${\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}}$ $\,=\,$ $e_{11} \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$ $\,-\,$ $e_{12} \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{13} \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{23} \\ \end{vmatrix}}$

According to the definition of the determinant of a matrix, a formula for the determinant of a 3 by 3 matrix can be derived in algebraic form by following four fundamental steps. The following mathematical expression represents the determinant of a square matrix of the order $3$ in algebraic form.

${\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}}$

Now, let’s learn how to derive the determinant formula for the matrix of the order $3 \times 3$ by following the four steps.

$\left|\begin{array}{c | c c} \color{red} e_{11} & e_{12} & e_{13} \\ \hline e_{21} & \color{blue} e_{22} & \color{blue} e_{23} \\ e_{31} & \color{blue} e_{32} & \color{blue} e_{33} \\ \end{array}\right|$

- The entry $e_{11}$ is selected from the first row.
- It is an element in the first row and first column. So, evaluate the $-1$ raised to the power of the sum of “the number of the row” and “the number of the column” for the element $e_{11}$. It is equal to $(-1)^{1+1}$.
- Evaluate the determinant of the elements by leaving the entries in the row and the column of the element $e_{11}$.
- Calculate the product by multiplying the entry $e_{11}$ with $(-1)^{1+1}$ and the determinant of the square matrix of order $2$.

$\implies$ $\left|\begin{array}{c | c c} \color{red} e_{11} & e_{12} & e_{13} \\ \hline e_{21} & \color{blue} e_{22} & \color{blue} e_{23} \\ e_{31} & \color{blue} e_{32} & \color{blue} e_{33} \\ \end{array}\right|$ $\,=\,$ $e_{11} \times (-1)^{1+1} \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$

$\left|\begin{array}{c|c|c} e_{11} & \color{red} e_{12} & e_{13} \\ \hline \color{blue} e_{21} & e_{22} & \color{blue} e_{23} \\ \color{blue} e_{31} & e_{32} & \color{blue} e_{33} \\ \end{array}\right|$

- The element $e_{12}$ is chosen from the first row of the matrix.
- It is an entry in the first row and second column. Hence, find the $-1$ raised to the power of the sum of “the number of the row” and “the number of the column” for the chosen entry $e_{12}$. It means $(-1)^{1+2}$.
- Calculate the determinant of a matrix of the order $2$ by leaving the elements in the row and column of the selected entry $e_{12}$.
- Find the product by multiplying the entry $e_{12}$ with $(-1)^{1+2}$ and the determinant of $2 \times 2$ square matrix.

$\implies$ $\left|\begin{array}{c|c|c} e_{11} & \color{red} e_{12} & e_{13} \\ \hline \color{blue} e_{21} & e_{22} & \color{blue} e_{23} \\ \color{blue} e_{31} & e_{32} & \color{blue} e_{33} \\ \end{array}\right|$ $\,=\,$ $e_{12} \times (-1)^{1+2} \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$

$\left|\begin{array}{cc|c} e_{11} & e_{12} & \color{red} e_{13} \\ \hline \color{blue} e_{21} & \color{blue} e_{22} & e_{23} \\ \color{blue} e_{31} & \color{blue} e_{32} & e_{33} \\ \end{array}\right|$

- Select the remaining element in the first row and it is $e_{13}$.
- This entry is an element in the first row and third column of the matrix. Therefore, calculate the $-1$ raised to the power of the sum of “the number of the row” and “the number of the column” for the chosen entry $e_{13}$, and it is written as $(-1)^{1+3}$.
- Find the determinant of a matrix of the order $2 \times 2$ by leaving the elements in the first row and third column.
- Evaluate the product by multiplying the entry $e_{13}$ with $(-1)^{1+3}$ and the determinant of the square matrix of order $2$.

