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Module 2/Determinants

The Determinant Function: 2×2, 3×3 & Axioms

The determinant condenses a square matrix into a single number that tells you whether the matrix is invertible and measures the area or volume its rows span. This lesson builds the determinant from the 2×22\times 2 formula ad−bcad-bc, its geometric meaning, the three defining axioms (Definition 6.1), and the Rule of Sarrus for 3×33\times 3 matrices.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

Find the area of the parallelogram spanned by the vectors (2,3)(2,3) and (4,1)(4,1).

Use the Rule of Sarrus to compute det⁡[123014210]\det\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 2 & 1 & 0 \end{bmatrix}.

What you’ll be able to do

  • Compute the determinant of a 2×22\times 2 matrix with the formula det⁡(A)=ad−bc\det(A)=ad-bc and use it to decide whether AA is invertible.
  • Interpret ∣det⁡(A)∣|\det(A)| as the area of the parallelogram (2D) or the volume of the parallelepiped (3D) spanned by the rows or columns of AA.
  • State and apply the three determinant axioms of Definition 6.1: (D1) det⁡(In)=1\det(I_n)=1, (D2) linearly dependent rows give det⁡=0\det=0, and (D3) linearity in each row.
  • Use the Rule of Sarrus (Remark 6.1) to evaluate the determinant of a 3×33\times 3 matrix.
  • Predict how elementary row operations change a determinant and recognize the conditions that force det⁡(A)=0\det(A)=0.

In your course

· MATH2015 · Linear Algebra & Probability
§6.1 The determinant function
  • Definition 6.1The determinant function
    A function det⁡:Rn×n→R\det:\mathbb{R}^{n\times n}\to\mathbb{R} with (D1) det⁡(In)=1\det(I_n)=1, (D2) linearly dependent rows ⇒det⁡=0\Rightarrow \det=0, and (D3) linearity in each row.
  • Theorem 6.1Effect of elementary row operations on determinants
    Scaling a row by λ\lambda scales det⁡\det by λ\lambda; adding a multiple of one row to another leaves det⁡\det unchanged; interchanging two rows reverses the sign.
  • Theorem 6.2Determinant, rank and invertibility
    det⁡(A)≠0  ⟺  rank⁡(A)=n  ⟺  ker⁡(A)={0}  ⟺  A\det(A)\neq 0 \iff \operatorname{rank}(A)=n \iff \ker(A)=\{0\} \iff A is invertible.
  • Proposition 6.1Determinant of a triangular matrix
    The determinant of a triangular matrix is the product of its diagonal entries.
  • Remark 6.1Rule of Sarrus
    det⁡(A)=a11a22a33+a12a23a31+a13a21a32−a13a22a31−a11a23a32−a12a21a33\det(A)=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33} for 3×33\times 3 matrices.
Cofactor/Laplace expansion (Definitions 6.2–6.3) and the uniqueness theorem (Theorem 6.3) extend this section but are beyond this lesson.
1

The 2×2 determinant and invertibility

For a 2×22\times 2 matrix A=[abcd]A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} the determinant is the scalar det⁡(A)=∣abcd∣=ad−bc.\det(A)=\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc. From Example 4.9 we know AA is invertible if and only if ad−bc≠0ad-bc\neq 0. When det⁡(A)≠0\det(A)\neq 0 the matrix has rank 22 and the system Ax=yAx=y has a unique solution for every y∈R2y\in\mathbb{R}^2; equivalently, the row (and column) vectors of AA are linearly independent.

2

Geometric meaning: area and volume

Up to sign, ad−bcad-bc is exactly the area of the parallelogram determined by the two row vectors (equivalently, the column vectors) of AA, so that area equals ∣det⁡(A)∣|\det(A)|; the sign records orientation. This picture extends: up to sign, the determinant of a 3×33\times 3 matrix is the volume of the parallelepiped spanned by its three row vectors, and in general ∣det⁡∣|\det| is the nn-dimensional volume of the 'hyper'-parallelogram spanned by nn vectors in Rn\mathbb{R}^n. In particular the vectors are linearly dependent (zero area/volume) exactly when det⁡=0\det=0.

3

Definition 6.1 — the axioms (D1)–(D3)

The determinant is defined axiomatically as a function det⁡:Rn×n→R\det:\mathbb{R}^{n\times n}\to\mathbb{R}, where A=(r1,…,rn)A=(r_1,\dots,r_n) is written by its rows, satisfying: (D1) det⁡(In)=1\det(I_n)=1; (D2) if the rows are linearly dependent then det⁡(A)=0\det(A)=0; (D3) det⁡\det is linear in each row separately — for every scalar λ\lambda and row vector vv, det⁡(r1,…,λri,…,rn)=λdet⁡(r1,…,ri,…,rn)\det(r_1,\dots,\lambda r_i,\dots,r_n)=\lambda\det(r_1,\dots,r_i,\dots,r_n) and det⁡(r1,…,ri+v,…,rn)=det⁡(r1,…,ri,…,rn)+det⁡(r1,…,v,…,rn)\det(r_1,\dots,r_i+v,\dots,r_n)=\det(r_1,\dots,r_i,\dots,r_n)+\det(r_1,\dots,v,\dots,r_n). These three properties generalize the three properties of the 2×22\times 2 determinant, and (Theorem 6.3) they determine the function uniquely.

