Determinants: Formula, Properties & Applications

The Determinant as a Scalar Assignment

One account is an area: the parallelogram spanned by the columns.

Take

A = \begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}

and draw its two columns as arrows from the origin,

(3, 2)

and

(1, 5)

. Together they span a parallelogram. Its area can be found with nothing but a ruler: the surrounding

4 \times 7

box has area

28

, and the triangles and rectangles left over around the parallelogram add up to

15

, so the parallelogram has area

13

.

The arrows

(3, 2)

and

(1, 5)

and the two copies that close the shape. The region between them has area

13

.

The other account is a short calculation on the four entries.

Multiply the two entries on the main diagonal, multiply the other two, and subtract:

ad - bc = 3 \cdot 5 - 1 \cdot 2 = 13

. No drawing, no geometry — four numbers and one subtraction. Section 2 takes this formula as its starting point.

Both accounts give the same number every time.

Try a few pairs of columns both ways.

columns	area of the parallelogram	ad − bc
(3, 2) and (1, 5)	28 − 15 = 13	3·5 − 1·2 = 13
(2, 0) and (0, 3)	a 2 × 3 rectangle: 6	2·3 − 0·0 = 6
(1, 2) and (2, 4)	on one line: 0	1·4 − 2·2 = 0

The rectangle is the easy check — its area is plainly

2 \times 3

. The last pair is the telling one: both columns lie on the same line, the parallelogram collapses to nothing, and the formula gives

0

as well.

They had to agree, because area and ad − bc react to the columns in the same way.

Change the columns and watch both quantities.

change to the columns	area	ad − bc
slide one column along the other (add a multiple of it)	unchanged	unchanged
stretch one column by a factor k	times k	times k
start from the unit square, columns (1, 0) and (0, 1)	1	1·1 − 0·0 = 1

Sliding a column along the other is a shear: the parallelogram leans but keeps its base and height, and

ad - bc

is unchanged by the same move. Stretching a column stretches both by the same factor, and the unit square starts both at

1

. Any two quantities that behave this way under every change of columns are the same quantity — the two accounts were describing one number all along. (The formula can also come out negative; that sign records which way round the columns go, an orientation the plain area ignores.)

The definition

The determinant of a square matrix

A

is the single number

\det(A)

that both accounts describe: the signed factor by which

A

scales area (volume, in higher dimensions), equal for a

2 \times 2

matrix to

ad - bc

.

For

A = \begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}

,

\det(A) = 13

.

A determinant is a number, and only square matrices have one.

It is one real number (or complex number, if the entries are complex) attached to the whole matrix. It is defined only for square matrices — a rectangular matrix has no determinant. The notation

|A|

is widespread but potentially confusing, since the same vertical bars denote absolute value for real numbers and modulus for complex numbers. When there is any risk of ambiguity,

\det(A)

is preferred.

Use the formula to compute and the area to understand.

The formula is what you calculate with, and for larger matrices it grows into the expansions of the sections below. The area is what explains the results. A zero determinant means the columns have collapsed onto a line (a plane, in 3D), so the map

x \mapsto Ax

squashes space flat and cannot be undone:

A

is invertible if and only if

\det(A) \neq 0

. The determinant also determines whether that map preserves or reverses orientation, and appears in explicit formulas for eigenvalues, matrix inverses, and solutions to linear systems.

The next several sections develop the determinant starting from the smallest cases — the

2 \times 2

formula, the

3 \times 3

rule — and building toward cofactor expansion, the general definition, its properties, and its geometric meaning.

The 2×2 Formula

For a

2 \times 2

matrix

The 2×2 Formula

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

the determinant is

The 2×2 Formula

\det(A) = ad - bc

One way to see where this formula comes from is to solve the system

Ax = \mathbf{v}

for a general right-hand side. Applying elimination, the solution involves dividing by

ad - bc

at every step. When

ad - bc = 0

, the system either has no solution or infinitely many, and the matrix cannot be inverted. When

ad - bc \neq 0

, there is a unique solution for every

\mathbf{v}

, and the inverse exists.

