Dadas as matrizes \(A = \left( \begin{array}{rr} 3 & -1 \\ 4 & 5 \end{array}\right)\) e \(B = \left( \begin{array}{rr} 2 & 3 \\ -8 & 10 \end{array}\right)\), marque a alternativa que fornece \(3A-\frac{1}{2}B\).
\(\left( \begin{array}{rr} 8 & -\frac{9}{2} \\ 16 & 10 \end{array}\right)\)
\(\left( \begin{array}{rr} 2 & -\frac{4}{2} \\ 16 & 10 \end{array}\right)\)
\(\left( \begin{array}{rr} -8 & \frac{9}{2} \\ -16 & -10 \end{array}\right)\)
\(\left( \begin{array}{rr} 3 & -\frac{9}{2} \\ 6 & 1 \end{array}\right)\)
Calcular a matriz \(X\) na equação \(3X-A+2B=0\), sabendo que \(A=\left( \begin{array}{rr} 8 & -5 \\ -6 & 1 \end{array}\right)\), \(B = \left( \begin{array}{rr} 1 & \frac{1}{2} \\ 3 & 4 \end{array}\right)\) e \(0\) é a matriz nula.
\(X=\left( \begin{array}{rr} 2 & -2 \\ -4 & -\frac{7}{3} \end{array}\right)\)
\(X=\left( \begin{array}{rr} -2 & 2 \\ 4 & -\frac{7}{3} \end{array}\right)\)
\(X=\left( \begin{array}{rr} 1 & -1 \\ -2 & -\frac{4}{3} \end{array}\right)\)
