(An exercise in the chapter: determinants) Let $$A = \left[ \begin{matrix} I_k & U \\ 0 & I_l \end{matrix} \right] $$ Find the inverse of this matrix
Since $A$ is upper triangular with unit diagonal, then $\det A = 1$. Moreover, $\det A_{ij} = 1$, since such a matrix will be upper triangular with unit diagonal too. This means that $\operatorname{adj} A = [(-1)^{i+j}] $. Hence $$A^{-1} = \operatorname{adj} A = [(-1)^{i+j}] $$
But this doesn't depend on $U$. This means two different matrices could have the same inverse. A contradiction.
What am I missing?
/edit: I have an idea but have trouble generalizing it. Thesis: $$A^{-1} = \pmatrix {I_k & -U \\ 0 & I_l}$$
In a concrete case, when the minor is
- in the left bottom corner, I'd calculate the minor and swap the rows to get an upper triangular matrix
In the general case, when the minor is
- in the upper right corner, the minor will be an upper triangular matrix with at least one zero at the diagonal, hence its determinant will be zero
- on the diagonal: it will be an upper triangular matrix with ones on the diagonal
- in the unit matrix part outside the diagonal - there will be at least one zero column, rendering the determinant 0
Can you help me generalize the concrete case?
You have calculated the $(i,j)$-th minor incorrectly when $(i,j)$ is an off-diagonal position (i.e. when $i\ne j$). For instance, consider the case where $A=\pmatrix{1&1\\ 0&1}$. The $(1,2)$-th minor is zero, not $(-1)^{1+2}$.