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