In short

The determinant of a square matrix is a single number that tells you two things at once: whether the matrix is invertible (determinant \neq 0) and how much the matrix scales areas or volumes. For a 2 \times 2 matrix it is ad - bc. For a 3 \times 3 matrix you expand along a row or column using minors and cofactors. The Sarrus rule gives a quick visual shortcut for 3 \times 3.

Take two vectors in the plane: \mathbf{u} = (3, 0) pointing along the x-axis, and \mathbf{v} = (1, 2) pointing diagonally. Together they span a parallelogram.

What is the area of that parallelogram? You can work it out from geometry: the base is |\mathbf{u}| = 3, the height is the perpendicular distance from \mathbf{v} to the x-axis, which is 2. So the area is 3 \times 2 = 6.

Now arrange the coordinates into a matrix:

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

Compute 3 \times 2 - 1 \times 0 = 6. The same number. Not a coincidence — the number you just computed is called the determinant of the matrix, and it always gives the signed area of the parallelogram spanned by the matrix's rows (or columns).

The sign matters: if the two vectors are arranged counterclockwise, the determinant is positive. If clockwise, negative. Zero means the two vectors are parallel — they collapse onto a line, and the "parallelogram" has no area at all.

This single number — positive, negative, or zero — turns out to encode exactly the information you need to know whether a system of equations has a unique solution, whether a matrix can be inverted, and how much a linear transformation stretches or compresses space.

The parallelogram spanned by $\mathbf{u} = (3, 0)$ and $\mathbf{v} = (1, 2)$. Its area is the determinant $\begin{vmatrix} 3 & 0 \\ 1 & 2 \end{vmatrix} = 6$. The area is positive because the two vectors are arranged counterclockwise.

The 2×2 determinant

Determinant of a 2×2 matrix

For a 2 \times 2 matrix A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is

\det(A) = |A| = ad - bc

Two notations are standard: \det(A) and the vertical-bar notation \begin{vmatrix} a & b \\ c & d \end{vmatrix}. Both mean the same thing. The vertical bars look like absolute value bars, but they are not — the determinant can be negative.

Reading the formula. The determinant is the product of the main diagonal (a \times d) minus the product of the off-diagonal (b \times c). Think of it as a cross: multiply \searrow minus multiply \nearrow.

The 2 by 2 determinant as diagonal productsA visual showing a 2 by 2 matrix with two diagonal arrows. The main diagonal from a to d is labelled with a plus sign and coloured in the accent colour. The off-diagonal from b to c is labelled with a minus sign and drawn dashed. a b c d + ad − bc
The $2 \times 2$ determinant: multiply along the main diagonal (solid), subtract the product along the off-diagonal (dashed). The result is $ad - bc$.

A few quick computations to build muscle memory:

\begin{vmatrix} 5 & 3 \\ 2 & 4 \end{vmatrix} = 5(4) - 3(2) = 20 - 6 = 14
\begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} = 1(6) - 2(3) = 6 - 6 = 0

The second one is zero — and notice that the second row (3, 6) is exactly 3 times the first row (1, 2). The two rows point in the same direction. The "parallelogram" is flat, with zero area. When the determinant is zero, the matrix is singular — it has no inverse, and any system of equations using this matrix as coefficients either has no solution or infinitely many.

Minors and cofactors

To extend the idea beyond 2 \times 2, you need two new terms.

Minor

The minor M_{ij} of an element a_{ij} in a square matrix is the determinant of the matrix obtained by deleting row i and column j.

Cofactor

The cofactor C_{ij} of an element a_{ij} is the minor with a sign attached:

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

The sign alternates in a checkerboard pattern: + when i + j is even, - when i + j is odd.

The checkerboard pattern for a 3 \times 3 matrix looks like this:

\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}

The sign of C_{ij} depends only on the position (i, j), not on the value of the entry.