Pitfall: Subtracting in the wrong order

Taking

bc - ad

for

\begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}

gives

1 \cdot 2 - 3 \cdot 5 = -13

, the right size with the wrong sign. The main diagonal comes first:

ad - bc = 15 - 2 = 13

.

The expression

ad - bc

also carries geometric meaning. The columns of

A

are the vectors

(a, c)

and

(b, d)

in

\mathbb{R}^2

. The signed area of the parallelogram these two vectors span equals exactly

ad - bc

. A positive value means the columns are arranged counterclockwise; a negative value means clockwise; zero means the columns are parallel and the parallelogram collapses to a line segment.

Worked Examples

For

A = \begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}

, the determinant is

3 \cdot 5 - 1 \cdot 2 = 13

. The matrix is invertible and its column vectors span a parallelogram of area

13

.

For

A = \begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix}

, the determinant is

4 \cdot 3 - 6 \cdot 2 = 0

. The second column is

\frac{3}{2}

times the first, so the columns are parallel and the matrix is singular.

For

A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

, the determinant is

0 \cdot 0 - (-1) \cdot 1 = 1

. This is a

90°

rotation matrix — it preserves areas and orientation.

ad, the first half of ad − bc

The 2 × 2 determinant is built from the two diagonals of the square. Here the main diagonal is lit: a₁₁ and a₂₂, whose product is the ad term. The anti-diagonal, a₁₂ and a₂₁, supplies bc, and the determinant is the first product minus the second. Step through both diagonals and the subtraction on the determinant visualizer.

Every larger determinant is assembled out of 2 × 2 pieces like this one.

The 3×3 Formula

For a

3 \times 3

matrix

The 3×3 Formula

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

the determinant is computed by expanding along the first row:

Determinant 3x3

\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})

Learn more about this formula: Determinant 3x3 →

Browse all linear algebra formulas →

Each first-row entry multiplies the

2 \times 2

determinant of the submatrix that remains after deleting that entry's row and column. The signs alternate:

+, -, +

.

The Sarrus Mnemonic

A shortcut for the

3 \times 3

case is the rule of Sarrus. Write the matrix and copy its first two columns to the right:

The 3×3 Formula

\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix}

Sum the three products along the downward diagonals, then subtract the three products along the upward diagonals. This gives the same result as the first-row expansion. The Sarrus rule works only for

3 \times 3

matrices — there is no analogous diagonal trick for

4 \times 4

or larger.

Worked Example

For

A = \begin{pmatrix} 2 & 3 & 1 \\ 0 & -1 & 4 \\ 5 & 2 & 3 \end{pmatrix}

, expanding along the first row:

The 3×3 Formula

\det(A) = 2[(-1)(3) - (4)(2)] - 3[(0)(3) - (4)(5)] + 1[(0)(2) - (-1)(5)]

The 3×3 Formula

= 2(-3 - 8) - 3(0 - 20) + 1(0 + 5) = 2(-11) - 3(-20) + 5 = -22 + 60 + 5 = 43

Pitfall: Adding all three terms

Adding the terms instead of alternating the signs gives

2(-11) + 3(-20) + 1(5) = -77

. The middle term is subtracted:

2(-11) - 3(-20) + 1(5) = 43

.

The same result follows from the Sarrus rule: downward diagonals give

2(-1)(3) + 3(4)(5) + 1(0)(2) = -6 + 60 + 0 = 54

; upward diagonals give

1(-1)(5) + 2(4)(2) + 3(0)(3) = -5 + 16 + 0 = 11

; the determinant is

54 - 11 = 43

.

The six signed products of a 3×3

Three diagonals running one way are added and three running the other are subtracted — six products in total, each using one entry from every row and every column. The pattern is a convenience for this size only: it does not extend to 4×4, where the count jumps to twenty-four. Watch the diagonals traced out on the determinant visualizer.

The general definition in the next section is what these six products are a special case of.

