Odpowiedź:
[tex]X=\left[\begin{array}{ccc}-\frac{1}{2} &-1\\\frac{1}{2} &\frac{4}{3} \end{array}\right] [/tex]
Szczegółowe wyjaśnienie:
Równanie:
[tex]2X+\left[\begin{array}{ccc}1&3\\1&2\end{array}\right] =\left[\begin{array}{ccc}3&3\\-2&2\end{array}\right] X[/tex]
Przekształcamy do macierzy [tex]X[/tex] :
[tex]\left[\begin{array}{ccc}2&0\\0&2\end{array}\right]X-\left[\begin{array}{ccc}3&3\\-2&2\end{array}\right] X=-\left[\begin{array}{ccc}1&3\\1&2\end{array}\right] [/tex]
[tex]\Big(\left[\begin{array}{ccc}2&0\\0&2\end{array}\right]-\left[\begin{array}{ccc}3&3\\-2&2\end{array}\right] \Big)X=-\left[\begin{array}{ccc}1&3\\1&2\end{array}\right] [/tex]
[tex]\left[\begin{array}{ccc}-1&-3\\2&0\end{array}\right] X=-\left[\begin{array}{ccc}1&3\\1&2\end{array}\right] [/tex]
[tex]X=-\left[\begin{array}{ccc}-1&-3\\2&0\end{array}\right] ^{-1} \cdot \left[\begin{array}{ccc}1&3\\1&2\end{array}\right] [/tex]
Skorzystamy ze wzoru na macierz odwrotną do danej macierzy [tex]2 \times 2[/tex]:
[tex]$A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] \wedge ad-bc \neq 0 \Rightarrow A^{-1}=\frac{1}{ad-bc} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] [/tex]
Zatem:
[tex]$\left[\begin{array}{ccc}-1&-3\\2&0\end{array}\right] ^{-1}=\frac{1}{0+6} \left[\begin{array}{ccc}0&3\\-2&-1\end{array}\right] =\left[\begin{array}{ccc}0&\frac{1}{2} \\-\frac{1}{3} &-\frac{1}{6} \end{array}\right][/tex]
Stąd:
[tex]X=-\left[\begin{array}{ccc}0&\frac{1}{2} \\-\frac{1}{3} &-\frac{1}{6} \end{array}\right] \cdot \left[\begin{array}{ccc}1&3\\1&2\end{array}\right] =\left[\begin{array}{ccc}-\frac{1}{2} &-1\\\frac{1}{2} &\frac{4}{3} \end{array}\right] [/tex]
