Question

prove that rank of a non singular matrix is equal to the rank of its reciprocal.​

Answer 1

Answer:

I answered true but the professor said it is false, he said that not all matrices have an inverse, but, I tough that since the statement says "and its inverse" it had an inverse since the statement say it; so my question would be, is the statement fine by being false or is it an error in the thinking of the solution of my professor, thanks

Step-by-step explanation:

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Answer 2

Answer:

Proved

Step-by-step explanation:

To Prove:- Rank of a non singular matrix is equal to the rank of its reciprocal.​

Proof:-

A square matrix A of order n is said to be non-singular 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

                                $A^{-1} = \frac{1}{|A|} (adj A)$,

where if |A| = 0 then A has no inverse. Then a is known as Singular matrix.

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 has rank 0.

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

Hence, Proved.

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