The General n×n Determinant

The pattern from the

3 \times 3

case extends to any dimension. For an

n \times n

matrix, expand along the first row:

The General n×n Determinant

\det(A) = \sum_{j=1}^{n} (-1)^{1+j} \, a_{1j} \, M_{1j}

where

M_{1j}

is the determinant of the

(n-1) \times (n-1)

submatrix obtained by deleting row

1

and column

j

. This is a recursive definition: each

n \times n

determinant reduces to

n

determinants of size

(n-1) \times (n-1)

, each of which reduces further, until reaching the base case of

1 \times 1

matrices where

\det(a) = a

.

A crucial fact is that the expansion need not use the first row.

THEOREM

Expanding along any row or any column gives the same result.

Fact: Drop the signs and it gets hard

Fully expanded, the determinant is a sum of

n!

signed products. Remove every minus sign and the result is the permanent: row additions now change its value, elimination fails, and the best known methods still take more than

2^n

steps against the determinant's

n^3

.

The signed sub-determinants

(-1)^{i+j} M_{ij}

are called cofactors, and the freedom to choose the expansion axis is what makes the formula practical: a row or column with many zeros dramatically reduces the number of terms.

The recursive nature of this definition means the computational cost grows factorially. An

n \times n

determinant via cofactor expansion requires on the order of

n!

arithmetic operations. For

n = 5

that is

120

operations; for

n = 10

it is over

3.6

million. This explosion is what motivates the row-reduction approach, which accomplishes the same task in

O(n^3)

operations by exploiting how row operations affect the determinant.

Cofactor expansion defines the determinant in every size, but its cost grows like

n!

, so large ones are computed by row reduction.

Singular and Nonsingular Matrices

The value

\det(A) = 0

marks a sharp dividing line. A square matrix with zero determinant is called singular; a square matrix with nonzero determinant is called nonsingular or invertible.

THEOREM

When

\det(A) = 0

, the columns of

A

are linearly dependent — at least one column can be written as a linear combination of the others.

The homogeneous system

Ax = \mathbf{0}

has nontrivial solutions, the rank of

A

is strictly less than

n

, and the matrix maps

\mathbb{R}^n

onto a lower-dimensional subspace. Geometrically, the transformation collapses at least one dimension, flattening regions of positive volume down to zero volume.

When

\det(A) \neq 0

, everything works. The columns of

A

form a basis for

\mathbb{R}^n

. The system

Ax = \mathbf{b}

has exactly one solution for every right-hand side

\mathbf{b}

. The matrix has an inverse

A^{-1}

, and the linear map

x \mapsto Ax

is a bijection from

\mathbb{R}^n

to itself that stretches or compresses volumes by the factor

|\det(A)|

without collapsing any dimension.

The determinant thus answers the single most important structural question about a square matrix in a single number.

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A sharp dividing line

The value

\det(A) = 0

splits square matrices in two. The determinant answers the single most important structural question about a square matrix in a single number.

Square matrix

\det(A)

,

A

is

n \times n

split by whether $\det(A) = 0$

0

Case 1

\det(A) = 0

Singular

the columns are linearly dependent,

Ax = \mathbf{0}

has nontrivial solutions, the rank is less than

n

, and at least one dimension collapses

More

See linear dependence.

✓

Case 2

\det(A) \neq 0

Nonsingular (invertible)

the columns form a basis,

Ax = \mathbf{b}

has exactly one solution for every

\mathbf{b}

, and

A^{-1}

exists

More

The map

x \mapsto Ax

is a bijection from

\mathbb{R}^n

to itself. See basis.

Takeaway

A nonsingular matrix stretches or compresses volumes by

|\det(A)|

without collapsing any dimension; a singular one flattens regions of positive volume down to zero volume.

Singular or nonsingular: what det(A) = 0 decides·/linear-algebra/determinantsLearn Math Class

Computing Small Determinants

The most direct way to build intuition is to compute by hand. The

2 \times 2

formula

ad - bc

is immediate. The

3 \times 3

case requires either first-row expansion or the Sarrus shortcut. For

4 \times 4

matrices, no shortcut exists — the computation goes through cofactor expansion, but choosing a good row or column makes a significant difference.

