Introduction to Determinants

A $2 \times 2$ matrix is invertible exactly when its determinant is nonzero, and for $A = [a c b d]$ that determinant is the number $a d - b c$ . Extending this quantity to a square matrix of any size is recursive: an $n \times n$ determinant is a signed sum of $(n - 1) \times (n - 1)$ determinants, each of which unwinds the same way down to the $2 \times 2$ base case. The single number that results decides invertibility, measures how a linear map scales area and volume, and appears in closed-form solution formulas.

Origin of the definition

The general definition comes from the condition for row reduction of a larger matrix to succeed. Take a $3 \times 3$ matrix $A = [a_{ij}]$ with $a_{11} \neq = 0$ . Scaling rows 2 and 3 by $a_{11}$ and clearing the first column produces an equivalent matrix whose lower-right $2 \times 2$ block, after one more elimination, has the single nonzero entry $a_{11} Δ$ in its corner, where¹

Δ = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} - a_{13} a_{22} a_{31} .

Because $A$ reduces to a triangular matrix whose corner entry carries this factor $Δ$ , the matrix is invertible exactly when $Δ \neq = 0$ . The quantity $Δ$ is the determinant of the $3 \times 3$ matrix. Grouping its six terms by the first-row entries $a_{11}, a_{12}, a_{13}$ rewrites $Δ$ using three $2 \times 2$ determinants:

Δ = a_{11} det [a_{22} a_{32} a_{23} a_{33}] - a_{12} det [a_{21} a_{31} a_{23} a_{33}] + a_{13} det [a_{21} a_{31} a_{22} a_{32}] .

Each $2 \times 2$ matrix here is $A$ with its first row and one column struck out. This is the pattern that generalizes.

Minors, cofactors, and the recursive definition

The submatrix is one row and one column smaller than $A$ . Deleting the crossing row and column is the mechanical step behind every determinant computation.

The submatrix

A_{32}

is what remains after striking out row 3 and column 2; its determinant is the

(3, 2)

-minor.

The determinant of a $1 \times 1$ matrix $[a_{11}]$ is $a_{11}$ ; the $2 \times 2$ case is $a d - b c$ . Everything larger is defined by reducing to these.

The sign $(- 1)^{1 + j}$ alternates $+, -, +, \dots$ across the row. Reading the definition top-down, an $n \times n$ determinant calls $n$ determinants of size $n - 1$ , each of which calls $n - 1$ of size $n - 2$ , and so on. The recursion bottoms out at the $2 \times 2$ or $1 \times 1$ rule.

3 \times 3

determinant expands into three signed

2 \times 2

minors, each a lower-order determinant to evaluate directly.

It is convenient to fold the alternating sign into the minor itself.

The cofactor is the minor with its position sign attached, so the expansion becomes an ordinary (unsigned) dot product of the first row with its cofactors.

Expansion along any row or column

The first row is not special. The following theorem, stated without proof to avoid a lengthy digression, is what makes determinants practical to compute.¹

Every row and every column gives the same number. The freedom to choose which one is the practical value of the theorem: pick the row or column with the most zeros, and the terms attached to those zeros vanish before any minor is computed.

The sign $(- 1)^{i + j}$ attached to position $(i, j)$ does not depend on the entry $a_{ij}$ itself, only on where it sits. Laid out over the whole matrix the signs form a checkerboard.

The cofactor sign

(- 1)^{i + j}

alternates like a checkerboard; the top-left corner is always

+

Triangular matrices

The chaining above always terminates in one step when the matrix is triangular, because every column below the diagonal is a column of zeros in the relevant submatrix. Expanding repeatedly down the first column peels off one diagonal entry at a time.

For a triangular matrix the cofactor expansion keeps only the diagonal entries; their product is the determinant.

A row or column of all zeros forces $det A = 0$ by the same reasoning: the cofactor expansion along that line is a sum of zeros. This is often the fastest way to spot a zero determinant by eye.

