Theory

Matrices, determinants and linear systems

How to work with matrices, how to compute the determinant, the inverse and the rank, when a linear system has solutions and how to find them with Cramer, Rouché-Capelli and Gauss, also in the presence of a parameter, with the equilibrium of two markets and the Leontief model.

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Matrices and operations
Square matrices

Matrices and operations

1. Matrix

An $m \times n$ matrix is a table of real numbers arranged in $m$ rows and $n$ columns:

A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} .

The number $a_{ij}$ is the entry in position $(i, j)$ , that is, in row $i$ and column $j$ , and one also writes $A = (a_{ij})$ . A $1 \times n$ matrix is a row vector, an $m \times 1$ matrix is a column vector. For example

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

is a $2 \times 3$ matrix with $a_{12} = 2$ and $a_{23} = 4$ . Matrices collect in a single object the coefficients of a linear system, or data arranged in a table.

2. Sum and product by a scalar

Let $A = (a_{ij})$ and $B = (b_{ij})$ be two $m \times n$ matrices and let $\lambda \in \mathbb{R}$ . The sum $A + B$ and the product by the scalar $\lambda$ are computed entry by entry:

A + B = (a_{ij} + b_{ij}), \qquad \lambda A = (\lambda\, a_{ij}) .

The sum is defined only between matrices with the same dimensions, and it is commutative and associative; the zero matrix $O$ is the neutral element. With the matrix $A$ of the previous definition and $B = \begin{pmatrix} 2 & 0 & 1 \\ -1 & 1 & 2 \end{pmatrix}$ we have

A + B = \begin{pmatrix} 3 & 2 & 1 \\ 2 & 0 & 6 \end{pmatrix}, \qquad 2A = \begin{pmatrix} 2 & 4 & 0 \\ 6 & -2 & 8 \end{pmatrix} .

3. Row-by-column product

Let $A$ be an $m \times n$ matrix and $B$ an $n \times p$ matrix, that is, the number of columns of $A$ equals the number of rows of $B$ . The product $AB$ is the $m \times p$ matrix whose entry in position $(i, j)$ is obtained by multiplying row $i$ of $A$ by column $j$ of $B$ , entry by entry, and adding:

(AB)_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + \dots + a_{in} b_{nj} = \sum_{k = 1}^{n} a_{ik}\, b_{kj} .

With $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \end{pmatrix}$ , of type $2 \times 3$ , and $C = \begin{pmatrix} 1 & 0 \\ 2 & 1 \\ -1 & 3 \end{pmatrix}$ , of type $3 \times 2$ , the product $AC$ is $2 \times 2$ :

AC = \begin{pmatrix} 1 + 4 + 0 & 0 + 2 + 0 \\ 3 - 2 - 4 & 0 - 1 + 12 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ -3 & 11 \end{pmatrix} .

The product $CA$ , instead, is a $3 \times 3$ matrix.

4. Properties of the product

For matrices with dimensions such that the products and sums are defined, and for every $\lambda \in \mathbb{R}$ , the following hold:

associativity: $(AB)C = A(BC)$ ;
distributivity: $A(B + C) = AB + AC$ and $(A + B)C = AC + BC$ ;
compatibility with scalars: $(\lambda A)B = A(\lambda B) = \lambda\,(AB)$ .

The properties follow by writing the generic entry of both sides. By associativity one writes $ABC$ without parentheses, and for a square matrix one defines the powers $A^2 = AA$ , $A^3 = AAA$ and so on.

Warning

The matrix product is not commutative, even when $AB$ and $BA$ are both defined and have the same dimensions. For example

\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} .

Moreover, the product of two nonzero matrices can be the zero matrix, since $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = O$ . Therefore $AB = O$ does not imply that $A = O$ or $B = O$ .

5. Transpose matrix

The transpose of a matrix $A$ of type $m \times n$ is the matrix $A^{T}$ of type $n \times m$ whose rows are the columns of $A$ , that is, $(A^{T})_{ij} = a_{ji}$ . The following hold:

(A + B)^{T} = A^{T} + B^{T}, \qquad (\lambda A)^{T} = \lambda A^{T}, \qquad (AB)^{T} = B^{T} A^{T} ,

and in the last formula the order of the factors is reversed. For example $\begin{pmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \end{pmatrix}^{T} = \begin{pmatrix} 1 & 3 \\ 2 & -1 \\ 0 & 4 \end{pmatrix}$ .

Square matrices

1. Square matrix and identity matrix

A matrix is square of order $n$ if it has $n$ rows and $n$ columns. The entries $a_{11}, a_{22}, \dots, a_{nn}$ form the main diagonal. The identity matrix $I_n$ is the square matrix of order $n$ with $1$ on the main diagonal and $0$ elsewhere:

I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} .

For every square matrix $A$ of order $n$ we have $A I_n = I_n A = A$ , because the product of $A$ by column $j$ of $I_n$ is column $j$ of $A$ ; in the matrix product the identity therefore plays the role of the number $1$ .

2. Diagonal and triangular matrices

Let $A$ be a square matrix of order $n$ .

$A$ is diagonal if $a_{ij} = 0$ for every $i \neq j$ , that is, if the entries off the main diagonal are zero.
$A$ is upper triangular if $a_{ij} = 0$ for every $i > j$ , that is, if the entries below the main diagonal are zero.
$A$ is lower triangular if $a_{ij} = 0$ for every $i < j$ .

The product of two diagonal matrices is diagonal and is obtained by multiplying the corresponding diagonal entries, so the powers of a diagonal matrix are computed at once:

\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}^{3} = \begin{pmatrix} 8 & 0 \\ 0 & -1 \end{pmatrix} .

Gauss’s method transforms a square matrix into an upper triangular matrix, whose determinant is the product of the diagonal entries.

3. Symmetric matrix

A square matrix $A$ is symmetric if $A^{T} = A$ , that is, if $a_{ij} = a_{ji}$ for every $i$ and $j$ ; the entries are arranged symmetrically with respect to the main diagonal. For example $\begin{pmatrix} 1 & 4 \\ 4 & -3 \end{pmatrix}$ is symmetric. For every matrix $B$ the matrix $B B^{T}$ is symmetric, since $(B B^{T})^{T} = (B^{T})^{T} B^{T} = B B^{T}$ . Symmetric matrices appear in the study of functions of two variables, where the Hessian matrix is symmetric by Schwarz’s theorem.

The theory continues with 8 more sections (determinant, properties of the determinant, inverse matrix…), plus all worked examples, the solutions and the PDF.

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