Let $$ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} $$ a 3x3 matrix, we want to calculate the determinant, that is to say det(A), we don't need too much operations to do that. We are going to develop on the first line, we can develop on any line or column, but this way works well, in the future you will choose the most appropriate line or column according to the number of 0.
we have : $$ det(A) =a\begin{vmatrix} e & f\\ h & i \end{vmatrix} - b\begin{vmatrix} d & f\\ g & i \end{vmatrix} +c\begin{vmatrix} d & e\\ g & h \end{vmatrix}$$
we have also : $$ det(A) =-d\begin{vmatrix} b & c\\ h & i \end{vmatrix} + e\begin{vmatrix} a & c\\ g & i \end{vmatrix} -f\begin{vmatrix} d & e\\ g & h \end{vmatrix}$$
we have also : $$ det(A) =g\begin{vmatrix} b & c\\ e & f \end{vmatrix} - h\begin{vmatrix} a & c\\ d & f \end{vmatrix} +i\begin{vmatrix} a & b\\ d & e \end{vmatrix}$$
But we can also calculate by developping on the columns :
we have also : $$ det(A) =a\begin{vmatrix} e & f\\ h & i \end{vmatrix} -d\begin{vmatrix} b & c\\ h & i \end{vmatrix} +g\begin{vmatrix} b & c\\ e & f \end{vmatrix}$$
we have also : $$ det(A) =-b\begin{vmatrix} d & f\\ g & i \end{vmatrix} +e\begin{vmatrix} a & c\\ g & i \end{vmatrix} -h\begin{vmatrix} a & c\\ d & f \end{vmatrix}$$
we have also : $$ det(A) =c\begin{vmatrix} d & e\\ g & h \end{vmatrix} -f\begin{vmatrix} a & b\\ g & h \end{vmatrix} +i\begin{vmatrix} a & b\\ d & e \end{vmatrix}$$
Let $$ A = \begin{pmatrix} 1 & 1 & 3 \\ 0 & 2 & 0 \\ 1 & 1 & 7 \\ \end{pmatrix} $$ a 3x3 matrix, the very first thing we notice is that the second line is the best candidate to develop into, so let's develop into the second line, following our formulas above we have : $$ det(A) =-0\begin{vmatrix} 1 & 3\\ 1 & 7 \end{vmatrix} + 2\begin{vmatrix} 1 & 3\\ 1 & 7 \end{vmatrix} -0\begin{vmatrix} 1 & 1\\ 1 & 1 \end{vmatrix}\\ = 2(7-3) \\ = 2*4 = 8 $$
When you develop into a line or a column, you might not know exactly the sign to use for the coefficient, this matrix, that we call H, the sign matrix is defined here and will help you to retrieve the sign of each coefficient : $$ H = \begin{pmatrix} + - +\\ - + -\\ + - + \end{pmatrix} $$
So we have everything with us now, we are able to calculate the determinant of any 3x3 matrix given its coefficients, if we want to go into more details we could go for triangular and diagonal matrices, if you begin in linear algebra these matrices will be obviously your favourite ones when it comes to calculate the determinant, it is nothing else than the product of each term in the diagonal, but we have now the general method,don't hesitate to contact me if you see any thing that you didn't understand, thanks...