What is the rank of a $5 \times 5$ matrix $A$ if $A^{-1}$ exists?

Question

Is it possible to determine with little or no calculation the rank of a $5 \times 5$ or any square matrix if it has an inverse matrix?

I found the requirements for an inverse:

• The matrix must be square (same number of rows and columns).
• The determinant of the matrix must not be zero. This is instead of the real number not being zero to have an inverse, the determinant must not be zero to have an inverse.

I don't see any word on the rank here.

Any help is appreciated.

Answer 1

Hint: $A$ doesn't have full rank $\implies$ $A^{-1}$ doesn't exist.

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Now, what is the contrapositive of the above statement?

Answer 2

A lot of these properties are intimately connected in strong equivalences and it's useful to have an overview of the most important ones. Here are some for a square $n \times n$-matrix $A$:

• $A$ is invertible
• $\det A \ne 0$
• $\mbox{rank}\, A = n$
• the linear map $X \mapsto AX$ is a bijection
• the columns of $A$ are linearly independent
• the rows of $A$ are linearly independent
• (...)

So for your specific question: $A$ is invertible if and only if $A$ has full rank.