For A=[a,b\c,d]; A is of size 2*2
det A = |a,b\c,d| = ad-cb and A^-1 exists & det A != 0
What if A is 3*3?
A=[a,b,c\d,e,f\g,h,i]
det A = det([a,b,c\d,e,f\g,h,i])
We visualize this as:
\begin{align*} \begin{bmatrix} a&b&c&|&a&b\ d&e&f&|&d&e\ g&h&i&|&g&h \end{bmatrix} \end{align*}
The right half isn’t the augmented matrix, but reptition of the other side.
\begin{align*} aei+bfg+cdh\ -ceg-afh-bdi \end{align*}
So addition goes down and right in a diagonal. Subtraction goes down and left.
Factor that out: \begin{align*} a(ei-fh) - b(di-fg)+ c(dh-eg)\ = 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} \end{align*}
Hence to compute the determinate of a 33 matrix multiply each entry in row 1 by a sign times the determinant of the 22 matrix obtained by deleting the row and column of that entry and add the result.
example
\begin{align*} det( \begin{bmatrix} 2&3&7\ -4&0&6\ 1&5&0\ \end{bmatrix} )\ =& 2[-30] - 3[-6] + 7[-20]\ =& -60 + 18 - 140\ =& -182\ \end{align*}
Cofactor of a matrix
Assume that determinates of (n-1)(n-1) matrices have been defined. Given the nn matrix A, let denote the (n-1)*(n-1) matrix obtained from A by deleting row i and column j. Then, (ij) cofactor $C_{ij}(A) is the scalar defined by