Determinants of a few standard matrices

The triangular-determinant rule fixes several determinants that recur constantly and are worth knowing on sight. Each follows by reading the diagonal of a triangular (indeed diagonal) matrix.

Identity. $det I_{n} = 1$ , the product of $n$ ones.
Diagonal. $det diag (d_{1}, \dots, d_{n}) = d_{1} d_{2} \dots d_{n}$ .
Scalar. For the scalar matrix $c I_{n}$ , every diagonal entry is $c$ , so $det (c I_{n}) = c^{n}$ .

The elementary matrices — the identity altered by one row operation — also have determinants readable at a glance. A matrix that adds a multiple of one row to another is triangular with ones on the diagonal, so its determinant is $1$ ; a matrix that scales a row by $k$ is diagonal with a single off-one entry $k$ , so its determinant is $k$ ; and a matrix that swaps two rows has determinant $- 1$ , a fact that takes one cofactor expansion to check.

Elementary matrix	Effect on $I$	Determinant
Replacement	add $k \cdot (row j)$ to row $i$	$1$
Scaling	multiply row $i$ by $k$	$k$
Interchange	swap rows $i$ and $j$	$- 1$

These three numbers are the factors by which the corresponding row operations rescale a determinant, which is the content of Properties of Determinants.

Base cases

The base cases recur inside every larger computation.

Size	Matrix	Determinant
$1 \times 1$	$[a]$	$a$
$2 \times 2$	$[a c b d]$	$a d - b c$
$3 \times 3$ (first-row)	$[a_{ij}]$	$a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}$

Every expansion ultimately resolves into the $2 \times 2$ rule, main-diagonal product minus anti-diagonal product. A diagonal-based mnemonic exists for the $3 \times 3$ case, but it does not extend to $4 \times 4$ or larger, so cofactor expansion remains the general method.

The cost of the definition

Cofactor expansion defines the determinant cleanly but computes it slowly. An $n \times n$ expansion generates $n$ subdeterminants, each of size $n - 1$ , so the recursion visits on the order of $n!$ terms. Even for a matrix that today counts as small, this is impractical.¹

Cofactor-expansion cost grows as

n!

, far outrunning the

\sim n^{3}

cost of the row-reduction method.

A $25 \times 25$ determinant by cofactor expansion needs more than $25! \approx 1.5 \times 1 0^{25}$ multiplications; a machine doing a trillion per second would run for over $500, 000$ years.¹ The definition is the right tool for small matrices and for proofs. The practical algorithm for large matrices comes from row reduction, which also settles the invertibility criterion $det A \neq = 0$ and the multiplicative law $det (A B) = (det A) (det B)$ .

Summary

The determinant of a square matrix is defined recursively by cofactor expansion: $det A = \sum_{j} (- 1)^{1 + j} a_{1 j} det A_{1 j}$ , bottoming out at the $2 \times 2$ rule $a d - b c$ .
The minor $det A_{ij}$ is the determinant of $A$ with row $i$ and column $j$ deleted; the cofactor $C_{ij} = (- 1)^{i + j} det A_{ij}$ folds in the position sign.
Cofactor expansion can run along any row or down any column, so choosing a line rich in zeros minimizes work.
Triangular determinant: a triangular matrix has determinant equal to the product of its diagonal; a zero row or column forces $det A = 0$ .
Cofactor expansion costs $\sim n!$ operations, unusable for large $n$ ; row reduction is the practical algorithm.

Lay, Lay & McDonald, Linear Algebra and Its Applications, 5th ed., §3.1 — Introduction to Determinants: the $3 \times 3$ motivation via row reduction, the recursive definition, Theorem 1 (expansion across any row or down any column), Theorem 2 (triangular matrices), and the numerical note on the $n!$ cost of cofactor expansion. ↩ ↩² ↩³ ↩⁴

Origin of the definition

Minors, cofactors, and the recursive definition

Expansion along any row or column

Triangular matrices

Determinants of a few standard matrices

Base cases

The cost of the definition

Summary

Footnotes