) products, against roughly \(333\) operations for row reduction. The formula earns its place by making properties provable — the transpose result in Section 9 falls straight out of it — not by being evaluated.
Dropping the signs is not a simplification either. The resulting function is the permanent, which is a genuinely different quantity and, notoriously, far harder to compute than the determinant.
This is remarkable given how unlike the two sides look. Matrix multiplication mixes every entry of \(A\) with every entry of \(B\); the determinants are computed separately and simply multiplied.
Fix \(B\) with \(\det B \neq 0\) and define \(D(A) = \det(AB)/\det(B)\). Checking that \(D\) satisfies D1, D2 and D3 is routine, and uniqueness then forces \(D = \det\), which is the statement. The singular case is handled separately. The Multiplicativity of the Determinant gives the argument in full.
The geometric reading makes it unsurprising: applying \(B\) then \(A\) scales volume by one factor and then the other, so the total factor is the product.
\(\det(A + B)\) has no useful expression in terms of \(\det A\) and \(\det B\). The product rule holds and the sum rule does not, and the reason is the one from Section 2: multilinearity is a statement about rows, not about matrices.
The proof is a one-line reindexing of the Leibniz sum. Each term \(\operatorname{sgn}(\sigma)\prod_i a_{\sigma(i),i}\) coming from \(A^{T}\) is a term of the sum for \(A\) indexed by \(\sigma^{-1}\), and a permutation and its inverse have the same sign. Summing over all \(\sigma\) is the same as summing over all \(\sigma^{-1}\), so the two sums have identical term sets. The Determinant of the Transpose spells it out.
Every axiom and every rule on this page was stated for rows. This theorem converts all of them into column statements at a stroke: \(\det\) is multilinear and alternating in the columns too, column swaps negate it, adding a multiple of one column to another changes nothing, and cofactor expansion works along columns. Without it, half the page would need proving twice.
\(\operatorname{adj}(A)\) is the transpose of the matrix of cofactors: \(\operatorname{adj}(A)_{ij} = C_{ji}\). The index swap is not optional.
The identity holds on both sides, and rearranging it when \(\det A \neq 0\) gives a closed formula for the inverse:
This is where the familiar \(\tfrac{1}{ad - bc}\) in the two-row inverse comes from — it is this formula with \(n = 2\).
Building \(\operatorname{adj}(A)\) as the cofactor matrix without transposing it is the standard slip, and the identity above then fails. The explorer computes \(A \operatorname{adj}(A)\) and checks it really is \((\det A) I\), so the failure is visible rather than theoretical.
For \(A\mathbf{x} = \mathbf{b}\) with \(\det A \neq 0\), let \(A_k\) be \(A\) with its \(k\)th column replaced by \(\mathbf{b}\). Then
\[ x_k = \frac{\det A_k}{\det A}. \]
Cramer’s Rule proves it through two lemmas about that companion matrix.
Solving \(A\mathbf{x} = (1, 2, 3)\) with \(\det A = -85\) gives \(x_1 = \tfrac{41}{85}\), \(x_2 = \tfrac{39}{85}\), \(x_3 = \tfrac{14}{85}\). Substituting back reproduces \((1, 2, 3)\) exactly — the explorer performs that substitution rather than asserting it.
Cramer needs \(n + 1\) determinants, so it costs far more than simply solving the system by elimination. Its value is theoretical: it shows the solution depends rationally on the entries, which matters when those entries are symbols rather than numbers.
Choose a matrix. The explorer computes its determinant three ways — by row reduction, by cofactor expansion along the first row, and by the full Leibniz permutation sum — then checks the adjugate identity and solves the system by Cramer, substituting the answer back. Every value is an exact fraction.
Nothing here is tabulated in advance: every determinant, cofactor, permutation sign and solution is computed live in exact rational arithmetic.
The determinant is defined by three properties, not by a formula. Everything else is a consequence.
The characteristic polynomial \(\det(A - \lambda I)\) turns the determinant into a detector of eigenvalues, which is the subject of the third tutorial in this unit. The equivalence “\(\det = 0\) exactly when the rows are dependent” is the entire reason that works.
| Result | Statement | Proof on this site |
|---|---|---|
| Existence and uniqueness | Exactly one function satisfies D1, D2, D3 | The Determinant: Existence and Uniqueness |
| Cofactor expansion | \(\det A = \sum_j a_{ij}C_{ij}\), any row or column | Cofactor Expansion and the Adjugate Formula |
| Leibniz formula | \(\det A = \sum_\sigma \operatorname{sgn}(\sigma)\prod_i a_{i,\sigma(i)}\) | The Leibniz Formula for the Determinant |
| Multiplicativity | \(\det(AB) = \det A \det B\) | The Multiplicativity of the Determinant |
| Transpose | \(\det(A^{T}) = \det(A)\) | The Determinant of the Transpose |
| Adjugate identity | \(A \operatorname{adj}(A) = (\det A) I\) | Cofactor Expansion and the Adjugate Formula |
| Cramer’s rule | \(x_k = \det(A_k)/\det(A)\) | Cramer’s Rule |
| Adjugate pitfall | Transpose the cofactor matrix, or the identity fails | Explorer, Section 10 |