Prove that $(AB)^T = B^TA^T$

verify that $|-B| = (-1)^n |B|$

let $E = e^A=Taylor(e,A)$, show that:

  1. $A^{-1} = e^{-A}$
  2. $e^B e^C = e^{B + C}$ (if $B$ and $C$ commute)
  3. $E$ is orthogonal if $A$ is antisymmetric

Matrix inverse

Find the inverse of \[ A=\begin{pmatrix}2& 2& 3\nl 2& 5& 3\nl 1& 0& 8\end{pmatrix}. \]

Sol:

We begin by forming the matrix \[ \begin{pmatrix} A & | & I_3 \end{pmatrix} =\left(\begin{array}{ccc|ccc}2 & 2 & 3 & 1 & 0 & 0\nl 2 & 5 & 3 & 0 & 1 & 0\nl 1 & 0 & 8 & 0 & 0 & 1\end{array}\right). \]

Interchanging the first and third rows of the matrix $\begin{pmatrix} A & | & I_3 \end{pmatrix}$, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 2 & 5 & 3 & 0 & 1 & 0\nl 2 & 2 & 3 & 1 & 0 & 0\end{array}\right). \]

Adding $(-2)$ times the first row of the matrix to its second row, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 0 & 5 & -13 & 0 & 1 & -2\nl 2 & 2 & 3 & 1 & 0 & 0\end{array}\right). \]

Multiplying the second row of the matrix by $\frac{1}{5}$, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\nl 2 & 2 & 3 & 1 & 0 & 0\end{array}\right). \]

Adding $(-2)$ times the first row of the matrix to its third row, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\nl 0 & 2 & -13 & 1 & 0 & -2\end{array}\right). \]

Adding $(-2)$ times the second row of the matrix to its third row, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\nl 0 & 0 & -\frac{39}{5} & 1 & -\frac{2}{5} & -\frac{6}{5}\end{array}\right). \]

Multiplying the third row of the matrix by $(-\frac{5}{39})$, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 0 & 1 & -\frac{13}{5} & 0 & \frac{1}{5} & -\frac{2}{5}\nl 0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right). \]

Adding $(\frac{13}{5})$ times the third row of the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 8 & 0 & 0 & 1\nl 0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\nl 0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right). \]

Adding $(-8)$ times the third row of the matrix to its first row, we obtain the matrix \[\left(\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{40}{39} & -\frac{16}{39} & -\frac{3}{13}\nl 0 & 1 & 0 & -\frac{1}{3} & \frac{1}{3} & 0\nl 0 & 0 & 1 & -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{array}\right). \]

Thus, \[ A^{-1}=\begin{pmatrix}\frac{40}{39} & -\frac{16}{39} & -\frac{3}{13}\nl -\frac{1}{3} & \frac{1}{3} & 0\nl -\frac{5}{39} & \frac{2}{39} & \frac{2}{13}\end{pmatrix}. \]