Repeat problem 3 given

$\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right)$

Thetransposeof matrix A is computed by interchange of rows and columns in A and is denoted by A^T. The inverse of matrix A is given by $A^{-1}=\frac{1}{detA}{C}^T$ , where C_ij represent thecofactorof A_ij. Thecomplex conjugateof the matrix A is computed by taking the complex conjugate of each element of the matrix A. The transpose conjugate of the matrix A is computed by taking the complex conjugate of each element of the matrix A and then transpose the new matrix. Unit matrix is represented by $I=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$.

Matrix A is given by $A=\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right)$. To find the transpose, the inverse, the complex conjugate, and the transpose conjugate of . Also, to Verify that $A{A}^{-1}=A^{-1}A=$the unit matrix.

The given matrix is $A=\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right)$.

The transpose of matrix A is computed by interchange of rows and columns in A.

Find the transpose of the matrix A.

$A^T=\left(\begin{array}{ccc}0& -i& 3\\ -2i& 2& 0\\ -1& 0& 0\end{array}\right)$

The inverse of matrix A is given by $A^{-1}=\frac{1}{detA}{C}^T=\frac{1}{detA}adj\left(A\right)$.

Find the determinant of the matrix A.

$$\begin{gathered} det\left(A\right)=\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right) \\ =\left(0\right)\left(\left(2\right)\left(0\right)-\left(0\right)\left(0\right)\right)-\left(2i\right)\left(\left(-i\right)\left(0\right)-\left(0\right)\left(3\right)\right)+\left(-1\right)\left(\left(-i\right)\left(0\right)-\left(2\right)\left(3\right)\right) \\ =0-0+\left(-1\right)\left(0-6\right) \\ =6 \end{gathered}$$

Find the adjoint of the matrix A.

$$\begin{gathered} adj\left(A\right)={\left(\begin{array}{ccc}\left|\begin{array}{cc}2& 0\\ 0& 0\end{array}\right|& -\left|\begin{array}{cc}-i& 0\\ 3& 0\end{array}\right|& \left|\begin{array}{cc}-i& 2\\ 3& 0\end{array}\right|\\ -\left|\begin{array}{cc}2i& -1\\ 0& 0\end{array}\right|& \left|\begin{array}{cc}0& -1\\ 3& 0\end{array}\right|& -\left|\begin{array}{cc}0& 2i\\ 3& 0\end{array}\right|\\ \left|\begin{array}{cc}2i& -1\\ 2& 0\end{array}\right|& -\left|\begin{array}{cc}0& -1\\ -i& 0\end{array}\right|& \left|\begin{array}{cc}0& 2i\\ -i& 2\end{array}\right|\end{array}\right)}^T \\ ={\left(\begin{array}{ccc}\left(\left(2\right)\left(0\right)-\left(0\right)\left(0\right)\right)& -\left[\left(-i\right)\left(0\right)-\left(0\right)\left(3\right)\right]& \left(\left(-i\right)\left(0\right)-\left(2\right)\left(3\right)\right)\\ -\left[\left(2i\right)\left(0\right)-\left(-1\right)\left(0\right)\right]& \left(\left(0\right)\left(0\right)-\left(-1\right)\left(3\right)\right)& -\left[\left(0\right)\left(0\right)-\left(2i\right)\left(3\right)\right]\\ \left(\left(2i\right)\left(0\right)-\left(-1\right)\left(2\right)\right)& -\left[\left(0\right)\left(0\right)-\left(-1\right)\left(-i\right)\right]& \left(\left(0\right)\left(2\right)-\left(-i\right)\left(2i\right)\right)\end{array}\right)}^T \\ ={\left(\begin{array}{ccc}0& 0& -6\\ 0& 3& 6i\\ 2& i& -2\end{array}\right)}^T \\ =\left(\begin{array}{ccc}0& 0& 2\\ 0& 3& i\\ -6& 6i& -2\end{array}\right) \end{gathered}$$

Find the inverse of the matrix A.

$$\begin{gathered} A^{-1}=\frac{1}{6}\left(\begin{array}{ccc}0& 0& 2\\ 0& 3& i\\ -6& 6i& -2\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& \frac{1}{3}\\ 0& \frac{1}{2}& \frac{1}{6}i\\ -1& i& \frac{-1}{3}\end{array}\right) \end{gathered}$$

The complex conjugate of the matrix A is computed by taking the complex conjugate of each element of the matrix A.

Find the complex conjugate of the matrix A .

$$\begin{gathered} A^{*}=\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& 2\times -i& -1\\ -1\times -i& 2& 0\\ 3& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& -2i& -1\\ i& 2& 0\\ 3& 0& 0\end{array}\right) \end{gathered}$$

The transpose conjugate of the matrix A is computed by taking the complex conjugate of each element of the matrix A and then, transpose the new matrix.