Computing the minor M₁₂ by deleting row 1 and column 2A 3 by 3 matrix with row 1 and column 2 crossed out. The remaining four entries form the 2 by 2 sub-matrix whose determinant is the minor M₁₂. 2 3 1 4 5 6 7 8 9 → 4 6 7 9 M₁₂ = 36 − 42 = −6
To find the minor $M_{12}$, delete row 1 and column 2 (shown crossed out). The remaining entries form a $2 \times 2$ matrix. Its determinant is the minor. The cofactor attaches the checkerboard sign: $C_{12} = (-1)^{1+2}(-6) = +6$.

Take a concrete 3 \times 3 matrix:

A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}

The minor M_{11} (delete row 1, column 1) is:

M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 45 - 48 = -3

The cofactor C_{11} = (-1)^{1+1} M_{11} = (+1)(-3) = -3.

The minor M_{12} (delete row 1, column 2) is:

M_{12} = \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = 36 - 42 = -6

The cofactor C_{12} = (-1)^{1+2} M_{12} = (-1)(-6) = 6.

The minor M_{13} (delete row 1, column 3) is:

M_{13} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = 32 - 35 = -3

The cofactor C_{13} = (-1)^{1+3} M_{13} = (+1)(-3) = -3.

These three cofactors are the building blocks of the 3 \times 3 determinant.

The 3×3 determinant

Determinant of a 3×3 matrix (expansion along the first row)

For a 3 \times 3 matrix 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

\det(A) = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}

where each C_{ij} is the cofactor defined above. Written out fully:

\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})

Apply this to the matrix above:

\det(A) = 2(-3) + 3(6) + 1(-3) = -6 + 18 - 3 = 9

Let's verify this step by step. The first term is a_{11} \cdot C_{11} = 2 \times (-3) = -6. The second term is a_{12} \cdot C_{12} = 3 \times 6 = 18. The third term is a_{13} \cdot C_{13} = 1 \times (-3) = -3. Summing: -6 + 18 - 3 = 9.

The determinant \det(A) = 9 \neq 0, so this matrix is invertible.

Sarrus' rule — a visual shortcut

For 3 \times 3 matrices only, there is a mnemonic that avoids computing cofactors explicitly.

Write out the 3 \times 3 matrix and copy the first two columns again to the right:

\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}

Now draw three diagonals going down-right (\searrow) and three diagonals going down-left (\swarrow). The determinant is the sum of the three down-right products minus the sum of the three down-left products.

Sarrus rule diagram for 3 by 3 determinantA 3 by 3 matrix with two extra columns copied on the right. Three solid diagonal lines go from top-left to bottom-right, representing the three positive products. Three dashed diagonal lines go from top-right to bottom-left, representing the three negative products. a₁₁ a₁₂ a₁₃ a₂₁ a₂₂ a₂₃ a₃₁ a₃₂ a₃₃ a₁₁ a₁₂ a₂₁ a₂₂ a₃₁ a₃₂ solid = positive products dashed = negative products
Sarrus' rule: copy the first two columns to the right of the matrix. Sum the three solid-line diagonal products (positive) and subtract the three dashed-line diagonal products (negative). This works only for $3 \times 3$ matrices — there is no analogous shortcut for $4 \times 4$ or larger.

Apply Sarrus to the same matrix A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}:

Positive diagonals (\searrow):

Negative diagonals (\swarrow):

\det(A) = (90 + 126 + 32) - (35 + 96 + 108) = 248 - 239 = 9

Same answer as the cofactor expansion. Sarrus' rule is faster when you're working by hand, but it is purely a 3 \times 3 trick. For 4 \times 4 and above, you must use cofactor expansion or row reduction.

Expansion along any row or column

The formula above expanded along the first row. But the determinant can be computed by expanding along any row or any column — you always get the same answer.

Expansion along row i:

\det(A) = \sum_{j=1}^{n} a_{ij} \, C_{ij} = a_{i1}C_{i1} + a_{i2}C_{i2} + \cdots + a_{in}C_{in}

Expansion along column j:

\det(A) = \sum_{i=1}^{n} a_{ij} \, C_{ij} = a_{1j}C_{1j} + a_{2j}C_{2j} + \cdots + a_{nj}C_{nj}

This flexibility is powerful. If a row or column has zeros in it, expand along that row or column — every zero entry kills the corresponding cofactor computation, saving work.

