Let's analyze the matrix to determine if it is "mitarymanix." Since the result is based on the given calculations, we'll lay out the solution step by step:
1. Define the matrix:
[tex]\[
A = \sqrt[1]{\sqrt{3}} \left[\begin{array}{cc}
1+i & -1 \\
1 & 1-i
\end{array}\right]
\][/tex]
However, the scalar factor [tex]\(\sqrt[1]{\sqrt{3}}\)[/tex] does not affect the properties of the matrix related to whether it is "mitarymanix." We can examine the inner matrix directly:
[tex]\[
A' = \left[\begin{array}{cc}
1+i & -1 \\
1 & 1-i
\end{array}\right]
\][/tex]
2. Calculate the determinant of [tex]\(A'\)[/tex]:
[tex]\[
\text{det}(A') = (1+i)(1-i) - (-1)(1)
\][/tex]
Simplifying the terms:
[tex]\[
\text{det}(A') = (1^2 - i^2) + 1 = (1 - (-1)) + 1 = 2 + 1 = 3
\][/tex]
Thus, the determinant is [tex]\(3\)[/tex].
3. Calculate the inverse of [tex]\(A'\)[/tex]:
The inverse of a 2x2 matrix [tex]\(\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]\)[/tex] is given by:
[tex]\[
\frac{1}{ad - bc} \left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]
\][/tex]
For [tex]\(A'\)[/tex]:
[tex]\[
a = 1+i, \; b = -1, \; c = 1, \; d = 1-i
\][/tex]
The inverse is:
[tex]\[
A'^{-1} = \frac{1}{3} \left[\begin{array}{cc} 1-i & 1 \\ -1 & 1+i \end{array}\right]
\][/tex]
Simplifying:
[tex]\[
A'^{-1} = \left[\begin{array}{cc} \frac{1-i}{3} & \frac{1}{3} \\ \frac{-1}{3} & \frac{1+i}{3} \end{array}\right]
\][/tex]
4. Calculate the conjugate transpose of [tex]\(A'\)[/tex]:
Conjugate transpose (also known as Hermitian transpose) is found by taking the transpose of the matrix and then taking the complex conjugate of each element:
[tex]\[
A'^{\dagger} = \left[\begin{array}{cc}
1-i & 1 \\
-1 & 1+i
\end{array}\right]
\][/tex]
5. Check if [tex]\(A'\)[/tex] is Hermitian:
A matrix is Hermitian if it is equal to its own conjugate transpose:
Compare [tex]\(A'\)[/tex] and [tex]\(A'^{\dagger}\)[/tex]:
[tex]\[
A' = \left[\begin{array}{cc}
1+i & -1 \\
1 & 1-i
\end{array}\right], \quad A'^{\dagger} = \left[\begin{array}{cc}
1-i & 1 \\
-1 & 1+i
\end{array}\right]
\][/tex]
Since [tex]\(A' \ne A'^{\dagger}\)[/tex], [tex]\(A'\)[/tex] is not Hermitian.
In conclusion, given the determinant, inverse, and Hermitian check, the matrix:
[tex]\[
\sqrt[1]{\sqrt{3}}\left[\begin{array}{cc}
1+i & -1 \\
1 & 1-i
\end{array}\right]
\][/tex]
is not a "mitarymanix" (assuming "mitarymanix" refers to being Hermitian).