4

The Rule of Sarrus for 3×3 matrices

For a 3×33\times 3 matrix, Remark 6.1 gives a quick scheme: copy the first two columns to the right of the matrix, then add the three products running down the diagonals parallel to the main diagonal and subtract the three products running up the anti-diagonals: det⁡(A)=a11a22a33+a12a23a31+a13a21a32−a13a22a31−a11a23a32−a12a21a33.\det(A)=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}. Warning: Sarrus works only for 3×33\times 3 matrices — it does not generalize to larger sizes.

Definition 6.1 — The determinant function

A determinant is a function det⁡:Rn×n→R\det:\mathbb{R}^{n\times n}\to\mathbb{R} such that (D1) det⁡(In)=1\det(I_n)=1; (D2) if the rows of AA are linearly dependent then det⁡(A)=0\det(A)=0; (D3) det⁡\det is linear in each row: det⁡(…,λri,… )=λdet⁡(…,ri,… )\det(\dots,\lambda r_i,\dots)=\lambda\det(\dots,r_i,\dots) and det⁡(…,ri+v,… )=det⁡(…,ri,… )+det⁡(…,v,… )\det(\dots,r_i+v,\dots)=\det(\dots,r_i,\dots)+\det(\dots,v,\dots).

Intuition. Think of these as the rules a signed-volume gauge must obey: the unit cube has volume 11 (D1); a flattened, degenerate box has volume 00 (D2); and stretching one edge by λ\lambda, or splitting one edge into a sum, scales or splits the volume accordingly (D3).
Theorem 6.1 — Effect of elementary row operations

(1) Multiplying a row by λ\lambda multiplies the determinant by λ\lambda. (2) Adding λ\lambda times one row to another row does not change the determinant. (3) Interchanging two rows reverses the sign of the determinant.

Intuition. These follow from the axioms and let you compute any determinant by Gaussian elimination, tracking a sign for each row swap and a factor for each row scaling until the matrix is triangular (whose determinant is the product of the diagonal entries, Proposition 6.1).
Theorem 6.2 — Determinant and invertibility

For A∈Rn×nA\in\mathbb{R}^{n\times n}: det⁡(A)=0  ⟺  rank⁡(A)<n  ⟺  dim⁡ker⁡(A)>0  ⟺  \det(A)=0 \iff \operatorname{rank}(A)<n \iff \dim\ker(A)>0 \iff the rows (columns) are linearly dependent. Equivalently, det⁡(A)≠0  ⟺  rank⁡(A)=n  ⟺  ker⁡(A)={0}  ⟺  A\det(A)\neq 0 \iff \operatorname{rank}(A)=n \iff \ker(A)=\{0\} \iff A is invertible.

Intuition. A single number certifies invertibility: a nonzero determinant means full rank, trivial kernel, independent rows/columns, and a unique solution to Ax=yAx=y. A zero determinant means the box the rows span is degenerate, with zero volume.
Remark 6.1 — Rule of Sarrus

For a 3×33\times 3 matrix, det⁡(A)=a11a22a33+a12a23a31+a13a21a32−a13a22a31−a11a23a32−a12a21a33\det(A)=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}.

Intuition. Augment the matrix with copies of its first two columns, then sum the three down-right diagonals and subtract the three down-left diagonals. It is a memory aid specific to 3×33\times 3 and gives the same answer as cofactor expansion along the first row.

Worked examples

Example 1

Compute det⁡(A)\det(A) for A=[3124]A=\begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix} and state whether AA is invertible.

  1. 1

    Identify the entries: a=3, b=1, c=2, d=4a=3,\ b=1,\ c=2,\ d=4.

  2. 2

    Apply the formula: det⁡(A)=ad−bc=(3)(4)−(1)(2)=12−2\det(A)=ad-bc=(3)(4)-(1)(2)=12-2.

  3. 3

    So det⁡(A)=10\det(A)=10.

  4. 4

    Since 10≠010\neq 0, by the invertibility criterion AA is invertible (it has rank 22).

Answer. det⁡(A)=10\det(A)=10, so AA is invertible.
Example 2

Use the Rule of Sarrus to compute det⁡(A)\det(A) for A=[210131021]A=\begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 2 & 1 \end{bmatrix}.

Example 3

Find the area of the parallelogram spanned by the vectors (1,2)(1,2) and (3,1)(3,1).