<s>[INST] finding determinants using different properties The equation is as follows:
$\operatorname{det}(2A^{-1} + 7\operatorname{adj}(A))$
Here I know that $\operatorname{det}(A^{-1}) = (\operatorname{det}(A))^{-1}$ and $\operatorname{det}(kA) = k^n \operatorname{det}(A)$
using these, we know that $\operatorname{det}((2A)^{-1} )= \frac{1}{4} \operatorname{det}(A)^{-1} =\frac{1}{8}$ 
I do not know a formula that relates to adjoints in this case besides maybe: 
$$A \operatorname{adj}(A) = \operatorname{det}(A)I$$
 [/INST] Look at this:
since
$A\operatorname{adj}(A) = \det(A)I, \tag{1}$
we have
$\operatorname{adj}(A) = \det(A) A^{-1} \tag{2}$
provided $A^{-1}$ exists.  Well, since we are given that (see comments above)
$\det(A) = 2, \tag{3}$
we know $A^{-1}$ does indeed exist, so using (2) we have
$2A^{-1} + 7\operatorname{adj}(A) = 2A^{-1} + 7\det(A)A^{-1} = (2 + 7\det(A))A^{-1}$
$= (2 + 7(2))A^{-1} = 16A^{-1}, \tag{4}$
whence
$\det(2A^{-1} + 7\operatorname{adj}(A)) = \det(16A^{-1}) = (16)^n (\det(A))^{-1},\tag{5}$
where $n = \operatorname{size}(A)$.  Now since
$\operatorname{det}((2A)^{-1} )= \frac{1}{4} \operatorname{det}(A)^{-1} =\frac{1}{8}, \tag{6}$ 
with
$\det(2A) = 2^n \det(A) \tag{7}$
and
$\det((2A)^{-1}) = (\det(2A))^{-1}, \tag{8}$
we see that
$2^{-n} = \dfrac{1}{4}, \tag{9}$
which shows that $n = 2$; thus (5) yields
$\det(2A^{-1} + 7\operatorname{adj}(A)) = \det(16A^{-1})$
$= (16)^2(\det(A))^{-1} = 256 (\det(A))^{-1} = \dfrac{256}{2} = 128.\tag{10}$
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