To determine if the given matrix is a Hermitian matrix, we need to verify whether the matrix is equal to its own conjugate transpose. A matrix [tex]\( A \)[/tex] is Hermitian if [tex]\( A = A^\dagger \)[/tex], where [tex]\( A^\dagger \)[/tex] represents the conjugate transpose of [tex]\( A \)[/tex].
Given matrix:
[tex]\[
A = \begin{pmatrix}
3i & 2 + 2i \\
2 - 2i & 1
\end{pmatrix}
\][/tex]
### Step 1: Calculate the Conjugate Transpose
To find the conjugate transpose [tex]\( A^\dagger \)[/tex], we first take the transpose of [tex]\( A \)[/tex] and then take the complex conjugate of each element.
#### Transpose of [tex]\( A \)[/tex]:
[tex]\[
A^T = \begin{pmatrix}
3i & 2 - 2i \\
2 + 2i & 1
\end{pmatrix}
\][/tex]
#### Conjugate of [tex]\( A^T \)[/tex]:
[tex]\[
A^\dagger = (A^T)^* =
\begin{pmatrix}
(3i)^ & (2 - 2i)^ \\
(2 + 2i)^ & 1^
\end{pmatrix}
=
\begin{pmatrix}
-3i & 2 + 2i \\
2 - 2i & 1
\end{pmatrix}
\][/tex]
Where [tex]\( (3i)^ = -3i \)[/tex], [tex]\( (2 - 2i)^ = 2 + 2i \)[/tex], [tex]\( (2 + 2i)^ = 2 - 2i \)[/tex], and [tex]\( 1^ = 1 \)[/tex].
### Step 2: Compare the Original Matrix with its Conjugate Transpose
Original matrix:
[tex]\[
A = \begin{pmatrix}
3i & 2 + 2i \\
2 - 2i & 1
\end{pmatrix}
\][/tex]
Conjugate transpose:
[tex]\[
A^\dagger = \begin{pmatrix}
-3i & 2 + 2i \\
2 - 2i & 1
\end{pmatrix}
\][/tex]
### Step 3: Check for Equality
We must check if each corresponding element in [tex]\( A \)[/tex] and [tex]\( A^\dagger \)[/tex] are equal:
- Is [tex]\( 3i = -3i \)[/tex]? No, they are not equal.
- Is [tex]\( 2 + 2i = 2 + 2i \)[/tex]? Yes, they are equal.
- Is [tex]\( 2 - 2i = 2 - 2i \)[/tex]? Yes, they are equal.
- Is [tex]\( 1 = 1 \)[/tex]? Yes, they are equal.
Since the first elements [tex]\( 3i \)[/tex] and [tex]\( -3i \)[/tex] are not equal, we conclude that [tex]\( A \neq A^\dagger \)[/tex].
Therefore, the matrix
[tex]\[
\begin{pmatrix}
3i & 2 + 2i \\
2 - 2i & 1
\end{pmatrix}
\][/tex]
is not a Hermitian matrix.
