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Determinants of matrices

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 $a$:

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

The determinant of a matrix $\mathbf{A}$ of order 2, $\mathbf{A}=\left[ \begin {array} {cc}

a & b \\

c & d \\

\end{array} \right]$ , is the number:

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

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

$\mathbf{A}=\left[ \begin {array} {ccc}

a & b & c \\

d & e & f \\

g & h & i \\

\end{array} \right]$ , is the number:

$$det \mathbf{A} = a \left| \begin {array} {cc}

e & f \\

h & i \\

\end{array} \right| – d \left| \begin {array} {cc}

b & c \\

h & i \\

\end{array} \right| + g \left | \begin {array} {cc}

b & c \\

e & f \\

\end{array} \right|.$$

$$\vdots$$

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}= \left[ \begin {array} {cc}

-1 & 5 \\

3 & -2 \\

\end{array} \right]$

b) $ \mathbf{B}= \left[ \begin {array} {ccc}

2 & 1 & -2 \\

3 & 4 & -6 \\

-3 & 0 & 9 \\

\end{array} \right]$.

Solution:

a)

$$\left| \begin {array} {cc}

-1 & 5 \\

3 & -2 \\

\end{array} \right| = (-1) \cdot (-2) – 5 \cdot 3 = 2 – 15 = -13$$

b)

$$\left| \begin {array} {ccc}

2 & 1 & -2 \\

3 & 4 & -6 \\

-3 & 0& 9 \\

\end{array} \right| = 2 \left| \begin {array} {cc}

4 & -6 \\

-3 & 9 \\

\end{array} \right| – 3 \left| \begin {array} {cc}

1 & -2 \\

0 & -9 \\

\end{array} \right| + (-3) \left| \begin {array} {cc}

1 & -2 \\

4 & -6 \\

\end{array} \right|=$$

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

$$2 \cdot  (36-18) + 3 \cdot (9 + 0) – 3 \cdot (-6 +8) = $$

$$2 \cdot 18 + 3 \cdot 9 – 3 \cdot 2 = 36 + 27 -6 = 57$$

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 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$.

Example 2.

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

$$\mathbf{A} =

\left[ \begin {array} {cccc}

1 & -4 & -2 & 0 \\

3 &  5 &  0 &-1 \\

7 & 0& -9 & 0 \\

-2 & 1& -3 & 2 \\

\end{array} \right].$$

Solution:

$$

\left| \begin {array} {cccc}

1 & -4 & -2 & 0 \\

3 &  5 &  0 &-1 \\

7 & 0& -9 & 0 \\

-2 & 1& -3 & 2 \\

\end{array} \right| = (-1)^{2+4} \cdot (-1) \cdot

\left| \begin {array} {ccc}

1 & -4 & -2  \\

7 & 0& -9  \\

-2 & 1& -3  \\

\end{array} \right| + 2 \cdot

\left| \begin {array} {ccc}

1 & -4 & -2  \\

3 & 5 & 0  \\

-2 & 1& -3  \\

\end{array} \right| = $$

$$=1 \cdot (-1) \cdot \left( 1 \cdot

\left| \begin {array} {cc}

0 & -9 \\

1 & -3   \\

\end{array} \right| – 7 \cdot

\left| \begin {array} {cc}

-4 & -2 \\

1 & -3 \\

\end{array} \right| + (-2) \cdot

\left| \begin {array} {cc}

-4 & -2 \\

0 & -9 \\

\end{array} \right| \right) +$$

$$+2 \cdot

\left( 1 \cdot

\left| \begin {array} {cc}

5 & 0 \\

1 & -3   \\

\end{array} \right| – 3 \cdot

\left| \begin {array} {cc}

-4 & -2 \\

1 & -3 \\

\end{array} \right| + (-2) \cdot

\left| \begin {array} {cc}

-4 & -2 \\

5 & 0   \\

\end{array} \right| \right)  \\

=

$$

$$= (-1) \cdot (-161) + 2 \cdot (-47) = 161 – 94 = 67.$$

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,

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

 

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} =\left[ \begin {array} {cc}

1 & -2 \\

3 & 4   \\

\end{array} \right]$ is a matrix $\mathbf{\tilde A} =\left[ \begin {array} {cc}

4 & -3 \\

2 & 1   \\

\end{array} \right]$.

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.