Question

Prove that rank of a nonsingural matrix is equal to rank of its reciprocal matrix

Answer 1

Answer:

A square matrix $A$ of order $n$ is said to be nonsingular if and only if there is a matrix $B =: A^{-1}$ such that $AA^{-1} = A^{-1}A = I_{n}$. The matrix

$A^{-1}$ is said to be the inverse of $A$.

From the inverse formula

$A^{-1} = (1/det A)adj A$.

We see that, if $det A = 0$ then $A$ has no inverse (singular matrix). And by definition of rank of matrix:

Definition. A non-zero matrix $A$ is said to have rank $r$ if at least one of its $r$-square minors is different from zero while every $(r + 1)$-square minor, if any, is zero. A zero matrix is said to have rank $0$.

Therefore A is nonsingular if and only if $rank A = n$. Equivalently, $A$ has no inverse if and only if $rank A$ less than $n$.