# Finding an inverse of a matrix with determinants

(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 $(-1)^{1+2}$.
• @marmistrz Huh? Why not? consider $A=I_4$ ($k=l=2,\ U=0$). The $(1,4)$ minor is zero. – user1551 Feb 20 '16 at 13:08
• @marmistrz Consider the scalar case first. Suppose $A=\pmatrix{1&u\\ 0&1}$. Can you find its inverse? Now, for a block-triangular $A$, try an inverse of a similar structure. – user1551 Feb 21 '16 at 5:48