# Determinants of matrices

## Determinant of a matrix

The determinant is a real function  such that each square matrix $\mathbf{A}$ joins a real number (the determinant of a matrix $\mathbf{A}$). The determinant of a square matrix $\mathbf{A}$ is denoted as $det \mathbf{A}$ or $|\mathbf{A}|$.

The determinant of matrices we define as inductive, that is, the determinant of a square matrix of the $n$-th order we define using the determinant of a square matrix of the $(n-1)$ -th order.

The determinant of a matrix $\mathbf{A}=[a]$ of order $1$ is the number $a$:

$$\mathbf{A}=a.$$

The determinant of a matrix $\mathbf{A}$ of order 2, $\mathbf{A}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , is the number:

$$det \mathbf{A} = ad-bc$$.

The determinant of a matrix $\mathbf{A}$ of order 3,

$\mathbf{A}=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ , is the number:

$$det \mathbf{A} = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} – d \begin {vmatrix} b & c \\ h & i \end{vmatrix} + g \begin {vmatrix} b & c \\ e & f \end{vmatrix} .$$

$$\vdots$$

The determinant of a matrix

$$\mathbf{A} = \left[ \begin{array} {ccccc} a_{11} & a_{12} & a_{13} & \ldots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \ldots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \ldots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & a_{n3} & \ldots & a_{nn} \\ \end{array} \right]$$

of the $n$-th order is the number

$$det \mathbf{A} = a_{11} det \mathbf{A}_{11} + a_{21} det \mathbf{A}_{21} + \ldots + (-1) ^{n-1} a_{n1} det \mathbf{A}_{n1}.$$

Example 1.

Calculate the determinant of the following matrices:

a) $\mathbf{A}= \begin{bmatrix} -1 & 5 \\ 3 & -2 \end{bmatrix}$

b) $\mathbf{B}= \begin{bmatrix} 2 & 1 & -2 \\ 3 & 4 & -6 \\ -3 & 0 & 9 \end{bmatrix}$.

Solution:

a)

$$\begin{vmatrix} -1 & 5 \\ 3 & -2 \end{vmatrix} = (-1) \cdot (-2) – 5 \cdot 3 = 2 – 15 = -13$$

b)

$$\begin{vmatrix} 2 & 1 & -2 \\ 3 & 4 & -6 \\ -3 & 0 & 9 \end{vmatrix} = 2 \begin {vmatrix} 4 & -6 \\ 0 & 9 \end{vmatrix} – 3 \begin {vmatrix} 1 & -2 \\ 0 & 9 \end{vmatrix} + (-3) \begin {vmatrix} 1 & -2 \\ 4 & -6 \end{vmatrix}=$$

$$= 2 \cdot (4\cdot 9 – (-6) \cdot 0) – 3\cdot (1 \cdot 9 – (-2) \cdot 0) -3 \cdot (1 \cdot (-6) – (-2) \cdot 4) =$$

$$2 \cdot 36 – 3 \cdot (9 + 0) – 3 \cdot (-6 +8) =$$

$$72 – 3 \cdot 9 – 3 \cdot 2 = 72 – 27 -6 = 39$$

As we can see, calculating the determinant it is not easy for matrices of a higher order. There are several methods for calculating the determinant. The general method is the Laplace expansion of a determinant along a given column or row.

Let $\mathbf{A}=[a_{ij}]$ be a square matrix of order $n$. If in a matrix $\mathbf{A}$ is removed $i$-th row and $j$-th column, we have thus obtained a matrix of order $(n-1)$ of  which the determinant is called a minor and is denoted as $M_{ij}$.

## The algebraic complement

The algebraic complement or cofactor of an element $a_{ij}$ is the number:

$$A_{ij}=(-1)^{i+j}M_{ij}.$$

Let $\mathbf{A}$ be an $n\times n$ matrix. For a matrix $\mathbf{A}$ we have the following expansions:

a) for a fixed $i$ : $$det \mathbf{A}= \sum_{j=1}^n (-1)^{i+j} a_{ij} M_{ij},$$

b) for a fixed $j$: $$det \mathbf{A}= \sum_{i=1}^n (-1)^{i+j} a_{ij} M_{ij}.$$

## The properties of determinants

Let $\mathbf{A}$ be a matrix of order $n$.