A 4×4 Example

Computing Small Determinants

A = \begin{pmatrix} 1 & 0 & 2 & -1 \\ 3 & 0 & 0 & 5 \\ 2 & 1 & 4 & -3 \\ 1 & 0 & 5 & 0 \end{pmatrix}

Column

2

has three zeros and a single

1

in position

(3,2)

, making it the best expansion axis. Expanding along column

2

:

Computing Small Determinants

\det(A) = (-1)^{3+2} \cdot 1 \cdot M_{32}

where

M_{32}

is the

3 \times 3

determinant obtained by deleting row

3

and column

2

:

Computing Small Determinants

M_{32} = \det\begin{pmatrix} 1 & 2 & -1 \\ 3 & 0 & 5 \\ 1 & 5 & 0 \end{pmatrix}

Expanding this along its first row:

Computing Small Determinants

M_{32} = 1(0 \cdot 0 - 5 \cdot 5) - 2(3 \cdot 0 - 5 \cdot 1) + (-1)(3 \cdot 5 - 0 \cdot 1)

Computing Small Determinants

= 1(-25) - 2(-5) + (-1)(15) = -25 + 10 - 15 = -30

So

\det(A) = (-1)^{5} \cdot 1 \cdot (-30) = (-1)(-30) = 30

.

Pitfall: Reusing the first-row signs

Applying the first-row pattern

+, -, +, -

down column

2

gives the entry at

(3,2)

the sign

+

and

\det(A) = -30

. The sign comes from the position,

(-1)^{3+2} = -1

, so

\det(A) = 30

.

The choice of column

2

reduced the problem from four

3 \times 3

determinants down to one. This illustrates why scanning for zeros before expanding is the first step in any hand computation.

To compute the determinant of a specific matrix, use the matrix calculator.

Expanding by Minors and Cofactors

The sub-determinants appearing in cofactor expansion have their own terminology. The

(i,j)

minor

M_{ij}

is the determinant of the submatrix formed by deleting row

i

and column

j

from

A

. The

(i,j)

cofactor is the signed version:

Cofactor Definition

C_{ij} = (-1)^{i+j} M_{ij}

Learn more about this formula: Cofactor Definition →

Browse all linear algebra formulas →

The sign factor

(-1)^{i+j}

follows a checkerboard pattern starting with

+

at position

(1,1)

:

Expanding by Minors and Cofactors

\begin{pmatrix} + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ + & - & + & - & \cdots \\ \vdots & & & & \ddots \end{pmatrix}

Using this notation, the Laplace expansion along row

i

takes the compact form

Expanding by Minors and Cofactors

\det(A) = \sum_{j=1}^{n} a_{ij} \, C_{ij}

and the expansion along column

j

is

Expanding by Minors and Cofactors

\det(A) = \sum_{i=1}^{n} a_{ij} \, C_{ij}

Both give the same result regardless of which row or column is chosen. Collecting all cofactors into a matrix and transposing produces the adjugate, which leads to an explicit formula for the matrix inverse. The full development of minors, cofactors, the adjugate, and the structural results they enable is on the cofactors page.

Expansion along a row, minors highlighted

For each entry of the chosen row, its own row and column are struck out and the determinant of what remains is taken, then signed and weighted. Any row or column gives the same answer, so the one with the most zeros is always the cheapest to pick. Choose a different row or column and compare on the determinant visualizer.

This is the definition that actually generalises, at the price of a cost that grows faster than almost anything else in the subject.

How Row Operations Affect the Determinant

The three elementary row operations interact with the determinant in simple, predictable ways. Swapping two rows multiplies the determinant by

-1

. Multiplying a row by a nonzero scalar

k

multiplies the determinant by

k

. Adding a scalar multiple of one row to a different row leaves the determinant unchanged.

Pitfall: Scaling the whole matrix once

Writing

\det(2A) = 2\det(A)

for

A = \begin{pmatrix} 2 & 4 \\ 3 & 7 \end{pmatrix}

gives

4

. Doubling

A

doubles both rows, so

\det(2A) = 2^2 \cdot 2 = 8

; in general

\det(kA) = k^n \det(A)

.

These three rules turn Gaussian elimination into a determinant-computing algorithm. Reduce

A

to upper triangular form using row operations, tracking every swap and every scaling. The determinant of the resulting triangular matrix is the product of its diagonal entries. Adjusting for the tracked sign flips and scale factors gives

\det(A)

.

