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(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?

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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 $(-1)^{1+2}$.

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  • $\begingroup$ But if the matrix is bigger than 2x2, then there is no non-zero minor, is it? $\endgroup$ – marmistrz Feb 20 '16 at 13:05
  • $\begingroup$ @marmistrz Huh? Why not? consider $A=I_4$ ($k=l=2,\ U=0$). The $(1,4)$ minor is zero. $\endgroup$ – user1551 Feb 20 '16 at 13:08
  • $\begingroup$ Can you give me a hint please? I have trouble solving this exercise. $\endgroup$ – marmistrz Feb 20 '16 at 20:49
  • $\begingroup$ @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. $\endgroup$ – user1551 Feb 21 '16 at 5:48
  • $\begingroup$ I have an idea, but have trouble generalizing. Can you take a look? $\endgroup$ – marmistrz Feb 23 '16 at 12:44

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