Exercício 3 Dados as seguintes matrizes: A=\left[\begin{array}{ccc}3 & 2 & -1 \\\\ 3 & 3 & -\frac{1}{7} \\\\ 1 & -1 & 2 \\\\ 3 & 2 & -1\end{array}\right] \quad B=\left[\begin{array}{ccc}\frac{1}{2} & -2 & 0 \\\\ 3 & -3 & \frac{1}{3} \\\\ 1 & 0 & 0 \\\\ -3 & -2 & -1\end{array}\right] C=\left[\begin{array}{cccc}3 & 3 & -\frac{1}{7} & 0 \\\\ 1 & -1 & 2 & -9 \\\\ 3 & 2 & -1 & 10\end{array}\right] \quad D=\left[\begin{array}{cccc}0 & -\frac{1}{7} & 0 & 1 \\\\ 2 & -9 & -2 & 1 \\\\ 8 & -2 & -1 & 1\end{array}\right] \quad E=\left[\begin{array}{cc}2 & 1 \\\\ 0 & -1 \\\\ 3 & 2\end{array}\right] \quad F=\left[\begin{array}{cc}0 & 1 \\\\ 7 & 3 \\\\ -4 & 5\end{array}\right] calcule: A+B ; D+2 C ; D-C ; \frac{1}{2} E+F ; E-F, Passo 1Estamos com matrizes compatíveis para soma em cada caso e vamos apenas somar ou subtrair elemento a elemento, lembrando que antes de somar um escalar vezes uma matriz (como em 2C ou \tfrac12E) precisamos multiplicar cada elemento. Passo 2Calcular A+B. Somando componente a componente, A+B =\begin{pmatrix} 3+\tfrac12 & 2-2 & -1+0\\\\ 3+3 & 3-3 & -\tfrac17+\tfrac13\\\\ 1+1 & -1+0 & 2+0\\\\ 3-3 & 2-2 & -1-1 \end{pmatrix} = \begin{pmatrix} \tfrac72 & 0 & -1\\\\ 6 & 0 & \tfrac4{21}\\\\ 2 & -1 & 2\\\\ 0 & 0 & -2 \end{pmatrix}. Passo 3Calcular D+2C. Primeiro 2C: 2C=\begin{pmatrix} 6 & 6 & -\tfrac27 & 0\\\\ 2 & -2 & 4 & -18\\\\ 6 & 4 & -2 & 20 \end{pmatrix}. Agora soma com D: D+2C =\begin{pmatrix} 0+6 & -\tfrac17+6 & 0-\tfrac27 & 1+0\\\\ 2+2 & -9-2 & -2+4 & 1-18\\\\ 8+6 & -2+4 & -1-2 & 1+20 \end{pmatrix} = \begin{pmatrix} 6 & \tfrac{41}7 & -\tfrac27 & 1\\\\ 4 & -11 & 2 & -17\\\\ 14 & 2 & -3 & 21 \end{pmatrix}. Passo 4Calcular D-C. Subtraindo componente a componente, D-C =\begin{pmatrix} 0-3 & -\tfrac17-3 & 0-(-\tfrac17) & 1-0\\\\ 2-1 & -9-(-1) & -2-2 & 1-(-9)\\\\ 8-3 & -2-2 & -1-(-1) & 1-10 \end{pmatrix} = \begin{pmatrix} -3 & -\tfrac{22}7 & \tfrac17 & 1\\\\ 1 & -8 & -4 & 10\\\\ 5 & -4 & 0 & -9 \end{pmatrix}. Passo 5Calcular \tfrac12E+F. Primeiro \tfrac12E: \tfrac12E =\begin{pmatrix} 1 & \tfrac12\\\\ 0 & -\tfrac12\\\ \tfrac32 & 1 \end{pmatrix}. Somando com F: \tfrac12E+F =\begin{pmatrix} 1+0 & \tfrac12+1\\\\ 0+7 & -\tfrac12+3\\\ \tfrac32-4 & 1+5 \end{pmatrix} = \begin{pmatrix} 1 & \tfrac32\\\\ 7 & \tfrac52\\\\ -\tfrac52 & 6 \end{pmatrix}. Passo 6Calcular E-F. Subtraindo componente a componente, E-F =\begin{pmatrix} 2-0 & 1-1\\\\ 0-7 & -1-3\\\\ 3-(-4) & 2-5 \end{pmatrix} = \begin{pmatrix} 2 & 0\\\\ -7 & -4\\\\ 7 & -3 \end{pmatrix}. Resposta A+B =\begin{pmatrix} \tfrac72 & 0 & -1\\\\ 6 & 0 & \tfrac4{21}\\\\ 2 & -1 & 2\\\\ 0 & 0 & -2 \end{pmatrix},\quad D+2C =\begin{pmatrix} 6 & \tfrac{41}7 & -\tfrac27 & 1\\\\ 4 & -11 & 2 & -17\\\\ 14 & 2 & -3 & 21 \end{pmatrix}, D-C =\begin{pmatrix} -3 & -\tfrac{22}7 & \tfrac17 & 1\\\\ 1 & -8 & -4 & 10\\\\ 5 & -4 & 0 & -9 \end{pmatrix},\quad \tfrac12E+F =\begin{pmatrix} 1 & \tfrac32\\\\ 7 & \tfrac52\\\\ -\tfrac52 & 6 \end{pmatrix},\quad E-F =\begin{pmatrix} 2 & 0\\\\ -7 & -4\\\\ 7 & -3 \end{pmatrix}. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosExercício 2 Determine a matriz oposta, matriz transposta das seguintes matrizes D=\left[\begin{array}{ccc}1 & -1 & 2 \\\\ 3 & 2 & -\sqrt{5} \\\\ 3 & 0 & 0 \\\\ 1 & -1 & -3 \\\\ 3 & 2 & -\sqrt{5} \\\\ 0 & 3 & -\frac{1}{7}\end{array}\right] Exercício 2 Determine a matriz oposta, matriz transposta das seguintes matrizes C=\left[\begin{array}{cccc}3 & 0 & 0 & -3 \\\\ 3 & 0 & 0 & 3 \\\\ 0 & 1 & -1 & 2\end{array}\right] Exercício 4 Determine a e b para que a matriz \left[\begin{array}{ccc}2 & 4 & 2 a-b \\\ a+b & 3 & 0 \\\\ -1 & 0 & 5\end{array}\right] seja simétrica Exercício 5 Determine a, b e c para que \left[\begin{array}{ccc}a & 3 & 2 a \\\ c & 0 & -2\end{array}\right]+\left[\begin{array}{ccc}b & -3 & -1 \\\\ 1 & 4 & 3\end{array}\right]=\left[\begin{array}{ccc}2 & 0 & 5 \\\\ 3 & 4 & 1\end{array}\right]