• Matéria: Matemática
  • Autor: iziscaroline17
  • Perguntado 9 anos atrás

Matrizes Ajudem... Se A=  \left[\begin{array}{cc}2&1\\3&-1\end{array}\right]  , B=  \left[\begin{array}{cc}-1&2\\1&0\end{array} \right] C=   \left[\begin{array}{cc}4&-1\\2&1\end{array}\right] determine a matriz X de ordem 2, tal que  \frac{X-A}{2} = \frac{B+X}{3} + C

Respostas

respondido por: fagnerdi
13
Oi. Bom dia Izis

 \frac{X-A}{2} = \frac{B+X}{3} +C \\  \\  \frac{3(X-A)=2(B+X)+6C}{6}  \\  \\ 3X-3A=2B+2X+6C \\ 3X-2X=3A+2B+6C \\ X=3.  \left[\begin{array}{ccc}2&1\\3&-1\end{array}\right] +2.  \left[\begin{array}{ccc}-1&2\\1&0\end{array}\right] +6.  \left[\begin{array}{ccc}4&-1\\2&1\end{array}\right]  \\  \\ X=  \left[\begin{array}{ccc}6&3\\9&-3\end{array}\right] + \left[\begin{array}{ccc}-2&4\\2&0\end{array}\right] + \left[\begin{array}{ccc}24&-6\\12&6\end{array}\right]

X= \left[\begin{array}{ccc}28&1\\23&3\end{array}\right]


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