how can I find the imaginary matrix inverse without using the calculator?

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I have non-graphical calculator and I am strgguling to find way around to solve the imaginary matrix inverse becuase the calculator does not accept imaginary matrix.

I fould a way for solving if the matrix has just imaginary part such as $$ \begin{bmatrix} 1j & 3j \\ 2j & 4j \\ \end{bmatrix} $$

The inverse for this matrix can be found by find the inverse for $$ \begin{bmatrix} 1 & 3 \\ 2 & 4 \\ \end{bmatrix} $$

then multiply the matrix by $$ -j$$

The inverse equal

$$ \begin{bmatrix} -2 & \frac32 \\ 1 & - \frac12 \\ \end{bmatrix} *-j $$

The final form is

$$ \begin{bmatrix} 2j & -j\frac32 \\ -1j & j\frac12 \\ \end{bmatrix} $$

I am looking for a way to solve the following matrix by similar way:

$$ \begin{bmatrix} 3+1j & 2+3j \\ 3+2j & 2+4j \\ \end{bmatrix} $$

and

$$ \begin{bmatrix} 3+1j & 2+3j & 2+5j \\ 3+2j & 2+4j & 4+5j \\ 5+1j & 4+4j & 1+2j \\ \end{bmatrix} $$

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The inverse of a $2\times 2$ matrix is given by $$ \begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}, $$ so you can calculate that explicitly. For a bigger matrix you could use row reduction, where again all you need to do are simple sums and multiplications.

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$$(A+iB)^{-1}=(A+BA^{-1}B)^{-1}+i(B+AB^{-1}A)^{-1}$$ where $A$, $B$ are invertible real $n\times n$ matrices. Reference