1.) A matrix $\mathbf{A}$ and its transpose matrix $\mathbf{A}^T$ have the same determinants, that is $$det \mathbf{A} = det \mathbf{A}^T.$$

2.) If $\mathbf{A}$ is a triangular matrix, then its determinant is equal to the product of all diagonal elements, that is

$$det \mathbf{A} = a_{11} \cdot a_{22}\cdot \ldots \cdot a_{nn}.$$

3.) If two columns or rows of the determinant interchange the location, then

$$det \mathbf{A}= – det \mathbf{A}.$$

4.) If a matrix $\mathbf{A}$ has two equal columns or rows, then $det \mathbf{A}=0$.

5.) If $\mathbf{A}$ is a matrix with a row or column where every element is equal to zero, then $det \mathbf{A}=0$.

6.) The determinant value does not change if to one column we add a linear combination of the remaining columns.

7.) If a matrix $\mathbf{B}$ is obtained from a matrix $\mathbf{A}$ by multiplying one its row (or column) with a scalar $\alpha$, then

$$det \mathbf{B} = \alpha \cdot det \mathbf{A}.$$

8.) A matrix $\mathbf{A}$ is a regular matrix iff $det \mathbf{A} \neq 0$.

9.) For a singular matrix $\mathbf{A}$ is $det\mathbf{A} =0$.

10.) For a regular matrix $\mathbf{A}$ is $det\mathbf{A} \neq 0$.

Example 2.

Using the Laplace expansion of a determinant along $4$-th column, calculate the determinant of the following matrix:

$$\mathbf{A} =\begin{bmatrix} 1 & -4 & -2 & 0 \\ 3 & 5 & 0 &-1 \\ 7 & 0& -9 & 0 \\ -2 & 1& -3 & 2 \end{bmatrix} .$$

Solution:

$$\begin{vmatrix} 1 & -4 & -2 & 0 \\ 3 & 5 & 0 &-1 \\ 7 & 0& -9 & 0 \\ -2 & 1& -3 & 2 \end{vmatrix} = (-1) \cdot (-1)^{2+4} \cdot \begin{vmatrix} 1 & -4 & -2 \\ 7 & 0& -9 \\ -2 & 1& -3 \end{vmatrix} + 2 \cdot (-1)^{4+4} \begin{vmatrix} 1 & -4 & -2 \\ 3 & 5 & 0 \\ 7 & 0 & -9 \end{vmatrix} =$$

$$=-1 \cdot \left( 1 \cdot \begin{vmatrix} 0 & -9 \\ 1 & -3 \end{vmatrix} – 7 \cdot \begin{vmatrix} -4 & -2 \\ 1 & -3 \end{vmatrix} + (-2) \cdot \begin{vmatrix} -4 & -2 \\ 0 & -9 \end{vmatrix}\right)+$$

$$2 \cdot\left( 1 \cdot \begin{vmatrix} 5 & 0 \\0 & -9 \end{vmatrix} – 3 \cdot \begin {vmatrix} -4 & -2 \\ 0 & -9 \end{vmatrix} + 7 \cdot \begin{vmatrix} -4 & -2 \\ 5 & 0 \end{vmatrix} \right)=$$

$$=161-166= -5.$$

In general, we choose the row or column that contains the most zeroes, because then we have a shorter calculation.

## The Binet – Cauchy theorem

If matrices $\mathbf{A}$ and $\mathbf{B}$ are square matrices of the same order, then

$$\det (\mathbf{A}\mathbf{B}) = det \mathbf{A} det \mathbf{B}$$

is valid.

