Dadas as matrizes A=\left[\begin{array}{ccc}1 & 2 & 3 \\\\ 0 & -1 & 4 \\\\ 5 & 6 & 1\end{array}\right] \quad B=\left[\begin{array}{ccc}-1 & 5 & 7 \\\\ 2 & 1 & -3 \\\\ 0 & 4 & 6\end{array}\right] \quad C=\left[\begin{array}{ccc}5 & -7 & 4 \\\\ 2 & 1 & -3 \\\\ 0 & 4 & 6\end{array}\right] Calcule: \frac{B-C}{2}-4 \mathrm{~A} Passo 1O enunciado pede para calcular \frac{B - C}{2} \;-\; 4A onde A, B e C são matrizes 3\times3. Vamos usar operações elementares de matrizes: subtração, multiplicação por escalar e depois subtração de matrizes. Passo 2Primeiro, subtraímos C de B elemento a elemento: B - C = \begin{pmatrix} -1 & 5 & 7\\\\ 2 & 1 & -3\\\\ 0 & 4 & 6 \end{pmatrix} - \begin{pmatrix} 5 & -7 & 4\\\\ 2 & 1 & -3\\\\ 0 & 4 & 6 \end{pmatrix} = \begin{pmatrix} -1-5 & 5-(-7) & 7-4\\\\ 2-2 & 1-1 & -3-(-3)\\\\ 0-0 & 4-4 & 6-6 \end{pmatrix} = \begin{pmatrix} -6 & 12 & 3\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix}. Passo 3Agora dividimos o resultado por 2, escalarmente: \frac{B - C}{2} = \frac{1}{2} \begin{pmatrix} -6 & 12 & 3\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} -3 & 6 & \tfrac{3}{2}\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix}. Passo 4Em seguida, calculamos 4A multiplicando cada elemento de A por 4: 4A =4\begin{pmatrix} 1 & 2 & 3\\\\ 0 & -1 & 4\\\\ 5 & 6 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 8 & 12\\\\ 0 & -4 & 16\\\\ 20 & 24 & 4 \end{pmatrix}. Passo 5Por fim, subtraímos 4A de \tfrac{B-C}{2} elemento a elemento: \frac{B - C}{2} - 4A = \begin{pmatrix} -3 & 6 & \tfrac{3}{2}\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix} - \begin{pmatrix} 4 & 8 & 12\\\\ 0 & -4 & 16\\\\ 20 & 24 & 4 \end{pmatrix} = \begin{pmatrix} -3-4 & 6-8 & \tfrac{3}{2}-12\\\\ 0-0 & 0-(-4) & 0-16\\\\ 0-20 & 0-24 & 0-4 \end{pmatrix} = \begin{pmatrix} -7 & -2 & -\tfrac{21}{2}\\\\ 0 & 4 & -16\\\\ -20 & -24 & -4 \end{pmatrix}. Resposta \frac{B - C}{2} - 4A = \begin{pmatrix} -7 & -2 & -\tfrac{21}{2}\\\\ 0 & 4 & -16\\\\ -20 & -24 & -4 \end{pmatrix}. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosDadas as matrizes A=\left[\begin{array}{ccc}1 & 2 & 3 \\\\ 0 & -1 & 4 \\\\ 5 & 6 & 1\end{array}\right] \quad B=\left[\begin{array}{ccc}-1 & 5 & 7 \\\\ 2 & 1 & -3 \\\\ 0 & 4 & 6\end{array}\right] \quad C=\left[\begin{array}{ccc}5 & -7 & 4 \\\\ 2 & 1 & -3 \\\\ 0 & 4 & 6\end{array}\right] Calcule: -A+4 B-1 / 3 C Dadas as matrizes A=\left[\begin{array}{ccc}1 & 2 & 3 \\\\ 0 & -1 & 4 \\\\ 5 & 6 & 1\end{array}\right] \quad B=\left[\begin{array}{ccc}-1 & 5 & 7 \\\\ 2 & 1 & -3 \\\\ 0 & 4 & 6\end{array}\right] \quad C=\left[\begin{array}{ccc}5 & -7 & 4 \\\\ 2 & 1 & -3 \\\\ 0 & 4 & 6\end{array}\right] Calcule: A+B-2 C Dadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha AB Dadas as matrizes A=\left[\begin{array}{ccc}0 & 1 & 3 \\\\ 4 & -2 & 5 \\\\ -1 & 2 & 7\end{array}\right], \quad B=\left[\begin{array}{ccc}2 & 4 & -1 \\\\ 3 & 0 & 1 \\\\ 1 & 2 & 5\end{array}\right] e C=\left[\begin{array}{ccc}-1 & 2 & 0 \\\\ 3 & 4 & 6 \\\\ 1 & 0 & -1\end{array}\right] Obtenha B A