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Let $ A$ be $n \times n$ matrix all of whose main diagonal entries are 0 and elsewhere $1,$ Find $A^{-1}$•

For such matrix we can write $A = B - I$ Where $B$ is a matrix with all its entries $1$ and $I$ is identity matrix. Then nullity$( A + I) = n-1$ So algebraic multiplicity corresponding to $-1$ is $≥ n-1$ Since the row sum for $A$ is $n-1$, so this must be an eigen value.

Characteristic equation I got is $ (x+1)^{n-1} (x-(n-1))$ now how to get $A^{-1}$ from here$?$

Mathaddict
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    Use the Cayley-Hamilton theorem and rearrange. – Joppy Sep 13 '19 at 09:03
  • Can you please tell how to arrange this polynomial to get $A^{-1}$ I can't split the polynomial to get $I$ at one side of equation. – Mathaddict Sep 13 '19 at 09:11
  • Martin Sleziak and Nick Alger have nice answers at the duplicate, with a formula for $A^{-1}$. – Dietrich Burde Sep 13 '19 at 09:15
  • @DietrichBurde https://math.stackexchange.com/questions/780160/inverse-of-this-3-by-3-matrix-using-the-cayley-hamilton-theorem is closer to being an actual duplicate, as is https://math.stackexchange.com/questions/897529/use-the-cayley-hamilton-theorem-to-find-the-inverse-of-this-3-by-3-matrix – Arthur Sep 13 '19 at 09:22
  • @Arthur Yes, they have Cayley-Hamilton. But the above duplicate has the very same $n\times n$-matrix. May be, both duplicates together are sufficient. – Dietrich Burde Sep 13 '19 at 09:30
  • @DietrichBurde I missed the description of the matrix in this question. You are right. – Arthur Sep 13 '19 at 09:47

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