Question

Given $$\mathrm{A}=\left(\begin{array}{rrr} 1 & 0 & 2 i \\ i & -3 & 0 \\ 1 & 0 & i \end{array}\right), \quad \text { find } \mathrm{A}^{\mathrm{T}}, \overline{\mathrm{A}}, \mathrm{A}^{\dagger}, \mathrm{A}^{-1}$$

Short answer

A^{\textrm{T}} = \begin{pmatrix}1 & i & 1 \ 0 & -3 & 0 \ 2i & 0 & i \end{pmatrix}, \bar{\textrm{A}} = \begin{pmatrix}1 & 0 & -2i \ -i & -3 & 0 \ 1 & 0 & -i \end{pmatrix}, \textrm{A}^\text{\textdagger} = A^{\textrm{T}} = \begin{pmatrix}1 & -i & 1 \ 0 & -3 & 0 \ -2i & 0 & -i \end{pmatrix}, \textrm{A}^{-1} = -\frac{1}{9} \begin{pmatrix}-1i & -3i - 2 & 3i \ 3 & 0 & 3 \ -3i & 0 & 1 \end{pmatrix}.

Step-by-step solution

Find the Transpose of A (\textrm{A}^{\textrm{T}})

To find the transpose of matrix \textrm{A}, switch the rows and columns. The original matrix \textrm{A} is: \[ \textrm{A} = \begin{pmatrix} 1 & 0 & 2i \ i & -3 & 0 \ 1 & 0 & i \end{pmatrix} \] The transpose \textrm{A}^{\textrm{T}} is: \[ \textrm{A}^{\textrm{T}} = \begin{pmatrix} 1 & i & 1 \ 0 & -3 & 0 \ 2i & 0 & i \end{pmatrix} \]

Find the Conjugate of A (\bar{\textrm{A}})

To find the conjugate of matrix \textrm{A}, take the complex conjugate of each element. The original matrix \textrm{A} is: \[ \textrm{A} = \begin{pmatrix} 1 & 0 & 2i \ i & -3 & 0 \ 1 & 0 & i \end{pmatrix} \] The conjugate \bar{\textrm{A}} is: \[ \bar{\textrm{A}} = \begin{pmatrix} 1 & 0 & -2i \ -i & -3 & 0 \ 1 & 0 & -i \end{pmatrix} \]

Find the Conjugate Transpose of A (\textrm{A}^\text{\textdagger})

To find the conjugate transpose (Hermitian adjoint) of matrix \textrm{A}, first transpose \textrm{A} and then take the complex conjugate of each element. Using the transpose \textrm{A}^{\textrm{T}} from Step 1: \[ \textrm{A}^{\textrm{T}} = \begin{pmatrix} 1 & i & 1 \ 0 & -3 & 0 \ 2i & 0 & i \end{pmatrix} \] Taking the complex conjugate of \textrm{A}^{\textrm{T}} gives us: \[ \textrm{A}^\text{\textdagger} = \begin{pmatrix} 1 & -i & 1 \ 0 & -3 & 0 \ -2i & 0 & -i \end{pmatrix} \]

Find the Inverse of A (\textrm{A}^{-1})

To find the inverse of \textrm{A}, compute the matrix of minors, then the matrix of cofactors, followed by the adjugate, and finally divide by the determinant. The determinant of \textrm{A} is: \[ \text{det}(\textrm{A}) = 1(-3i - 0) - 0(0) + 2i(-3) = -6i - 3i = -9i \] Thus, \textrm{A}^{-1} is calculated as follows: \[ \textrm{A}^{-1} = \frac{1}{-9i} \begin{pmatrix} 3i & 0 & 0 \ 3i - 2i^2 & -3 & -3 \ 3i & 0 & -1 \end{pmatrix} \] Simplified, we get: \[ \textrm{A}^{-1} = -\frac{1}{9} \begin{pmatrix}-1i & -3i - 2 & 3i \ 3 & 0 & 3 \ -3i & 0 & 1 \end{pmatrix} \]

Key concepts

Transpose of a Matrix

To compute the transpose of a matrix, you switch its rows and columns. This operation is commonly denoted by the symbol \( \mathrm{A}^{\mathrm{T}} \).
For the given matrix A:
\[ \mathrm{A} = \begin{pmatrix} 1 & 0 & 2i \ i & -3 & 0 \ 1 & 0 & i \end{pmatrix} \] The transpose \( \mathrm{A}^{\mathrm{T}} \) will look like this:
\[ \mathrm{A}^{\mathrm{T}} = \begin{pmatrix} 1 & i & 1 \ 0 & -3 & 0 \ 2i & 0 & i \end{pmatrix} \]
Notice how the first row \( (1, 0, 2i) \) becomes the first column and so on. Transposing complex matrices is essential in various applications, such as solving systems of equations and finding eigenvalues.

Complex Conjugate

The complex conjugate of a matrix involves taking the complex conjugate of each individual element within the matrix.
This means changing the sign of the imaginary part. For our matrix A:
\[ \mathrm{A} = \begin{pmatrix} 1 & 0 & 2i \ i & -3 & 0 \ 1 & 0 & i \end{pmatrix} \] The complex conjugate, denoted as \( \overline{\mathrm{A}} \), will be:
\[ \overline{\mathrm{A}} = \begin{pmatrix} 1 & 0 & -2i \ -i & -3 & 0 \ 1 & 0 & -i \end{pmatrix} \]
Conjugating a matrix often helps in solving certain types of equations and is crucial in fields like quantum mechanics.

Conjugate Transpose

The conjugate transpose, also known as the Hermitian adjoint (denoted as \( \mathrm{A}^{\dagger} \)), involves two steps. First, you transpose the matrix, then take the complex conjugate of each element.
Let's start with the transpose from step one:
\[ \mathrm{A}^{\mathrm{T}} = \begin{pmatrix} 1 & i & 1 \ 0 & -3 & 0 \ 2i & 0 & i \end{pmatrix} \] Now, take the complex conjugate:
\[ \mathrm{A}^{\dagger} = \begin{pmatrix} 1 & -i & 1 \ 0 & -3 & 0 \ -2i & 0 & -i \end{pmatrix} \]
The conjugate transpose is extremely useful in complex vector spaces and is a key operation in quantum mechanics. It allows us to define concepts such as Hermitian and unitary matrices.

Matrix Inverse

Finding the inverse of a matrix (denoted as \( \mathrm{A}^{-1} \)) requires several steps: calculating minors, cofactors, adjugate, and finally dividing by the determinant.
For the matrix A, the determinant is:
\[ \mathrm{det}(\mathrm{A}) = 1(-3i) - 0 + 2i(-3) = -6i - 3i = -9i \]
Next, we need the matrix of minors, and then cofactors. The inverse is then calculated as:
\[ \mathrm{A}^{-1} = \frac{1}{-9i} \begin{pmatrix} 3i & 0 & 0 \ 3i - 2i^{2} & -3 & -3 \ 3i & 0 & -1 \end{pmatrix} \]
Simplified:
\[ \mathrm{A}^{-1} = -\frac{1}{9} \begin{pmatrix} -1i & -3i - 2 & 3i \ 3 & 0 & 3 \ -3i & 0 & 1 \end{pmatrix} \]
The matrix inverse is crucial for solving linear systems of equations, as it provides a way to 'reverse' the matrix's effect.