Rozwiązać równanie macierzowe

[tex]\left[\begin{array}{ccc}4&0\\0&4\end{array}\right]X+\left[\begin{array}{ccc}5&3\\3&2\end{array}\right] =\left[\begin{array}{ccc}3&3\\2&2\end{array}\right] X[/tex]

[tex]\left[\begin{array}{ccc}4&0\\0&4\end{array}\right]X-\left[\begin{array}{ccc}3&3\\2&2\end{array}\right] X=-\left[\begin{array}{ccc}5&3\\3&2\end{array}\right][/tex]

[tex]\Big(\left[\begin{array}{ccc}4&0\\0&4\end{array}\right]-\left[\begin{array}{ccc}3&3\\2&2\end{array}\right]\Big)X=-\left[\begin{array}{ccc}5&3\\3&2\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}1&-3\\-2&2\end{array}\right]X=-\left[\begin{array}{ccc}5&3\\3&2\end{array}\right][/tex]

[tex]X=-\left[\begin{array}{ccc}1&-3\\-2&2\end{array}\right]^{-1}\left[\begin{array}{ccc}5&3\\3&2\end{array}\right][/tex]

Najpierw niech:

[tex]A=\left[\begin{array}{ccc}1&-3\\-2&2\end{array}\right][/tex]

Odwracamy macierz:

[tex]$A^{-1}=(A^{D})^{T} \cdot \frac{1}{\det(A)} [/tex]

[tex]\det(A)=\left|\begin{array}{ccc}1&-3\\-2&2\end{array}\right|=2-6=-4[/tex]

Minory:

[tex]M_{11}=(-1)^{1+1} \cdot 2=2[/tex]

[tex]M_{12}=(-1)^{1+2} \cdot (-2)=2[/tex]

[tex]M_{21}=(-1)^{2+1} \cdot (-3)=3[/tex]

[tex]M_{22}=(-1)^{2+2} \cdot 1=1[/tex]

Macierz dopełnień algebraicznych:

[tex]A^{D}=\left[\begin{array}{ccc}2&2\\3&1\end{array}\right][/tex]

Transponowana macierz dopełnień:

[tex](A^{D})^{T}=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] [/tex]

Macierz odwrotna:

[tex]$A^{-1}=-\left[\begin{array}{ccc}\frac{1}{2} &\frac{3}{4} \\\frac{1}{2} &\frac{1}{4} \end{array}\right] [/tex]

Teraz obliczamy macierz [tex]X[/tex] :

[tex]X=\left[\begin{array}{ccc}\frac{1}{2} &\frac{3}{4} \\\frac{1}{2} &\frac{1}{4} \end{array}\right] \cdot \left[\begin{array}{ccc}5&3\\3&2\end{array}\right]=\left[\begin{array}{ccc}\frac{19}{4} &3\\\frac{13}{4} &2\end{array}\right] [/tex]