$If A= If A= \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}a\\d\end{array}}&{\begin{array}{*{20}{c}}c\\b\end{array}}\end{array}} \right], then {A^{ - 1}}=$
A. $\dfrac{1}{{ab - cd}}\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}b\\{ - d}\end{array}}&{\begin{array}{*{20}{c}}{ - c}\\a\end{array}}\end{array}} \right]$
B. $\dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}b\\{ - d}\end{array}}&{\begin{array}{*{20}{c}}{ - c}\\a\end{array}}\end{array}} \right]$
C. $\dfrac{1}{{ab - cd}}\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}b\\c\end{array}}&{\begin{array}{*{20}{c}}d\\a\end{array}}\end{array}} \right]$
D. None of these

Hint: We have to find ${A^{ - 1}}$, so for that firstly we will solve $adjA$ and the formula of this is mentioned below and then |A|which has also the formula below. And then dividing both of them will get our required solution. As for finding ${A^{ - 1}}$ the formula suggests ${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$.

Formula Used: ${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$
If $A= \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]$
Then $adjA= \left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{{a_{22}}}\\{ - {a_{21}}}\end{array}}&{\begin{array}{*{20}{c}}{ - {a_{12}}}\\{{a_{11}}}\end{array}}\end{array}} \right]$. As for 2×2 matrix we have a shortcut for $adjA$ which is we have to interchange element ${a_{11}}$ to element ${a_{22}}$
And change signs of element ${a_{12}}$ and ${a_{21}}$
And if $A= \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]$
So |A| is cross multiplication of on diagonal elements and off diagonal elements
On diagonal elements are ${a_{{{11}_{}}}}and {a_{22}}$ and off diagonal elements are ${a_{12}}and {a_{21}}$
Then $|A|={a_{11}}{a_{22}} - {a_{12}}{a_{21}}$


Complete step by step solution: we are given $A$ = $\left[ {\begin{array}{*{20}{c}}a&c\\d&b\end{array}} \right]$ and we have to find ${A^{ - 1}}$
so first of all as the formula suggests we have to find $adjA$
so $adjA$ of A=$\left[ {\begin{array}{*{20}{c}}a&c\\d&b\end{array}} \right]$ is
$adjA$ = $\left[ {\begin{array}{*{20}{c}}b&{ - c}\\{ - d}&a\end{array}} \right]$. As we have to interchange element ${a_{11}}$ to element $ {a_{22}}$
And change signs of element ${a_{12}}$ and ${a_{21}}$
Now $|A|$ of matrix $\left[ {\begin{array}{*{20}{c}}a&c\\d&b\end{array}} \right]$ is
$|A| = ab - cd$
Now as ${A^{ - 1}} = \dfrac{{adjA}}{{|A|}}$
 $\Rightarrow {A^{ - 1}} = \dfrac{{\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}b\\{ - d}\end{array}}&{\begin{array}{*{20}{c}}{ - c}\\a\end{array}}\end{array}} \right]}}{{ab - cd}} = \dfrac{1}{{ab - cd}}\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}b\\{ - d}\end{array}}&{\begin{array}{*{20}{c}}{ - c}\\a\end{array}}\end{array}} \right]$

Option ‘A’ is correct

Note: ${A^{ - 1}}$ exists only when $|A| \ne 0$. As in this question if ab - cd=0 then {A^{ - 1}}will not exist. We have to remember the formula for ${A^{ - 1}}$and for that we have to remember $adjA$ and $|A|$ for a 2×2 matrix. Sometimes students make mistakes while solving $adjA$ and $|A|$ as they don’t interchange elements in the formula of $adjA$ and in $|A|$ also we have to cross multiply very carefully and take care of signs as well.