Answer
A is invertible
Work Step by Step
If we don't want many calculations, we test a criterion from Th.8:
$\mathrm{e}.\quad $ The columns of $A$ form a linearly independent set.
(then: $\mathrm{a}. \quad A$ is an invertible matrix.)
$7= 5\displaystyle \times\frac{7}{5}$ , but $-6\displaystyle \neq-3\times\frac{7}{5}$,
so the columns are not multiples, meaning they are linearly independent.
A is invertible
(Another way for 2$\times$2 matrices is to test $ad-bc$)
$5(-6)-(7)(-3)\neq 0$ , A is invertible