We can observe that for a regular matrix $\mathbf{A}$

$$\mathbf{I} = \mathbf{A} \mathbf{A}^{-1}$$

is valid if we know the determinant of the matrix $\mathbf{A}$ and through using the Binet – Cauchy theorem we can calculate $\mathbf{A}^{-1}$. Namely,

$$det \mathbf{A}^{-1}= \frac{1}{det \mathbf{A}}.$$

The adjugate or adjoint of a square matrix $\mathbf{A}$ is a matrix $\mathbf{\tilde A} = [x_{ij}]$, where $x_{ij} = a_{ji}$, $\forall i, j = 1, \ldots, n$. This means that the adjugate of a square matrix $\mathbf{A}$ is the transpose of a cofactor matrix of a matrix $\mathbf{A}$.

The adjugate of a matrix $\mathbf{A}$ can also be denoted as $adj (\mathbf{A})$.

For instance, the adjugate matrix of a matrix $\mathbf{A} =\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$ is a matrix $\mathbf{\tilde A} =\begin{bmatrix} 4 & -3 \\ 2 & 1 \end{bmatrix}$.

As was mentioned previously, a square matrix $\mathbf{A}$ is a regular matrix iff $det \mathbf{A} \neq 0$. In this case, an inverse matrix $\mathbf{A}^{-1}$ of a matrix $\mathbf{A}$ is given by the following formula:

$$A^{-1} = \frac{1}{det \mathbf{A}} \mathbf{\tilde A}.$$

Now we can find, using the formula above, an inverse matrix of a given square matrix of any order.

Example 3: Use the formula above to find an inverse matrix of a matrix

$$\mathbf{A}=\begin{bmatrix} 1 & 2 & 1\\ 2 & 5 & 2 \\ -1 & -2 & 0 \end{bmatrix}.$$

Solution:

First, we will calculate the determinant of the given matrix:

$$\begin{vmatrix} 1 & 2 & 1\\ 2 & 5 & 2 \\ -1 & -2 & 0 \end{vmatrix}= 1 \cdot (-1)^{1+1} \begin{vmatrix} 5 & 2 \\ -2 & 0 \end{vmatrix} +2 \cdot (-1)^{1+2}\begin{vmatrix} 2 & 2 \\ -1 & 0 \end{vmatrix}+1 \cdot (-1)^{1+3} \begin{vmatrix} 2 & 5 \\ -1 & -2 \end{vmatrix}$$

$$= 0+4-2 \cdot 2 -4 + 5 = 1.$$

Now we have to calculate the adjugate matrix of matrix $A$. To do so, we need to find the cofactor of each element:

$x_{11}= \begin{vmatrix} 5 & 2 \\ -2 & 0 \end{vmatrix}= 4$

$x_{12}=-\begin{vmatrix} 2 & 2 \\ -1 & 0 \end{vmatrix}= -2$

$x_{13}=\begin{vmatrix}2 & 5 \\ -1 & -2\end{vmatrix}= 1$

$x_{21}=-\begin{vmatrix}2 & 1 \\ -2 & 0\end{vmatrix}= -2$

$x_{22}=\begin{vmatrix}1 & 1 \\ -1 & 0\end{vmatrix}= 1$

$x_{23}=-\begin{vmatrix}1 & 2 \\ -1 & -2\end{vmatrix}= 0$

$x_{31}=\begin{vmatrix}2 & 1 \\ 5 & 2\end{vmatrix}= -1$

$x_{32}=-\begin{vmatrix}1 & 1 \\ 2 & 2\end{vmatrix}= 0$

$x_{33}=\begin{vmatrix}1 & 2 \\ 2 & 5\end{vmatrix}= 1.$

The cofactor matrix of matrix $A$ is:

$$\begin{bmatrix} 4 & -2 & 1 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}.$$

The adjugate matrix of given matrix $A$ is the transpose of the cofactor matrix:

$$\begin{bmatrix} 4 & -2 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}.$$

Finally, we use formula $A^{-1} = \frac{1}{det \mathbf{A}} \mathbf{\tilde A}$ to find an inverse of matrix $A$:

$$A^{-1}=\frac{1}{1} \begin{bmatrix} 4 & -2 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$

$$=\begin{bmatrix} 4 & -2 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$