Properties of determinants

The determinants have certain properties that should be known. These properties are very useful to convert the determinants calculation into something a little less slow and tedious.

Let's see some of these properties:

Any matrix and its transpose (the transpose matrix is the result of rotating the rows of a matrix to turn them into columns) have the same determinant.

$| A | = | A^{t} |$

The determinant of a matrix is zero, $| A | = 0$ , if:
1. The matrix has two equal rows. It is easy to prove this in an exercise for a $3 \times 3$ case, for example:
$\begin{matrix} | \begin{matrix} a & b & c \\ d & e & f \\ a & b & c \end{matrix} | \end{matrix} = a \cdot e \cdot c + d \cdot b \cdot c + a \cdot b \cdot f - c \cdot e \cdot a - f \cdot b \cdot a - c \cdot b \cdot d = 0$
1. All the elements of a row are zeros.
2. The elements of a row are a linear combination of other rows. That is:
$| \begin{matrix} 2 & 3 & 2 \\ 1 & 2 & 4 \\ 3 & 5 & 6 \end{matrix} |$
The 3rd row is a linear combination of the other two ( $f_{3} = f_{1} + f_{2}$ ). Without calculating anything, we know that the determinant will be zero.
If we swap two parallel rows the determinant changes its sign:

$| \begin{matrix} 0 & 5 & 1 \\ 1 & 2 & 7 \\ 3 & 1 & 2 \end{matrix} | = - | \begin{matrix} 1 & 2 & 7 \\ 0 & 5 & 1 \\ 3 & 1 & 2 \end{matrix} |$

If we add the elements of a row to the elements of a parallel row that have previously been multiplied by a real number, the value of the determinant does not change.

$| \begin{matrix} 1 & 2 & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 2 \end{matrix} | \to C_{3} = 2 \cdot C_{1} + C_{3} \to | \begin{matrix} 1 & 2 & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 2 \end{matrix} | = | \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 3 & 1 & 8 \end{matrix} | = | \begin{matrix} 1 & 2 & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 2 \end{matrix} |$

Multiplying a determinant by a real number is the same as multiplying one of its rows by that real number.
The determinant of a product is equal to the product of determinants.

$| A \cdot B | = | A | \cdot | B |$

Knowing these properties the determinants calculation can be faster. Bearing in mind the 4th property, we can go on modifying our determinant by means of linear combinations in such a way that we can get the largest number of possible $0$ or $1$ , which would reduce the calculations a lot.

Example

$| \begin{matrix} 1 & 3 & 3 & 6 \\ 1 & 3 & 6 & 7 \\ 2 & 4 & 0 & 3 \\ 1 & 5 & 2 & 3 \end{matrix} | \to \begin{matrix} f_{1} \to f_{1} \\ f_{2} \to f_{2} - f_{1} \\ f_{3} \to f_{3} - 2 f_{1} \\ f_{4} \to f_{4} - f_{1} \end{matrix} \to | \begin{matrix} 1 & 3 & 3 & 6 \\ 0 & 0 & 3 & 1 \\ 0 & - 2 & - 6 & - 9 \\ 0 & 2 & - 1 & - 3 \end{matrix} |$

And as the first column is zero, except for the first element, we will have to calculate the determinant $| \begin{matrix} 0 & 3 & 1 \\ - 2 & - 6 & - 9 \\ 2 & - 1 & - 3 \end{matrix} |$ because the other contributions would be zero.