How to compute the last diagonal element of a matrix using determinants? I actually want to verify the following statement. Please note that I am not even sure if it is correct. I tried out some numerical examples in R and it seems that the statement is correct and can someone show the proof about why this is correct.
Suppose $\Gamma_m$ is a $m \times m$ matrix, then the $(m,m)$ component of $\Gamma^{-1}_m$ is 
$$\frac{(det\; \Gamma_{m-1})}{(det\; \Gamma_{m})}$$

One way to find $\Gamma^{-1}$ is
$$\Gamma^{-1}=\frac{1}{\det{\Gamma}}\begin{bmatrix}\Gamma_{11}&...&\Gamma_{1m}\\
...&...&...\\\Gamma_{m1}&...&\Gamma_{mm}\end{bmatrix}$$
where $\Gamma_{ij}$ is the determinant of the $(m-1)\times(m-1)$ matrix obtained by removing the $i$th row and $j$th column from $\Gamma$, multiplied by $(-1)^{i+j}$. 
If you look at the last element of $\Gamma^{-1}$, it is exactly what you want, assuming your $\Gamma_{m-1}$ is the matrix obtained by removing the last row and last column from $\Gamma_m$. Also, they have to be nonsingular.