Find the transpose of the complex conjugate of the matrix A.

$$\begin{gathered} {\left(A^{*}\right)}^T={\left(\begin{array}{ccc}0& -2i& -1\\ i& 2& 0\\ 3& 0& 0\end{array}\right)}^T \\ =\left(\begin{array}{ccc}0& i& 3\\ -2i& 2& 0\\ -1& 0& 0\end{array}\right) \end{gathered}$$

Verify the equation${\text{AA}}^{\text{- 1}}={\text{A}}^{\text{- 1}}\text{A}=\text{I}$.

$$\begin{gathered} A{A}^{-1}=\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right)\left(\begin{array}{ccc}0& 0& \frac{1}{3}\\ 0& \frac{1}{2}& \frac{1}{6}i\\ -1& i& -\frac{1}{3}\end{array}\right) \\ =\left(\begin{array}{ccc}\left(0\right)\left(0\right)+\left(2i\right)\left(0\right)+\left(-1\right)\left(-1\right)& \left(0\right)\left(0\right)+\left(2i\right)\left(\frac{1}{2}\right)+\left(-1\right)\left(i\right)& \left(0\right)\left(\frac{1}{3}\right)+\left(2i\right)\left(\frac{i}{6}\right)+\left(-1\right)\left(-\frac{1}{3}\right)\\ \left(-i\right)\left(0\right)+\left(2\right)\left(0\right)+\left(0\right)\left(-1\right)& \left(i\right)\left(0\right)+\left(2\right)\left(\frac{1}{2}\right)+\left(0\right)\left(i\right)& \left(-i\right)\left(\frac{1}{3}\right)+\left(2\right)\left(\frac{i}{6}\right)+\left(0\right)\left(-\frac{1}{3}\right)\\ \left(3\right)\left(0\right)+\left(0\right)\left(0\right)+\left(0\right)\left(-1\right)& \left(3\right)\left(0\right)+\left(0\right)\left(\frac{1}{2}\right)+\left(0\right)\left(i\right)& \left(3\right)\left(\frac{1}{3}\right)+\left(0\right)\left(\frac{i}{6}\right)+\left(0\right)\left(-\frac{1}{3}\right)\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \end{gathered}$$

Solve further.

$$\begin{gathered} A^{-1}A=\left(\begin{array}{ccc}0& 0& \frac{1}{3}\\ 0& \frac{1}{2}& \frac{1}{6}i\\ -1& i& \frac{-1}{3}\end{array}\right)\left(\begin{array}{ccc}0& 2i& -1\\ -i& 2& 0\\ 3& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}\left(0\right)\left(0\right)+\left(0\right)\left(-i\right)+\left(\frac{1}{3}\right)\left(3\right)& \left(0\right)\left(2i\right)+\left(0\right)\left(2\right)+\left(\frac{1}{3}\right)\left(0\right)& \left(0\right)\left(-1\right)+\left(0\right)\left(0\right)+\left(\frac{1}{3}\right)\left(0\right)\\ \left(0\right)\left(0\right)+\left(\frac{1}{2}\right)\left(-i\right)+\left(\frac{i}{6}\right)\left(3\right)& \left(0\right)\left(2i\right)+\left(\frac{1}{2}\right)\left(2\right)+\left(\frac{i}{6}\right)\left(0\right)& \left(0\right)\left(-1\right)+\left(\frac{1}{2}\right)\left(0\right)+\left(\frac{i}{6}\right)\left(0\right)\\ \left(-1\right)\left(0\right)+\left(i\right)\left(-i\right)+\left(-\frac{1}{3}\right)\left(3\right)& \left(-1\right)\left(2i\right)+\left(i\right)\left(2\right)+\left(-\frac{1}{3}\right)\left(0\right)& \left(-1\right)\left(-1\right)+\left(i\right)\left(0\right)+\left(-\frac{1}{3}\right)\left(0\right)\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right) \end{gathered}$$

Therefore, it is shown that the transpose of matrix A is $A^T=\left(\begin{array}{ccc}0& -i& 3\\ 2i& 2& 0\\ -1& 0& 0\end{array}\right)$,the inverse of the matrix A is $A^{-1}=\left(\begin{array}{ccc}0& 0& \frac{1}{3}\\ 0& \frac{1}{2}& \frac{1}{6}i\\ -1& i& -\frac{1}{3}\end{array}\right)$, the complex conjugate of the matrix A is $A^{*}=\left(\begin{array}{ccc}0& -2i& -1\\ i& 2& 0\\ 3& 0& 0\end{array}\right)$ and the transpose conjugate of the matrix A is ${\left(A^{*}\right)}^T=\left(\begin{array}{ccc}0& i& 3\\ -2i& 2& 0\\ -1& 0& 0\end{array}\right)$. It is shown that $A{A}^{-1}=A^{-1}A=I$, where I represent the identity(unit) matrix.