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Matrizes 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 \( \frac{X-A}{2} = \frac{B+X}{3} + C \)


RESOLVENDO

Oi.
\( \frac{X-A}{2} = \frac{B+X}{3} +C \\ \\ \frac{3(X-A)=2(B+X)+6C}{6} \\ \\ 3X-3A=2B+2X+6C \\ 3X-2X=3A+2B+6C \\ X=3. \left[\begin{array}{ccc}2&1\\3&-1\end{array}\right] +2. \left[\begin{array}{ccc}-1&2\\1&0\end{array}\right] +6. \left[\begin{array}{ccc}4&-1\\2&1\end{array}\right] \\ \\ X= \left[\begin{array}{ccc}6&3\\9&-3\end{array}\right] + \left[\begin{array}{ccc}-2&4\\2&0\end{array}\right] + \left[\begin{array}{ccc}24&-6\\12&6\end{array}\right] \)
\( X= \left[\begin{array}{ccc}28&1\\23&3\end{array}\right] \)



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