The sum of the products of first row entries (or elements) with negative one raised to the power of the sum of “the number of the row” and “the number of the column” for the chosen first row element, and the determinant of second order square matrix by leaving the elements in the row and the column of the selected first row element is called the determinant of a third order matrix.
The matrix $M$ is a square matrix of order $3$. In this $3 \times 3$ matrix, the nine entries (or elements) are arranged in three rows and three columns as follows.
$M$ $\,=\,$ ${\begin{bmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{bmatrix}}$
The determinant of a square matrix $M$ is written as $\operatorname{det}(M)$ or $|M|$ in mathematics. So, it can be represented mathematically in any of the following two forms.
$(1).\,\,\,$ $det(M)$ $\,=\,$ $\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}$
$(2).\,\,\,$ $|M|$ $\,=\,$ $\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}$
It is essential to know how to find the determinant of a 2 by 2 matrix for finding the determinant of a third order matrix.
There are four fundamental steps involved in finding the determinant of $3 \times 3$ square matrix.
In a third order matrix, there are three elements in the first row. So, we have to use the above four steps three times to find the products. After that, add the products to find the determinant of the matrix of the order $3 \times 3$.
${\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}}$ $\,=\,$ $e_{11} \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$ $\,-\,$ $e_{12} \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{13} \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{23} \\ \end{vmatrix}}$
It can be used as a formula for finding the determinant of any square matrix of the order $3 \times 3$. Now, let’s learn how to derive the determinant of a $3$ by $3$ matrix formula in mathematics.
Find ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$
Let’s use the above steps for calculating the determinant of any square matrix of the order $3 \times 3$.
$\implies$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $1 \times \begin{vmatrix} 5 & 6 \\ 8 & 9 \\ \end{vmatrix}$ $-$ $2 \times \begin{vmatrix} 4 & 6 \\ 7 & 9 \\ \end{vmatrix}$ $+$ $3 \times \begin{vmatrix} 4 & 5 \\ 7 & 8 \\ \end{vmatrix}$
$\implies$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $1 \times (5 \times 9-6 \times 8)$ $-$ $2 \times (4 \times 9-6 \times 7)$ $+$ $3 \times (4 \times 8-5 \times 7)$
$\implies$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $1 \times (45-48)$ $-$ $2 \times (36-42)$ $+$ $3 \times (32-35)$
$\implies$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $1 \times (-3)$ $-$ $2 \times (-6)$ $+$ $3 \times (-3)$
$\implies$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $-3+12-9$
$\implies$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $12-12$
$\,\,\,\therefore\,\,\,\,\,\,$ ${\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{vmatrix}}$ $\,=\,$ $0$
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