Since A is an invertible matrix,
As matrix
Then,
*** 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}\] Missing $ inserted. leading text: \end{array}\] Missing } inserted. leading text: \end{array}\] Missing \cr inserted. leading text: \end{array}\] Missing $ inserted. leading text: \end{array}\] Missing } inserted. leading text: \end{array}\] \begin{array} on input line 8 ended by \end{document}. leading text: \end{document} Improper \prevdepth. leading text: \end{document} Missing $ inserted. leading text: \end{document} Missing } inserted. leading text: \end{document} Missing } inserted. leading text: \end{document} Missing \cr inserted.
Hence, the correct answer is B.