CX=b wie sollte ich nach C^-1 lösen?

Question 1

Aufgabe:

Text erkannt:

\( \boldsymbol{C}=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right), \quad \boldsymbol{b}=\left(\begin{array}{c}1 \\ -1\end{array}\right) \)

Question 2

CX=b

==> X = C^(-1) * b

=\( \left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)^{-1} \cdot \left(\begin{array}{c}1 \\ -1\end{array}\right) \)

=\( \left(\begin{array}{ll}-2 & 1 \\ 1,5 & -0,5\end{array}\right)^{-1} \cdot \left(\begin{array}{c}1 \\ -1\end{array}\right) \)

=\( \left(\begin{array}{c}-3 \\ 2\end{array}\right) \)

mathef · Answer 1 · 2022-09-07T09:40:22+0000

CX=b

==> X = C^(-1) * b

=\( \left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)^{-1} \cdot \left(\begin{array}{c}1 \\ -1\end{array}\right) \)

=\( \left(\begin{array}{ll}-2 & 1 \\ 1,5 & -0,5\end{array}\right)^{-1} \cdot \left(\begin{array}{c}1 \\ -1\end{array}\right) \)

=\( \left(\begin{array}{c}-3 \\ 2\end{array}\right) \)

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