If A = [ * 20 c i&0 0& i/2 ] (i = - 1 ),then A^ - 1 = - Vidyadip

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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