${(aA)^{ - 1}} = \frac{1}{a}{A^{ - 1}}$, where $a$ is any real number and $A$ is a square matrix.

Correct Answer False

Since, we know that, if $A$ is a non-singular square matrix, then for any scalar a (non-zero), aA is invertible such that

$(aA)\left( {\frac{1}{a}{A^{ - 1}}} \right) = \left( {a \cdot \frac{1}{a}} \right)\left( {A \cdot {A^{ - 1}}} \right)$

$= I$
i.e., $(aA)$ is inverse of $\left( {\frac{1}{a}{A^{ - 1}}} \right)$ or ${(aA)^{ - 1}} = \frac{1}{a}{A^{ - 1}}$,

where $a$ is any non-zero scalar.
In the above statement $a$ is any real number.

So, we can conclude that above statement is false.