Odpowiedź:
[tex]X=\left[\begin{array}{ccc}1&2\\0&0\end{array}\right][/tex]
Szczegółowe wyjaśnienie:
Równanie:
[tex]\left[\begin{array}{ccc}3&1\\2&1\end{array}\right] X\left[\begin{array}{ccc}-3&-1\\2&1\end{array}\right] =\left[\begin{array}{ccc}3&3\\2&2\end{array}\right] [/tex]
Niech:
[tex]A=\left[\begin{array}{ccc}3&1\\2&1\end{array}\right][/tex]
[tex]B=\left[\begin{array}{ccc}-3&-1\\2&1\end{array}\right][/tex]
[tex]C=\left[\begin{array}{ccc}3&3\\2&2\end{array}\right] [/tex]
Zatem [tex]AXB=C[/tex]. Przekształcamy do macierzy [tex]X[/tex] :
[tex]AX=CB^{-1}[/tex]
[tex]X=A^{-1}CB^{-1}[/tex]
Skorzystamy ze wzoru na macierz odwrotną do danej macierzy [tex]2 \times 2[/tex]:
[tex]$A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] \wedge ad-bc \neq 0 \Rightarrow A^{-1}=\frac{1}{ad-bc} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] [/tex]
Mamy:
[tex]$A^{-1}=\frac{1}{3-2} \left[\begin{array}{ccc}1&-1\\-2&3\end{array}\right] = \left[\begin{array}{ccc}1&-1\\-2&3\end{array}\right][/tex]
[tex]$B^{-1}=\frac{1}{-3+2} \left[\begin{array}{ccc}1&1\\-2&-3\end{array}\right]= \left[\begin{array}{ccc}-1&-1\\2&3\end{array}\right][/tex]
Zatem:
[tex]X=\left[\begin{array}{ccc}1&-1\\-2&3\end{array}\right] \cdot \left[\begin{array}{ccc}3&3\\2&2\end{array}\right] \cdot \left[\begin{array}{ccc}-1&-1\\2&3\end{array}\right]=\left[\begin{array}{ccc}1&1\\0&0\end{array}\right] \cdot \left[\begin{array}{ccc}-1&-1\\2&3\end{array}\right]=[/tex]
[tex]=\left[\begin{array}{ccc}1&2\\0&0\end{array}\right][/tex]