Let's verify by expanding the same matrix along the second column:

A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}

Expanding along column 2:

\det(A) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}

C_{12} = (-1)^{1+2}\begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = (-1)(36 - 42) = (-1)(-6) = 6

C_{22} = (-1)^{2+2}\begin{vmatrix} 2 & 1 \\ 7 & 9 \end{vmatrix} = (+1)(18 - 7) = 11

C_{32} = (-1)^{3+2}\begin{vmatrix} 2 & 1 \\ 4 & 6 \end{vmatrix} = (-1)(12 - 4) = -8

\det(A) = 3(6) + 5(11) + 8(-8) = 18 + 55 - 64 = 9

Same answer — 9 — as before. The row or column you choose does not affect the result. It affects only the amount of arithmetic you have to do.

Strategy. When computing a 3 \times 3 determinant by hand, scan the matrix for the row or column with the most zeros. Expand along that one to minimise the number of 2 \times 2 determinants you have to compute.

Why does expansion along any row or column give the same answer? Think of it this way: the determinant is a property of the matrix as a whole, not of any particular row. The cofactor expansion is one recipe for computing it. The fact that you can start the recipe from any row or column — and always arrive at the same number — is a consequence of the multilinear, alternating nature of the determinant function. You will see this formalised in Properties of Determinants. For now, just use the freedom: pick the row or column that makes your life easiest.

Here is a matrix where the right choice of row saves real effort:

B = \begin{pmatrix} 3 & 0 & 0 \\ 1 & 4 & 2 \\ 5 & 1 & 7 \end{pmatrix}

Row 1 has two zeros. Expanding along row 1:

\det(B) = 3 \cdot C_{11} + 0 \cdot C_{12} + 0 \cdot C_{13} = 3 \begin{vmatrix} 4 & 2 \\ 1 & 7 \end{vmatrix} = 3(28 - 2) = 3(26) = 78

One 2 \times 2 determinant instead of three. The zeros in row 1 killed two of the three terms before any computation happened.

Worked examples

Example 1: Area of a triangle using determinants

Three vertices of a triangle are P = (1, 2), Q = (4, 6), and R = (7, 3). Find the area using a determinant.

Step 1. Set up the matrix. The area formula for a triangle with vertices (x_1, y_1), (x_2, y_2), (x_3, y_3) is

\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|

Why: each row represents a vertex, with a 1 appended. The determinant computes twice the signed area of the triangle. The absolute value and the \frac{1}{2} convert it to the actual area.

Step 2. Substitute the coordinates.

\text{Area} = \frac{1}{2} \left| \begin{vmatrix} 1 & 2 & 1 \\ 4 & 6 & 1 \\ 7 & 3 & 1 \end{vmatrix} \right|

Step 3. Expand along the third column (it has all 1's, which simplifies the arithmetic to just computing the cofactors).

\det = 1 \cdot C_{13} + 1 \cdot C_{23} + 1 \cdot C_{33}

C_{13} = (+1)\begin{vmatrix} 4 & 6 \\ 7 & 3 \end{vmatrix} = 12 - 42 = -30

C_{23} = (-1)\begin{vmatrix} 1 & 2 \\ 7 & 3 \end{vmatrix} = -(3 - 14) = 11

C_{33} = (+1)\begin{vmatrix} 1 & 2 \\ 4 & 6 \end{vmatrix} = 6 - 8 = -2

Why: expanding along the third column because all entries are 1, so the cofactors are not multiplied by varying factors — each just contributes its cofactor value directly.

Step 4. Sum and take the absolute value.

\det = -30 + 11 + (-2) = -21
\text{Area} = \frac{1}{2}|-21| = \frac{21}{2} = 10.5 \text{ square units}

Why: the negative sign of the determinant tells you the vertices were listed in clockwise order. The absolute value discards the orientation and gives the geometric area.

Result: The area of the triangle is \dfrac{21}{2} square units.

The triangle with vertices $P(1, 2)$, $Q(4, 6)$, and $R(7, 3)$. The determinant formula gives its area as $\frac{21}{2} = 10.5$ square units — a value you can roughly verify by counting grid squares enclosed by the triangle.

