Theory

Determinant and inverse matrix

How to compute the determinant of a square matrix, which properties simplify the computation, and how the determinant decides invertibility and gives the formula for the inverse.

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Determinant of order 2 and 3
Laplace expansion

Determinant of order 2 and 3

1. Determinant of order 2

Let $A = (a c b d) \in M_{2} (R)$ . The determinant of $A$ is the real number

det A = a c b d = a d - b c,

that is, the product of the entries on the main diagonal minus the product of the entries on the antidiagonal. The determinant assigns a single number to every square matrix, and by the invertibility criterion this number decides whether the matrix is invertible. For example

35 - 2 4 = 3 \cdot 4 - (- 2) \cdot 5 = 22.

Geometric meaning

Let $u = (u_{1}, u_{2})$ and $v = (v_{1}, v_{2})$ be two vectors in the plane. The area of the parallelogram with two adjacent sides along $u$ and $v$ is the absolute value of the determinant of the matrix with rows $u$ and $v$ :

Area = det (u_{1} v_{1} u_{2} v_{2}) .

The sign indicates the orientation: the determinant is positive when the shortest rotation taking $u$ to $v$ is counterclockwise, negative when it is clockwise. For $u = (3, 1)$ and $v = (1, 2)$ one gets $det = 3 \cdot 2 - 1 \cdot 1 = 5$ , so the parallelogram has area $5$ . The triangle with sides $u$ and $v$ is half of the parallelogram and has area $\frac{5}{2}$ .

The parallelogram with sides

u = (3, 1)

and

v = (1, 2)

has area

det (3112) = 5

Warning

The determinant is zero exactly when the parallelogram degenerates into a segment, that is, when $u$ and $v$ are parallel. For $u = (2, 1)$ and $v = (4, 2)$ we have $v = 2 u$ , so the rows of the matrix are proportional and

2412 = 4 - 4 = 0.

The vectors

u = (2, 1)

and

v = (4, 2)

lie on the same line: the parallelogram has area

0

and the determinant is zero

2. Determinant of order 3 and Sarrus' rule

Let $A = (a_{ij}) \in M_{3} (R)$ . The determinant of $A$ is

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} .

Sarrus’ rule is a way to remember the six products: copy the first two columns to the right of the matrix,

a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} a_{11} a_{21} a_{31} a_{12} a_{22} a_{32}

then add the products of the three diagonals going down to the right and subtract the products of the three diagonals going down to the left. For example

210131012 = (12 + 0 + 0) - (0 + 2 + 2) = 8.

Warning

Sarrus’ rule holds only for matrices of order $3$ . For the matrix $J \in M_{4} (R)$ with $1$ on the antidiagonal and $0$ elsewhere, the analogous scheme with the first three columns copied gives $- 1$ , whereas Laplace expansion gives $det J = 1$ . For order $n \geq 4$ one uses Laplace expansion and elementary operations.

Laplace expansion

1. Minor and cofactor

Let $A = (a_{ij}) \in M_{n} (R)$ with $n \geq 2$ . For each pair of indices $(i, j)$ , $A_{ij}$ denotes the matrix of order $n - 1$ obtained from $A$ by deleting row $i$ and column $j$ ; its determinant $det A_{ij}$ is the minor of the entry $a_{ij}$ . The cofactor of $a_{ij}$ is

C_{ij} = (- 1)^{i + j} det A_{ij} .

The sign $(- 1)^{i + j}$ is $+$ when $i + j$ is even and $-$ when it is odd, and it is arranged in a checkerboard pattern with $+$ in the top left corner:

+ - + - + - + - + .

For $A = 210131012$ we have $C_{12} = - 1012 = - 2$ .

2. Determinant of order n

The determinant of a matrix of order $1$ is its only entry: $det (a_{11}) = a_{11}$ . For $n \geq 2$ the determinant of $A \in M_{n} (R)$ is defined by expanding along the first row:

det A = a_{11} C_{11} + a_{12} C_{12} + \dots + a_{1 n} C_{1 n} = j = 1 \sum n a_{1 j} C_{1 j} .

The definition is recursive: a determinant of order $n$ reduces to $n$ determinants of order $n - 1$ . For $n = 2$ and $n = 3$ the formula returns $a d - b c$ and the six products of Sarrus’ rule.

3. Laplace's theorem

Let $A \in M_{n} (R)$ with $n \geq 2$ . For every row $i$ and every column $j$ ,

det A = k = 1 \sum n a_{ik} C_{ik} and det A = k = 1 \sum n a_{k j} C_{k j} .

Hence the determinant can be expanded along any row or any column, with the same result. In computations one chooses the row or column with the most zeros, because the terms with $a_{ik} = 0$ do not require computing the cofactor. The proof uses the expression of the determinant as a sum over the permutations of the indices and is omitted.

Expansion of a determinant of order 4

det A = 1020211000303212 = 2 \cdot (- 1)^{4 + 4} 102211003 = 2 \cdot 3 \cdot (- 1)^{3 + 3} 1021 = 6.

The fourth row has only one nonzero entry, therefore the expansion along that row leaves a single determinant of order $3$ . In this one the third column has only one nonzero entry, and the expansion along the third column reduces to a determinant of order $2$ .

The theory continues with 6 more sections (properties of the determinant, computation with elementary operations, determinant and invertibility…), plus all worked examples, the solutions and the PDF.

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