Inverse of a matrix having zeroes in diagonal and one elsewhere

Question

Could any one help me to find inverse of such matrix? I observed that $A= J-I$, where J is a matrix having all entries 1. Thanks for helping.

Answer 1

Note that the rank of $J$ is 1 because its range is spanned by $(1, \cdots, 1)$. Moreover $(1, \cdots, 1)$ is an eigenvector with eigenvalue $n$. Thus eigenvalues of $J$ are $n$ and $0$ (with multiplicity $n-1$). So the eigenvalues of $A$ are $n-1$ and $-1$ (with multiplicity $n-1$). The minimal polynomial of $A$ is then \begin{align*} (A-(n-1)I)(A+I)&=0\\ A^2-(n-2)A-(n-1)I&=0\\ A(A-(n-2)I)&=(n-1)I \\ A^{-1} &=\frac{1}{n-1}(A-(n-2)I)\end{align*}

Answer 2

You can use Gauss Jordon elimination. Take a representation of the following form: [A|I] Now using elementary linear row transformations, try convert A to I(Identity matrix) and simultaneously apply the same transformations to I in each step. The moment you convert A to I , the I in [A|I] will be converted to inverse of A. Thus, [A|I] transforms to [I|inv(A)]

Answer 3

If your $A$ is an $n\times n$ matrix, for $n\geq2$, the inverse is $$ \frac{1}{n-1}(A-(n-2)I) $$