Rozwiązać równanie macierzowe:
\(\displaystyle{ \left[\begin{array}{cc}1&3\\2&0\end{array}\right] + X = \left[\begin{array}{cc}1&3\\2&0\end{array}\right] \cdot X}\)
Za \(\displaystyle{ X}\) podstawiłem macierz \(\displaystyle{ \left[\begin{array}{ccc}a&b\\c&d\end{array}\right]}\) i wyszło mi:
\(\displaystyle{ \left[\begin{array}{cc}a+1&3+b\\2+c&d\end{array}\right] = \left[\begin{array}{cc}a+3c&b+3d\\2a&2b\end{array}\right]}\)
Co dalej robić?
Równanie macierzowe
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1991Kamil
- Użytkownik
- Posty: 23
- Rejestracja: 19 gru 2011, o 01:41
- Płeć: Mężczyzna
- Lokalizacja: md
- Podziękował: 4 razy
Równanie macierzowe
W ten sposób:
\(\displaystyle{ a+1 = a+3c}\)
\(\displaystyle{ a+1 = -a - 3c = 0}\)
\(\displaystyle{ 1-3c = 0}\)
\(\displaystyle{ -3c = -1 \setminus : (-3)}\)
\(\displaystyle{ c = \frac{1}{3}}\)
\(\displaystyle{ 3+b = b+3d}\)
\(\displaystyle{ 3+b-b-3d=0}\)
\(\displaystyle{ 3-3d=0}\)
\(\displaystyle{ -3d = -3 \setminus : (-3)}\)
\(\displaystyle{ d = 3}\)
\(\displaystyle{ 2+c = 2a}\)
\(\displaystyle{ 2+c -2a =0}\)
\(\displaystyle{ d=2b}\)
\(\displaystyle{ d-2b=0}\)
???
\(\displaystyle{ a+1 = a+3c}\)
\(\displaystyle{ a+1 = -a - 3c = 0}\)
\(\displaystyle{ 1-3c = 0}\)
\(\displaystyle{ -3c = -1 \setminus : (-3)}\)
\(\displaystyle{ c = \frac{1}{3}}\)
\(\displaystyle{ 3+b = b+3d}\)
\(\displaystyle{ 3+b-b-3d=0}\)
\(\displaystyle{ 3-3d=0}\)
\(\displaystyle{ -3d = -3 \setminus : (-3)}\)
\(\displaystyle{ d = 3}\)
\(\displaystyle{ 2+c = 2a}\)
\(\displaystyle{ 2+c -2a =0}\)
\(\displaystyle{ d=2b}\)
\(\displaystyle{ d-2b=0}\)
???