Finding the inverse of a Matrix using determinants Question 1
Lets consider a $3 \times 3$ matrix, A.
The inverse of this matrix can be found by finding the adjoint of the matrix and dividing it by its determinant. Can someone please explain why this process work and how it can be visualized?
I know that the determinant can be thought of as the volume spanned by the column vectors and that the adjoint of matrix A is $A^T$'s cofactors.
Question 2
Let's consider a $3\times 3$ matrix, A.
The cofactors of this matrix can be found by the cross product of each of its column vectors. So what does the cofactor matrix tell us?
Thank you.
 A: There is some explanation in the Geometric Interpretation here, but I don't really like it too much.
I think it's easier to think this way: The cross product of $v$ and $w$ is a vector $x$ orthogonal to the two given ones, whose length is twice the area of the triangle. Hence the volume of the tetrahedron on $u, v, w$ is one sixth of the the dot product of $x$ with the remaining vector $u$ (one should be careful with signs, but it works out). The dot product of $x$ with $v$ or $w$ is zero. Similar things are true about the cross product of $u$ and $w$ dotted with $v$ (and $u$ and $w$) and the cross product of $u$ and $v$ dotted with $w$ (and $u$ and $v$).  What all of this implies is that the matrix of cross product vectors (transposed) multiplied with the matrix of $u$, $v$, $w$ is diagonal, with all diagonal entries equal to six times the volume of the tetrahedron, i.e. the determinant of the $u$, $v$, $w$ matrix. The result follows.
Higher dimensional versions are fairly similar, with volumes of $n$ and $n-1$ simplices.