This approach requires on the order of

n^3

arithmetic operations — a dramatic improvement over the

n!

cost of cofactor expansion. For any matrix larger than

4 \times 4

, row reduction is the practical method.

A Quick Illustration

Pitfall: Forgetting to undo a scaling

Dividing row

1

of

\begin{pmatrix} 2 & 4 \\ 3 & 7 \end{pmatrix}

by

2

and then clearing below the pivot gives

\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

, with diagonal product

1

. That division halved the determinant, so

\det(A) = 2 \cdot 1 = 2

.

Example

Starting from

A =\begin{pmatrix} 2 & 4 \\ 3 & 7 \end{pmatrix}

, subtract

\frac{3}{2}

times row

1

from row

2

(this does not change the determinant) to get

\begin{pmatrix} 2 & 4 \\ 0 & 1 \end{pmatrix}

. The product of the diagonal is

2 \cdot 1 = 2

, which matches

\det(A) = 2 \cdot 7 - 4 \cdot 3 = 2

.

The complete set of algebraic properties — including the multiplicative rule

\det(AB) = \det(A)\det(B)

, transpose invariance, and the full invertibility equivalence — is developed on the properties page. The three rules above collect into a compact reference:

Row operation	Effect on det(A)	Adjustment when reading off det from triangular form
Swap two rows	multiplies det by −1	flip the sign once per swap
Multiply a row by k ≠ 0	multiplies det by k	divide the final product by k
Add a multiple of one row to another	leaves det unchanged	no adjustment

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Before you read

Row operations change the determinant predictably

Each of the three row operations has a fixed effect on

\det(A)

. That is what makes reduction to triangular form a fast way to compute it.

Row operations and the determinant3 effects

Swap two rows

Effect

\det

changes sign

Scale a row by k

Effect

\det

is multiplied by

k

Add a multiple of a row

Effect

\det

unchanged

Takeaway

Track the swaps and the scalings

Reduce to triangular form with row additions, which leave the determinant alone, and account for every swap and scaling along the way. The cost is about

n^3

.

How row operations change the determinant·/linear-algebra/determinantsLearn Math Class

Area, Volume, and Orientation

The determinant has a direct geometric meaning: it measures how a matrix, viewed as a linear transformation, distorts size and orientation.

In two dimensions,

|\det(A)|

equals the area of the parallelogram spanned by the columns of

A

. In three dimensions,

|\det(A)|

equals the volume of the parallelepiped spanned by the three column vectors, which also equals the scalar triple product

\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})

.

THEOREM

In

n

dimensions,

|\det(A)|

is the factor by which the map

x \mapsto Ax

scales

n

-dimensional volumes.

The sign carries its own meaning. A positive determinant means the transformation preserves orientation — counterclockwise stays counterclockwise in

\mathbb{R}^2

, right-handed stays right-handed in

\mathbb{R}^3

. A negative determinant means orientation is reversed. A zero determinant means the image is lower-dimensional: a

3 \times 3

transformation with

\det = 0

maps all of

\mathbb{R}^3

onto a plane, a line, or a point.

Rotation matrices always have determinant

+1

. Reflection matrices always have determinant

-1

. These are the cleanest examples of orientation-preserving and orientation-reversing maps. The full geometric treatment, including the change-of-variables formula from multivariable calculus and explicit area and volume formulas, is on the geometry page.

A grid stretched, then flattened

Above, the unit square has become a parallelogram and its area is exactly the determinant. Below, the matrix has flattened the plane onto a line: the parallelogram has no area left and the determinant is zero. Singularity and zero area are the same event seen from two sides. Change the entries and watch the area respond on the linear transformation explorer.

Every geometric use of the determinant — volume, orientation, change of variables — is this one fact applied in more dimensions.

Determinant-Based Formulas

The determinant is not only a diagnostic tool — it provides closed-form expressions for quantities that might otherwise require iterative procedures.

Fact: Determinants came before matrices

Seki Takakazu in Japan in 1683 and Leibniz in Europe in 1693 each found them while eliminating unknowns from systems of equations, as the expression that decides whether the equations can hold together. Matrices became objects of their own only in the 1850s, more than 150 years later.

