For a given matrix P = $\left[ {\begin{array}{*{20}{c}} {4+ 3i}&-i\ { i}&{4 - 3i} \end{array}} \right]$, where $i = \sqrt { - 1}$, the inverse of matrix P is

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For a given matrix P = $\left[ {\begin{array}{*{20}{c}} {4+ 3i}&-i\ { i}&{4 - 3i} \end{array}} \right]$, where $i = \sqrt { - 1}$, the inverse of matrix P is

TuteeHUB · Accepted Answer

$\frac{1}{{25}}\left[ {\begin{array}{*{20}{c}} i&{4 - 3i}\ {4 + 3i}&{ - i} \end{array}} \right]$

TuteeHUB · Answer

$\frac{1}{{24}}\left[ {\begin{array}{*{20}{c}} {4 - 3i}&i\ { - i}&{4 + 3i} \end{array}} \right]$

TuteeHUB · Answer

$\frac{1}{{24}}\left[ {\begin{array}{*{20}{c}} {4 + 3i}&{ - i}\ i&{4 - 3i} \end{array}} \right]$

TuteeHUB · Answer

$\frac{1}{{25}}\left[ {\begin{array}{*{20}{c}} {4 + 3i}&{ - i}\ i&{4 - 3i} \end{array}} \right]$

1.	For a given matrix P = \(\left[ {\begin{array}{*{20}{c}} {4+ 3i}&-i\\ { i}&{4 - 3i} \end{array}} \right]\), where \(i = \sqrt { - 1}\), the inverse of matrix P is
A.	\(\frac{1}{{24}}\left[ {\begin{array}{*{20}{c}} {4 - 3i}&i\\ { - i}&{4 + 3i} \end{array}} \right]\)
B.	\(\frac{1}{{25}}\left[ {\begin{array}{*{20}{c}} i&{4 - 3i}\\ {4 + 3i}&{ - i} \end{array}} \right]\)
C.	\(\frac{1}{{24}}\left[ {\begin{array}{*{20}{c}} {4 + 3i}&{ - i}\\ i&{4 - 3i} \end{array}} \right]\)
D.	\(\frac{1}{{25}}\left[ {\begin{array}{*{20}{c}} {4 + 3i}&{ - i}\\ i&{4 - 3i} \end{array}} \right]\)
Answer» B. \(\frac{1}{{25}}\left[ {\begin{array}{*{20}{c}} i&{4 - 3i}\\ {4 + 3i}&{ - i} \end{array}} \right]\)

For a given matrix P = \(\left[ {\begin{array}{*{20}{c}} {4+ 3i}&-i\\ { i}&{4 - 3i} \end{array}} \right]\), where \(i = \sqrt { - 1}\), the inverse of matrix P is

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