Odpowiedź:
[tex]x=\left[\begin{array}{ccc}\frac{1}{c-12} &\frac{3}{12-c} \\\frac{4}{12-c} &\frac{c}{c-12} \end{array}\right] , \ \ c\neq 12[/tex]
Rozwiązanie:
Równanie:
[tex]\left[\begin{array}{ccc}4&1\\c&3\end{array}\right] x=\left[\begin{array}{ccc}0&1\\1&0\end{array}\right][/tex]
Przekształcenie do macierzy [tex]x[/tex] :
[tex]x=\left[\begin{array}{ccc}4&1\\c&3\end{array}\right] ^{-1} \cdot \left[\begin{array}{ccc}0&1\\1&0\end{array}\right][/tex]
Macierz odwrotna:
[tex]A=\left[\begin{array}{ccc}4&1\\c&3\end{array}\right][/tex]
Wyznacznik:
[tex]\det(A)=\left|\begin{array}{ccc}4&1\\c&3\end{array}\right|=12-c[/tex]
Minory:
[tex]$M_{11}=3[/tex]
[tex]M_{12}=-c[/tex]
[tex]M_{21}=-1[/tex]
[tex]M_{22}=4[/tex]
Macierz dopełnień:
[tex]A^{D}=\left[\begin{array}{ccc}3&-c\\-1&4\end{array}\right][/tex]
Transponowana macierz dopełnień:
[tex](A^{D})^{T}=\left[\begin{array}{ccc}3&-1\\-c&4\end{array}\right][/tex]
Macierz odwrotna:
[tex]$A^{-1}=\frac{1}{\det(A)} \cdot (A^{D})^{T}=\frac{1}{12-c} \left[\begin{array}{ccc}3&-1\\-c&4\end{array}\right]=\left[\begin{array}{ccc}\frac{3}{12-c} &\frac{1}{c-12} \\\frac{c}{c-12} &\frac{4}{12-c} \end{array}\right][/tex]
A zatem:
[tex]x=\left[\begin{array}{ccc}\frac{3}{12-c} &\frac{1}{c-12} \\\frac{c}{c-12} &\frac{4}{12-c} \end{array}\right] \cdot \left[\begin{array}{ccc}0&1\\1&0\end{array}\right] =\left[\begin{array}{ccc}\frac{1}{c-12} &\frac{3}{12-c} \\\frac{4}{12-c} &\frac{c}{c-12} \end{array}\right][/tex]