Cramer's rule solves a linear system

Ax = \mathbf{b}

by expressing each component of the solution as a ratio of two determinants:

x_i = \det(A_i)/\det(A)

, where

A_i

is

A

with its

i

-th column replaced by

\mathbf{b}

. The adjugate formula gives the inverse of

A

as

A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)

, writing every entry of the inverse explicitly in terms of cofactors. In the

2 \times 2

case, this reduces to the familiar swap-and-negate formula.

The cross product

\mathbf{a} \times \mathbf{b}

in

\mathbb{R}^3

can be computed as a symbolic

3 \times 3

determinant with the unit vectors

\hat{i}, \hat{j}, \hat{k}

in the first row. The characteristic polynomial

\det(A - \lambda I)

defines the eigenvalues of

A

— its roots are precisely the scalars

\lambda

for which

A - \lambda I

is singular. The Wronskian, a determinant built from functions and their derivatives, tests linear independence in the setting of differential equations.

Each of these formulas is developed with full worked examples on the applications page. The five collect into a single reference card:

Formula	Expression	What it computes
Cramer's rule	x_i = det(A_i) / det(A)	each component of the unique solution to Ax = b
Adjugate formula for A⁻¹	A⁻¹ = adj(A) / det(A)	explicit closed form for the matrix inverse
Cross product in ℝ³	a × b = symbolic 3×3 det with î, ĵ, k̂ on top	vector perpendicular to a and b with prescribed length
Characteristic polynomial	det(A − λI)	polynomial whose roots are the eigenvalues of A
Wronskian	determinant of functions and their derivatives	linear independence test for solutions of an ODE

Summary: How to Compute a Determinant

Across the methods covered above, four routes lead to

\det(A)

— the closed-form

2\times 2

formula, the closed-form

3\times 3

expansion (or Sarrus), general cofactor expansion, and row reduction. Each is the right choice in a different regime, distinguished by matrix size, sparsity of the matrix, and whether the work needs to scale to large

n

. The table below collects all four side by side on the underlying formula, the situation it suits best, and the asymptotic cost.

Method	Formula / approach	Best for	Cost
2×2 formula	ad − bc	every 2×2 case — direct, no tricks needed	O(1)
3×3 (first-row expansion or Sarrus)	sum of three 2×2 dets with alternating signs; or downward minus upward diagonal products	3×3 only; Sarrus does not generalize	O(1)
Cofactor expansion	Σ (−1)^i+j a_ij M_ij along any row or column	small matrices, especially when a row or column has many zeros	O(n!)
Row reduction	reduce to triangular form, multiply the diagonal, adjust for tracked swaps and scalings	any matrix 4×4 or larger; the standard general-purpose method	O(n³)

Common Pitfalls

Six mistakes account for many wrong determinants. Five are marked with a box where they happen in the sections above, and the sixth, a diagonal shortcut used on a

4 \times 4

matrix, is flagged in the section on the

3 \times 3

formula. Here each is worked through in full, with the wrong and the right working side by side.

Subtracting in the wrong order

The

2 \times 2

formula is applied with its two products the wrong way round: the product of the other diagonal comes first, and the main-diagonal product is subtracted from it.

Wrong working: for

A = \begin{pmatrix} 3 & 1 \\ 2 & 5 \end{pmatrix}

,

bc - ad = 1 \cdot 2 - 3 \cdot 5 = 2 - 15 = -13

Right working: the main diagonal

a, d

comes first.

ad - bc = 3 \cdot 5 - 1 \cdot 2 = 15 - 2 = 13

Why it happens: both products are formed correctly and only their order is swapped. The two results differ only in sign, so the wrong one looks just as plausible.

How to check: the identity matrix must give

1

. With

ad - bc

,

\det(I) = 1 \cdot 1 - 0 \cdot 0 = 1

; with the order reversed it gives

-1

.

Adding all three terms

In the first-row expansion of a

3 \times 3

determinant, the three terms are added as if every sign were

+

. The middle term should be subtracted.

