A = \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\)
First, we need to find the characteristic polynomial of A.
\(\begin{bmatrix} a - \lambda & b & c \\ d & e - \lambda & f \\ g & h & i - \lambda \end{bmatrix}\)
(a - \(\lambda\))(e - \(\lambda\))(i - \(\lambda\)) + bfg + cdh - g(e - \(\lambda\))c - hf(a - \(\lambda\)) - db(i - \(\lambda\))
(ae - e\(\lambda\) - a\(\lambda\) + \(\lambda^2\))(i - \(\lambda\)) + bfg + cdh - gec + g\(\lambda\)c - hfa + hf\(\lambda\) - dbi + db\(\lambda\)
aei - ei\(\lambda\) - ai\(\lambda\) + i\(\lambda^2\) - ae\(\lambda\) + e\(\lambda^2\) + a\(\lambda^2\) - \(\lambda^3\) + bfg + cdh - gec + g\(\lambda\)c - hfa + hf\(\lambda\) - dbi + db\(\lambda\)
I will rearrange the above polynomial to make the constant the last term.
-\(\lambda^3\) + e\(\lambda^2\) + a\(\lambda^2\) + i\(\lambda^2\) - ei\(\lambda\) - ai\(\lambda\) - ae\(\lambda\) + g\(\lambda\)c + hf\(\lambda\) + db\(\lambda\) + aei + bfg + cdh - gec - hfa - dbi
As you can see, the constant term of the characteristic polynomial is aei + bfg + cdh - gec - hfa - dbi.
Now, we will find the determinant of A using the diagonal method.
det( \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\) )
aei + bfg + cdh - gec - hfa - dbi
Now to compare the two:
aei + bfg + cdh - gec - hfa - dbi = aei + bfg + cdh - gec - hfa - dbi
These are equivalent. The same would be true for other square matrices of different sizes.