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A square matrix A is said to be invertible or regular if there exists a square matrix A{-1} (called the inverse matrix) such that $A \cdot A^{-1} = A^{-1}\cdot A = I$ where I is the unitmatrix : $ \LARGE \begin{bmatrix} \color{black}1 & \color{black}0 \\\color{black}0 &\color{black}1 \end{bmatrix} $ We can show (see later): $ \LARGE A = \begin{bmatrix} \color{red}a & \color{red}b \\\color{red}c &\color{red}d \end{bmatrix} \rightarrow A^{-1} = \frac{1}{det A}\cdot \begin{bmatrix} \color{red}d & \color{red}-b \\\color{red}-c &\color{red}a\end{bmatrix} $ | 
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Got it !
Inverse einer 2 x 2 Matrix
Example: $\LARGE A = \begin{bmatrix} \color{black}2 & \color{black}1 \\\color{black}6 &\color{black}4 \end{bmatrix} \rightarrow A^{-1} = \frac{2}{2\cdot 4- 6 \cdot 1}\cdot \begin{bmatrix} \color{black}4 & \color{black}-1 \\\color{black}-6 &\color{black}2\end{bmatrix} = \begin{bmatrix} \color{black}2 & \color{black}-\frac{1}{2} \\\color{black}-3 &\color{black}1 \end{bmatrix}$ In fact: $\LARGE \begin{bmatrix} \color{black}2 & \color{black}1 \\\color{black}6 &\color{black}4 \end{bmatrix} \cdot \begin{bmatrix} \color{black}2 & \color{black}-\frac{1}{2} \\\color{black}-3 &\color{black}1 \end{bmatrix} = \begin{bmatrix} \color{black}2\cdot 2 + 1 \cdot -3 & \color{black}2\cdot -\frac{1}{2} + 1 \cdot 1\ \\\color{black}6\cdot 2 + 4\cdot -3 &\color{black}6\cdot -\frac{1}{2} + 4\cdot 1 \end{bmatrix} = I$ $\LARGE \begin{bmatrix} \color{black}2 & \color{black}-\frac{1}{2} \\\color{black}-3 &\color{black}1 \end{bmatrix} \cdot \begin{bmatrix} \color{black}2 & \color{black}1 \\\color{black}6 &\color{black}4 \end{bmatrix} = \begin{bmatrix} \color{black}2\cdot 2 + 6 \cdot -\frac{1}{2} & \color{black}2 \cdot 1 +4 \cdot -\frac{1}{2} \ \\\color{black}-3 \cdot 2 + 1\cdot 6 &\color{black}-3\cdot 1+ 1\cdot 4 \end{bmatrix} = I$| $\LARGE \begin{bmatrix}1 & 2\\3 & 5\end{bmatrix}^{-1} = $ | |
| $\LARGE \begin{bmatrix}1 & 1 \\2 & 2 \end{bmatrix}^{-1} = $ | |
| $\LARGE \begin{bmatrix}8 & -2 \\4 & -\frac{1}{2} \end{bmatrix}^{-1} = $ |