MathewsMD said:
For a 2x2 matrix, does the general definition hold?
If so, how exactly is the minor ## M_{i,j} ## computed in this case? If A is a 2x2 matrix, is det(A) only defined as ad - bc?
No, that definition does not hold for
any matrix because the determinant is a number where, as you have it, it is a function of "i".
What you should have is a specific number in place of "i". Typically it is "1" but any number from 1 to n would do.
In the case of a 2 by 2 matrix, n= 2 so, with your "i" replaced by "1", \sum_{j= 1}^N (-1)^{1+ j}a_{1j}M_{1j} is (-1)^{1+ 1}a_{11}a_{22}+ (-1)^{1+ 2}a_{12}a_{21}= a_{11}a_{22}- a_{12}a_{21}. In the case that
\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}= \begin{bmatrix}a & b \\ c & d \end{bmatrix}
that would be ad- bc.
Notice that if we replaced "1" above by "2" we would have \sum_{j= 1}^N (-1)^{2+ j}a_{2j}M_{2j} is (-1)^{2+ 1}a_{21}a_{12}+ (-1)^{2+ 2}a_{22}a_{11}= -a_{12}a_{21}+ a_{11}a_{22} the same as before.
The "minor", M_{ij} is, by definition, the determinant of the matrix you have after removing the "ith row" and "jth column". In the case that
A= \begin{bmatrix} a & b \\ c & d\end{bmatrix}
a_{11}= a and M_{11}= a_{22}= d since we remove the first row, \begin{bmatrix}a & b\end{bmatrix} and first column, \begin{bmatrix}a \\ c \end{bmatrix} we are left with the "1 by 1 matrix", d.
Similarly, a_{12}= b and M_{12}= a_{21}= c.