$\implies$ $\left|\begin{array}{cc|c} e_{11} & e_{12} & \color{red} e_{13} \\ \hline \color{blue} e_{21} & \color{blue} e_{22} & e_{23} \\ \color{blue} e_{31} & \color{blue} e_{32} & e_{33} \\ \end{array}\right|$ $\,=\,$ $e_{13} \times (-1)^{1+3} \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{23} \\ \end{vmatrix}}$

Finally, add the products for calculating the determinant of a square matrix of the order $3$.

${\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}}$ $\,=\,$ $e_{11} \times (-1)^{1+1} \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{12} \times (-1)^{1+2} \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{13} \times (-1)^{1+3} \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{32} \\ \end{vmatrix}}$

Now, let’s evaluate the product in each term in the right hand side expression.

$=\,\,\,$ $e_{11} \times (-1)^{2} \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{12} \times (-1)^{3} \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{13} \times (-1)^{4} \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{32} \\ \end{vmatrix}}$

$=\,\,\,$ $e_{11} \times 1 \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{12} \times (-1) \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{13} \times 1 \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{32} \\ \end{vmatrix}}$

$\therefore\,\,\,$ ${\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}}$ $\,=\,$ $e_{11} \times {\begin{vmatrix} e_{22} & e_{23} \\ e_{32} & e_{33} \\ \end{vmatrix}}$ $\,-\,$ $e_{12} \times {\begin{vmatrix} e_{21} & e_{23} \\ e_{31} & e_{33} \\ \end{vmatrix}}$ $\,+\,$ $e_{13} \times {\begin{vmatrix} e_{21} & e_{22} \\ e_{31} & e_{32} \\ \end{vmatrix}}$

It can used as a formula for calculating the determinant of a third order square matrix. It can also be simplified further by the determinant of a second order matrix.

$=\,\,\,$ $e_{11} \times \Big(e_{22} \times e_{33}-e_{23} \times e_{32}\Big)$ $\,-\,$ $e_{12} \times \Big(e_{21} \times e_{33}-e_{23} \times e_{31}\Big)$ $\,+\,$ $e_{13} \times \Big(e_{21} \times e_{32}-e_{22} \times e_{31}\Big)$

$=\,\,\,$ $e_{11} \times \Big(e_{22}e_{33}-e_{23}e_{32}\Big)$ $\,-\,$ $e_{12} \times \Big(e_{21}e_{33}-e_{23}e_{31}\Big)$ $\,+\,$ $e_{13} \times \Big(e_{21}e_{32}-e_{22}e_{31}\Big)$

Each multiplying factor can be distributed to the difference of the terms in the expression by the distributive property of multiplication over subtraction.

$=\,\,\,$ $e_{11} \times e_{22}e_{33}$ $\,-\,$ $e_{11} \times e_{23}e_{32}$ $\,-\,$ $e_{12} \times e_{21}e_{33}$ $\,+\,$ $e_{12} \times e_{23}e_{31}$ $\,+\,$ $e_{13} \times e_{21}e_{32}$ $\,-\,$ $e_{13} \times e_{22}e_{31}$

$=\,\,\,$ $e_{11}e_{22}e_{33}$ $\,-\,$ $e_{11}e_{23}e_{32}$ $\,-\,$ $e_{12}e_{21}e_{33}$ $\,+\,$ $e_{12}e_{23}e_{31}$ $\,+\,$ $e_{13}e_{21}e_{32}$ $\,-\,$ $e_{13}e_{22}e_{31}$

$\therefore\,\,\,$ ${\begin{vmatrix} e_{11} & e_{12} & e_{13} \\ e_{21} & e_{22} & e_{23} \\ e_{31} & e_{32} & e_{33} \\ \end{vmatrix}}$ $\,=\,$ $e_{11}e_{22}e_{33}$ $\,+\,$ $e_{12}e_{23}e_{31}$ $\,+\,$ $e_{13}e_{21}e_{32}$ $\,-\,$ $e_{11}e_{23}e_{32}$ $\,-\,$ $e_{12}e_{21}e_{33}$ $\,-\,$ $e_{13}e_{22}e_{31}$

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