Wrong working: for the matrix of the worked example,

A = \begin{pmatrix} 2 & 3 & 1 \\ 0 & -1 & 4 \\ 5 & 2 & 3 \end{pmatrix}

,

2(-11) + 3(-20) + 1(5) = -22 - 60 + 5 = -77

Right working: the signs alternate

+, -, +

.

2(-11) - 3(-20) + 1(5) = -22 + 60 + 5 = 43

Why it happens: the three

2 \times 2

determinants are computed carefully, and the sign in front of the middle one is lost when the terms are put together. A negative middle value, as here, makes the slip easier, because two minus signs have to combine into a plus.

How to check: compute the same determinant with the rule of Sarrus, which has no position signs to forget. For this matrix it gives

54 - 11 = 43

.

Using the diagonal rule on a 4×4 matrix

The rule of Sarrus is stretched to a

4 \times 4

matrix: the first three columns are copied to the right, and the four downward diagonal products minus the four upward ones are taken as the determinant.

Wrong working: for the

4 \times 4

matrix of the section on small determinants,

A = \begin{pmatrix} 1 & 0 & 2 & -1 \\ 3 & 0 & 0 & 5 \\ 2 & 1 & 4 & -3 \\ 1 & 0 & 5 & 0 \end{pmatrix}

, the downward diagonals give

0

,

0

,

0

and

(-1)(3)(1)(5) = -15

, and all four upward diagonals give

0

. The result would be

-15 - 0 = -15

.

Right working: expand along column

2

, which has a single nonzero entry.

\det(A) = (-1)^{3+2} \cdot 1 \cdot M_{32} = (-1)(-30) = 30

Why it happens: the diagonal picture is easy to remember, and nothing in it seems tied to the size

3

. But a

4 \times 4

determinant has

4! = 24

products, and the diagonals reach only

8

of them.

How to check: count the products. A determinant of size

n

has

n!

of them, a diagonal rule of this kind produces

2n

, and the two counts agree only for

n = 3

.

Reusing the first-row signs

When a determinant is expanded along a row or column other than the first row, the signs are taken from the first-row pattern

+, -, +, -

in the order the entries are used, instead of from their positions.

Wrong working: in the

4 \times 4

example, the expansion along column

2

has a single term, from the entry

1

at position

(3,2)

. Running

+, -, +, -

down the column gives row

3

the sign

+

:

\det(A) = +1 \cdot M_{32} = 1 \cdot (-30) = -30

Right working: the sign of position

(i,j)

is

(-1)^{i+j}

. For

(3,2)

it is

(-1)^{5} = -1

, so

\det(A) = (-1) \cdot 1 \cdot (-30) = 30

Why it happens: most worked expansions use the first row, where the pattern does start with

+

. Column

2

starts with

-

, and so does row

2

.

How to check: read the sign off the checkerboard pattern in the section on minors and cofactors. Position

(3,2)

sits on a

-

square.

Scaling the whole matrix once

The rule for scaling one row is applied to a multiple of the whole matrix:

\det(kA)

is taken to be

k\det(A)

.

Wrong working: for

A = \begin{pmatrix} 2 & 4 \\ 3 & 7 \end{pmatrix}

, with

\det(A) = 2

,

\det(2A) = 2\det(A) = 4

Right working:

2A

multiplies each of the

2

rows by

2

, and each row contributes its own factor.

\det(2A) = \det\begin{pmatrix} 4 & 8 \\ 6 & 14 \end{pmatrix} = 4 \cdot 14 - 8 \cdot 6 = 56 - 48 = 8 = 2^2 \cdot 2

For an

n \times n

matrix,

\det(kA) = k^n \det(A)

.

Why it happens: multiplying a matrix by

k

looks like one operation, but it scales all

n

rows at once, and the row rule counts each of them.

How to check: try

A = I

. The matrix

kI

is diagonal with

n

entries equal to

k

, so its determinant is

k^n

, not

k

.

Forgetting to undo a scaling

During row reduction a row is divided by its pivot to make the pivot

1

, and the diagonal product of the final triangular matrix is reported as the determinant without accounting for that division.