The determinant formula for area is not just a trick. It works because the determinant measures the signed area of the parallelogram formed by two edge vectors of the triangle, and a triangle is exactly half a parallelogram.

Example 2: Solving a system using Cramer's rule (preview)

Solve the system \begin{cases} 2x + y = 5 \\ 3x - 2y = 4 \end{cases} using determinants.

Step 1. Write the coefficient matrix and compute its determinant.

D = \begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} = 2(-2) - 1(3) = -4 - 3 = -7

Why: D \neq 0, so the system has a unique solution. If D were zero, the two lines would be parallel (no solution) or identical (infinitely many).

Step 2. Compute D_x by replacing the first column with the constants.

D_x = \begin{vmatrix} 5 & 1 \\ 4 & -2 \end{vmatrix} = 5(-2) - 1(4) = -10 - 4 = -14

Why: Cramer's rule says x = D_x / D. To form D_x, you replace the column of x-coefficients with the right-hand side.

Step 3. Compute D_y by replacing the second column with the constants.

D_y = \begin{vmatrix} 2 & 5 \\ 3 & 4 \end{vmatrix} = 2(4) - 5(3) = 8 - 15 = -7

Why: similarly, y = D_y / D. Replace the y-coefficient column with the right-hand side.

Step 4. Divide to get the solution.

x = \frac{D_x}{D} = \frac{-14}{-7} = 2, \qquad y = \frac{D_y}{D} = \frac{-7}{-7} = 1

Why: each variable equals the ratio of a modified determinant to the original determinant. This is Cramer's rule — a direct application of determinants to solving systems of linear equations.

Result: x = 2, y = 1.

The two lines $2x + y = 5$ (dark) and $3x - 2y = 4$ (lighter) intersect at exactly $(2, 1)$. The determinant $D = -7 \neq 0$ guaranteed a unique intersection, and Cramer's rule found it without elimination or substitution.

Verify: 2(2) + 1 = 5 and 3(2) - 2(1) = 4. Both equations check out. The answer matches the intersection point visible in the graph.

Common confusions

Going deeper

If you came here to learn how to compute 2 \times 2 and 3 \times 3 determinants, you have it — you can stop here. The rest explores the geometric meaning more carefully and connects determinants to volumes in higher dimensions.

The geometric meaning: signed area and volume

For a 2 \times 2 matrix, the determinant is the signed area of the parallelogram spanned by the row vectors. For a 3 \times 3 matrix, the determinant is the signed volume of the parallelepiped (a 3D parallelogram) spanned by the three row vectors.

Take A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}. This matrix stretches space by a factor of 1 along x, 2 along y, and 3 along z. The parallelepiped formed by the rows is a rectangular box with dimensions 1 \times 2 \times 3 = 6. And \det(A) = 1 \cdot 2 \cdot 3 = 6. The determinant is the volume.

If a transformation scales areas by a factor of k, its determinant is k. If it reflects (flips orientation), the determinant is negative. If it collapses a dimension — squashing 3D space into a plane, or a plane into a line — the determinant is zero.

Recursive definition for n×n

The cofactor expansion works for any n \times n matrix. Expanding along the first row:

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

where A_{1j} is the (n-1) \times (n-1) matrix obtained by deleting row 1 and column j. Each step reduces the problem by one dimension: a 4 \times 4 determinant becomes four 3 \times 3 determinants, each of which becomes three 2 \times 2 determinants.

This recursive structure is mathematically elegant but computationally expensive — an n \times n determinant via cofactor expansion requires roughly n! multiplications. For large n, row reduction methods (Gaussian elimination) compute determinants in O(n^3) operations instead.

Historical note

The concept of a determinant appeared in Japanese mathematician Seki Takakazu's work in 1683, independently of Leibniz who used them around the same time in Europe. In India, systems of linear equations appear in Brahmagupta's Brahmasphutasiddhanta (628 CE), and while Brahmagupta did not use the determinant notation, the underlying ideas — solving simultaneous equations by systematic elimination — are recognisably the same computational structure. The modern notation with vertical bars was introduced in the 19th century.

Where this leads next