MCQ
If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$ $(i = \sqrt { - 1} ),$then ${A^{ - 1}}$=
  • A
    $\left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$
  • $\left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&{ - 2i}\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}i&0\\0&{2i}\end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}0&i\\{2i}&0\end{array}} \right]$

Answer

Correct option: B.
$\left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&{ - 2i}\end{array}} \right]$
b
(b) For $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$,

$adj\,(A) = \left[ {\begin{array}{*{20}{c}}{i/2}&0\\0&i\end{array}} \right]$ and $|A| = - \frac{1}{2}$.

$\therefore $ ${A^{ - 1}} = \frac{1}{\Delta }(adj\,A) = \frac{1}{{ - 1/2}}\,\left[ {\begin{array}{*{20}{c}}{i/2}&0\\0&i\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&{ - 2i}\end{array}} \right]$.

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