Sejam: A=\left[\begin{array}{rrr}1 & 2 & 3 \\\\ 2 & 1 & -1\end{array}\right], B=\left[\begin{array}{rrr}-2 & 0 & 1 \\\\ 3 & 0 & 1\end{array}\right], C=\left[\begin{array}{r}-1 \\\\ 2 \\\\ 4\end{array}\right], D=\left[\begin{array}{ll}2 & -1\end{array}\right], E=\left[\begin{array}{rr}1 & 0 \\\\ 3 & -1 \\\\ 4 & 2\end{array}\right] e F=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 1\end{array}\right]. Calcule, quando possível: E^{t}+(-A) Passo 1O enunciado dá A\in\mathbb{R}^{2\times3} e E\in\mathbb{R}^{3\times2} e pede E^{t}+(-A). Como E^{t} será 2\times3 e A já é 2\times3, a soma faz sentido e na prática é E^{t}-A. Passo 2Vamos primeiro calcular a transposta de E. Como E = \begin{pmatrix} 1 & 0\\\\ 3 & -1\\\\ 4 & 2 \end{pmatrix}, temos E^{t} = \begin{pmatrix} 1 & 3 & 4\\\\ 0 & -1 & 2 \end{pmatrix}. Passo 3Em seguida calculamos -A. Como A = \begin{pmatrix} 1 & 2 & 3\\\\ 2 & 1 & -1 \end{pmatrix}, segue que -\,A = \begin{pmatrix} -1 & -2 & -3\\\\ -2 & -1 & 1 \end{pmatrix}. Passo 4Agora somamos E^{t} e -A: E^{t} + (-A) = \begin{pmatrix} 1 & 3 & 4\\\\ 0 & -1 & 2 \end{pmatrix} + \begin{pmatrix} -1 & -2 & -3\\\\ -2 & -1 & 1 \end{pmatrix} = \begin{pmatrix} 1-1 & 3-2 & 4-3\\\\ 0-2 & -1-1 & 2-(-1) \end{pmatrix} = \begin{pmatrix} 0 & 1 & 1\\\\ -2 & -2 & 3 \end{pmatrix}. Resposta E^{t}+(-A) = \begin{pmatrix} 0 & 1 & 1\\\\ -2 & -2 & 3 \end{pmatrix}. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosSejam: A=\left[\begin{array}{rrr}1 & 2 & 3 \\\\ 2 & 1 & -1\end{array}\right], B=\left[\begin{array}{rrr}-2 & 0 & 1 \\\\ 3 & 0 & 1\end{array}\right], C=\left[\begin{array}{r}-1 \\\\ 2 \\\\ 4\end{array}\right], D=\left[\begin{array}{ll}2 & -1\end{array}\right], E=\left[\begin{array}{rr}1 & 0 \\\\ 3 & -1 \\\\ 4 & 2\end{array}\right] e F=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 1\end{array}\right]. Calcule, quando possível: E^{t}+(-A) Sejam: A=\left[\begin{array}{rrr}1 & 2 & 3 \\\\ 2 & 1 & -1\end{array}\right], B=\left[\begin{array}{rrr}-2 & 0 & 1 \\\\ 3 & 0 & 1\end{array}\right], C=\left[\begin{array}{r}-1 \\\\ 2 \\\\ 4\end{array}\right], D=\left[\begin{array}{ll}2 & -1\end{array}\right], E=\left[\begin{array}{rr}1 & 0 \\\\ 3 & -1 \\\\ 4 & 2\end{array}\right] e F=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 1\end{array}\right]. Calcule, quando possível: C \cdot A Sejam: A=\left[\begin{array}{rrr}1 & 2 & 3 \\\\ 2 & 1 & -1\end{array}\right], B=\left[\begin{array}{rrr}-2 & 0 & 1 \\\\ 3 & 0 & 1\end{array}\right], C=\left[\begin{array}{r}-1 \\\\ 2 \\\\ 4\end{array}\right], D=\left[\begin{array}{ll}2 & -1\end{array}\right], E=\left[\begin{array}{rr}1 & 0 \\\\ 3 & -1 \\\\ 4 & 2\end{array}\right] e F=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 1\end{array}\right]. Calcule, quando possível: C \cdot D+2 E-A^{t} Sejam: A=\left[\begin{array}{rrr}1 & 2 & 3 \\\\ 2 & 1 & -1\end{array}\right], B=\left[\begin{array}{rrr}-2 & 0 & 1 \\\\ 3 & 0 & 1\end{array}\right], C=\left[\begin{array}{r}-1 \\\\ 2 \\\\ 4\end{array}\right], D=\left[\begin{array}{ll}2 & -1\end{array}\right], E=\left[\begin{array}{rr}1 & 0 \\\\ 3 & -1 \\\\ 4 & 2\end{array}\right] e F=\left[\begin{array}{ll}1 & 0 \\\\ 0 & 1\end{array}\right]. Calcule, quando possível: C \cdot D+2 E-A^{t}