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

Answer

Correct option: A.
$\frac{1}{{ab - cd}}\left[ {\begin{array}{*{20}{c}}b&{ - c}\\{ - d}&a\end{array}} \right]$
a
(a) ${A^{ - 1}} = \frac{{adj\,A}}{{|A|}}$
But $|A| = \left| {\,\begin{array}{*{20}{c}}a&c\\d&b\end{array}\,} \right| = ab - cd$ and $adj\,A = \left[ {\begin{array}{*{20}{c}}b&{ - c}\\{ - d}&a\end{array}} \right]$
therefore ${A^{ - 1}} = \frac{1}{{ab - cd}}\left[ {\begin{array}{*{20}{c}}b&{ - c}\\{ - d}&a\end{array}} \right]$.

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