Do the determinant of a matrix equal to the determinant of submatrices?
$$
det\begin{pmatrix}
a_{11} & a_{12} & a_{13} & \dots & a_{1k} \\
a_{21} & a_{22} & a_{23} & \dots & a_{2k} \\
a_{31} & a_{32} & a_{33} & \dots & a_{3k} \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & \dots & a_{nk} \\
\end{pmatrix}
$$
equals
$$
det\begin{pmatrix}
det\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix} & det\begin{pmatrix}a_{13} & a_{14}\\ a_{23} & a_{24}\end{pmatrix} & \dots\\
\ddots & \vdots \\
det\begin{pmatrix}a_{(n-1)1} & a_{(n-1)2}\\ a_{(n-2)1} & a_{(n-2)2}\end{pmatrix} & det\begin{pmatrix}a_{(n-1)3} & a_{(n-1)4}\\ a_{(n-2)3} & a_{(n-2)4}\end{pmatrix} & \dots\\
\end{pmatrix}
$$
And of course I would like to know if it can be generalized from 2×2 to mxm division.
Best Answer
As others point out, your attempted formula is not correct. However, there is a rather amazing formula that is true for the determinant of an $n \times n$ matrix $M$. Let $M^{NW}$ denote the $(n-1) \times (n-1)$ submatrix of $M$ obtained by using the upperleft most entries (i.e. crossing out the bottom row and last column). The NW stands for northwest. Similarly define $M^{NE}$, $M^{SW}$, $M^{SE}$. Finally, denote $M^C$ the $(n-2) \times (n-2)$ submatrix obtained by taking the center submatrix (i.e. cross out the top and bottom rows, and the first and last columns). Then $$ \det(M) = \frac{\det(M^{NW}) \det(M^{SE}) - \det(M^{NE}) \det(M^{SW})}{\det(M^C)}. $$