The inverse of the matrix \[\left[ {\begin{array}{*{20}{c}} {3 + 2{\text{i}}}&{\text{i}} \\ { – {\text{i}}}&{3 – 2{\text{i}}} \end{array}} \right]\] is A. \[\frac{1}{{12}}\left[ {\begin{array}{*{20}{c}} {3 + 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 – 2{\text{i}}} \end{array}} \right]\] B. \[\frac{1}{{12}}\left[ {\begin{array}{*{20}{c}} {3 – 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 + 2{\text{i}}} \end{array}} \right]\] C. \[\frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} {3 + 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 – 2{\text{i}}} \end{array}} \right]\] D. \[\frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} {3 – 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 + 2{\text{i}}} \end{array}} \right]\]

”[ rac{1}{{12}}left[
\]” option2=”\[\frac{1}{{12}}\left[ {\begin{array}{*{20}{c}} {3 – 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 + 2{\text{i}}} \end{array}} \right]\]” option3=”\[\frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} {3 + 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 – 2{\text{i}}} \end{array}} \right]\]” option4=”\[\frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} {3 – 2{\text{i}}}&{ – {\text{i}}} \\ {\text{i}}&{3 + 2{\text{i}}} \end{array}} \right]\]” correct=”option3″]

The correct answer is $\boxed{\frac{1}{14}\left[\begin{array}{cc}3-2i&-i\\i&3+2i\end{array}\right]}$.

To find the inverse of a 2×2 matrix, we can use the formula $A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}a_{2,2}&-a_{1,2}\\a_{2,1}&a_{1,1}\end{array}\right]$, where $|A|$ is the determinant of $A$.

In this case, $A=\left[\begin{array}{cc}3+2i&i\-i&3-2i\end{array}\right]$. Therefore,

\begin{align}
|A|&=(3+2i)(3-2i)-i(-i) \\
&=9-4i+i^2 \\
&=12
\end{align
}

Therefore, the inverse of $A$ is $\frac{1}{|A|}\left[\begin{array}{cc}3-2i&-i\\i&3+2i\end{array}\right]=\frac{1}{12}\left[\begin{array}{cc}3-2i&-i\\i&3+2i\end{array}\right]$.