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$$A=\left(\begin{array}{c}2 & 0 \\-1 & 2\end{array}\right)\quad;\quad B=\left(\begin{array}{c}1 & -2 \\-2 & -3\end{array}\right)\quad;\quad C=\left(\begin{array}{c}11 & 12 \\-3 & 0\end{array}\right)$$
Es ist \(\text{det}(A)=2\cdot2-(-1)\cdot0=4\).
Da \(\text{det}(A)\ne0\) ist, kann man die Matrix \(A\) invertieren:
$$\begin{array}{r}2 & 0 &|& 1 & 0 & | &:2\\-1 & 2 &|& 0 & 1 & | &\end{array}$$$$\begin{array}{r}1 & 0 &|& \frac{1}{2} & 0 & | &\\-1 & 2 &|& 0 & 1 & | &+\text{Zeile}\,1\end{array}$$$$\begin{array}{r}1 & 0 &|& \frac{1}{2} & 0 & | &\\0 & 2 &|& \frac{1}{2} & 1 & | &:2\end{array}$$$$\begin{array}{r}1 & 0 &|& \frac{1}{2} & 0 & | &\\0 & 1 &|& \frac{1}{4} & \frac{1}{2} & | &\end{array}$$$$\Rightarrow\quad A^{-1}=\left(\begin{array}{c}\frac{1}{2} & 0 \\\frac{1}{4} & \frac{1}{2}\end{array}\right)$$Damit können wir die Matrixgleichung lösen:
$$A\cdot X+B=C\quad\Leftrightarrow\quad A\cdot X=C-B\quad\Leftrightarrow\quad X=A^{-1}\cdot(C-B)$$$$X=\left(\begin{array}{c}\frac{1}{2} & 0 \\\frac{1}{4} & \frac{1}{2}\end{array}\right)\cdot\left[\left(\begin{array}{c}11 & 12 \\-3 & 0\end{array}\right)-\left(\begin{array}{c}1 & -2 \\-2 & -3\end{array}\right)\right]=\left(\begin{array}{c}\frac{1}{2} & 0 \\\frac{1}{4} & \frac{1}{2}\end{array}\right)\cdot\left(\begin{array}{c}10 & 14 \\-1 & 3\end{array}\right)$$$$X=\left(\begin{array}{c}5 & 7 \\2 & 1\end{array}\right)$$
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