Since A is an invertible matrix,
As matrix is of order 2, let
Then, and .

*** QuickLaTeX cannot compile formula:
\[\begin{array}{l}
A^{-1}=\frac{1}{|A|} a d j=\left[\begin{array}{ll}
\frac{d}{|A|} & \frac{-b}{|A|} \\
\frac{-c}{\mid A} & \frac{a}{|A|}
\end{array}\right] \\
{ \therefore | A ^ { - 1 } | = | \begin{array}{ll}
\frac{d}{|A|} &\left|A^{-1}\right|=\mid \begin{array} { l l } 
{ \frac { d } { | A | } } &{\frac{-b}{|A|}\left|=\frac{1}{|A|}\right| \begin{array}{cc}
d & -b \mid \\
-c & a \mid
\end{array} \mid}=\frac{1}{|A|^{2}}(a d-b c)=\frac{1}{|A|^{2}} \cdot|A|=\frac{1}{|A|}
\end{array}{ \therefore | A ^ { - 1 } | = | \begin{array} { l l } 
{ \frac { d } { | A | } } & { \frac { - b } { | A | } | = \frac { 1 } { | A | } | \begin{array} { c c } 
{ d } & { - b | } \\
{ - c } & { a | }
\end{array} | } = \frac { 1 } { | A | ^ { 2 } } ( a d - b c ) = \frac { 1 } { | A | ^ { 2 } } \cdot | A | = \frac { 1 } { | A | } } \\
\therefore \operatorname{det}\left(A^{-1}\right)=\frac{\mid}{\operatorname{det}(A)}
\end{array}\]

*** Error message:
Extra }, or forgotten $.
leading text: ... } } \cdot | A | = \frac { 1 } { | A | } }
Missing } inserted.
leading text: \end{array}\]
\begin{array} on input line 13 ended by \end{equation*}.
leading text: \end{array}\]
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leading text: \end{array}\]
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leading text: \end{array}\]
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leading text: \end{array}\]
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leading text: \end{array}\]
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leading text: \end{array}\]
\begin{array} on input line 8 ended by \end{document}.
leading text: \end{document}
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leading text: \end{document}
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leading text: \end{document}
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leading text: \end{document}
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leading text: \end{document}
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Hence, the correct answer is B.