Wrong working: for

A = \begin{pmatrix} 2 & 4 \\ 3 & 7 \end{pmatrix}

, divide row

1

by

2

to get

\begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}

, then subtract

3

times row

1

from row

2

:

\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \quad 1 \cdot 1 = 1

Right working: dividing row

1

by

2

divided the determinant by

2

, so the product is multiplied back by

2

.

\det(A) = 2 \cdot (1 \cdot 1) = 2

This matches

2 \cdot 7 - 4 \cdot 3 = 2

.

Why it happens: making every pivot equal to

1

is a habit from solving linear systems, where scaling a row does not change the solution. For a determinant it does change the value.

How to check: keep a running factor beside the reduction and multiply by it at the end. Or avoid scaling altogether: row additions alone, as in the illustration of the row-operations section, leave the determinant unchanged.

Determinants FAQ

Why does $ad - bc$ always equal the area of the parallelogram?

+

Because the two react to the columns in the same ways. Sliding one column along the other is a shear: the parallelogram leans but keeps its base and height, and

ad - bc

does not change either. Stretching one column stretches both by the same factor, and the unit square gives

1

for both. Quantities that agree under every such change are one quantity. For columns

(3, 2)

and

(1, 5)

both give

13

. The formula can also be negative; the sign records orientation, which plain area ignores.Read more →

Where does the formula $ad - bc$ come from?

+

From solving a system. Eliminate in

ax + by = v_1

,

cx + dy = v_2

and the solution comes out as

x = \frac{dv_1 - bv_2}{ad - bc}

and

y = \frac{av_2 - cv_1}{ad - bc}

, with the same number in every denominator. When

ad - bc \neq 0

there is exactly one solution for every right-hand side and the matrix has an inverse. When

ad - bc = 0

there is no solution or there are infinitely many, and the matrix cannot be inverted.Read more →

What is the permanent of a matrix?

+

It is the determinant with every minus sign removed. Fully expanded, a determinant is a sum of

n!

signed products, one entry from each row and each column; the permanent adds the same products with all signs

+

. That small change breaks the tools: adding a multiple of one row to another now changes the value, so elimination fails. The determinant takes about

n^3

steps by row reduction, while the best known methods for the permanent still take more than

2^n

.Read more →

Which came first, determinants or matrices?

+

Determinants, by more than 150 years. Seki Takakazu in Japan in 1683 and Leibniz in Europe in 1693 each found them while eliminating unknowns from systems of equations, as the expression that decides whether the equations can hold together. Matrices became objects of their own only in the 1850s. That order still shows in the formulas built on determinants, such as Cramer's rule, which solves a system

A\mathbf{x} = \mathbf{b}

with ratios

\det(A_i)/\det(A)

.Read more →

Does the rule of Sarrus work for a $4 \times 4$ matrix?

+

No. A

4 \times 4

determinant is a sum of

4! = 24

products, and the downward and upward diagonals reach only

8

of them. A diagonal rule of this kind gives

2n

products against the

n!

needed, and the two agree only for

n = 3

. On the page's

4 \times 4

example the diagonals give

-15

, while expanding along column

2

, which has a single nonzero entry, gives the true value

30

. From

4 \times 4

up, use cofactor expansion or row reduction.Read more →

Determinants

A Single Number from a Square Matrix

The Determinant as a Scalar Assignment

The 2×2 Formula

Worked Examples

The 3×3 Formula

The Sarrus Mnemonic

Worked Example

The General n×n Determinant

Singular and Nonsingular Matrices

Computing Small Determinants

A 4×4 Example

Expanding by Minors and Cofactors

How Row Operations Affect the Determinant

A Quick Illustration

Area, Volume, and Orientation

Determinant-Based Formulas

Summary: How to Compute a Determinant

Common Pitfalls

Determinants FAQ

Why does ad−bcad - bcad−bc always equal the area of the parallelogram?

Where does the formula ad−bcad - bcad−bc come from?

What is the permanent of a matrix?

Which came first, determinants or matrices?

Does the rule of Sarrus work for a 4×44 \times 44×4 matrix?

Why does $ad - bc$ always equal the area of the parallelogram?

Where does the formula $ad - bc$ come from?

Does the rule of Sarrus work for a $4 \times 4$ matrix?