Question

Exercícios: 4) Dadas as matrizes A 1 51 2 4), B = -1 3 1-2 -31 1 0e C= 4 2 oc- 16 -11 3 -2), calcule: 10 1
a) A-B c)
C - A+B
b) B-C d) A-C-B 1 -1 0 6) Dada a matrizA = 2 3 4 lo 1 , obtenha: - 213x3 a) At b) At – A Obs: A significa matriz A transposta, onde devemos trocar suas linhas pelas colunas. 7 8 Ex: A = 9 10 At = = 7 9 8 10​

Answer 1

Resposta:

Explicação passo a passo:

$4)$

$A=\begin{pmatrix}1&5\\ \:2&4\\ \:-1&3\end{pmatrix}B=\begin{pmatrix}-2&-3\\ \:1&0\\ \:4&2\end{pmatrix}C=\begin{pmatrix}6&-1\\ 3&-2\\ 0&1\end{pmatrix}\\\\\\a)A-B=\begin{pmatrix}1&5\\ \:2&4\\ \:-1&3\end{pmatrix}-\begin{pmatrix}-2&-3\\ \:1&0\\ \:4&2\end{pmatrix}=\begin{pmatrix}1-\left(-2\right)&5-\left(-3\right)\\ 2-1&4-0\\ \left(-1\right)-4&3-2\end{pmatrix}=\begin{pmatrix}3&8\\ 1&4\\ -5&1\end{pmatrix}$

$b)B-C=\begin{pmatrix}-2&-3\\ 1&0\\ 4&2\end{pmatrix}-\begin{pmatrix}6&-1\\ 3&-2\\ 0&1\end{pmatrix}=\begin{pmatrix}\left(-2\right)-6&\left(-3\right)-\left(-1\right)\\ 1-3&0-\left(-2\right)\\ 4-0&2-1\end{pmatrix}=\begin{pmatrix}-8&-2\\ -2&2\\ 4&1\end{pmatrix}$

$c) C-A+B=\begin{pmatrix}6&-1\\ \:\:3&-2\\ \:\:0&1\end{pmatrix}-\begin{pmatrix}1&5\\ \:\:\:\:2&4\\ \:\:\:\:-1&3\end{pmatrix}+\begin{pmatrix}-2&-3\\ \:\:\:\:1&0\\ \:\:\:\:4&2\end{pmatrix}=\begin{pmatrix}5&-6\\ 1&-6\\ 1&-2\end{pmatrix}+\begin{pmatrix}-2&-3\\ 1&0\\ 4&2\end{pmatrix}=\begin{pmatrix}3&-9\\ 2&-6\\ 5&0\end{pmatrix}$

$d)A-C-B=\begin{pmatrix}1&5\\ \:\:\:\:\:2&4\\ \:\:\:\:\:-1&3\end{pmatrix}-\begin{pmatrix}6&-1\\ \:\:\:\:3&-2\\ \:\:\:\:0&1\end{pmatrix}-\begin{pmatrix}-2&-3\\ \:\:\:\:\:1&0\\ \:\:\:\:\:4&2\end{pmatrix}=\begin{pmatrix}-5&6\\ -1&6\\ -1&2\end{pmatrix}-\begin{pmatrix}-2&-3\\ 1&0\\ 4&2\end{pmatrix}=\begin{pmatrix}-3&9\\ -2&6\\ -5&0\end{pmatrix}$

$5)$

$\begin{pmatrix}1&-1&0\\ 2&3&4\\ 0&1&-2\end{pmatrix}$

Obter a matriz transposta transformando as filas em colunas

$a)A^t=\begin{pmatrix}1&-1&0\\ 2&3&4\\ 0&1&-2\end{pmatrix}^T=\begin{pmatrix}1&2&0\\ -1&3&1\\ 0&4&-2\end{pmatrix}$

$b)A^t-A=\begin{pmatrix}1&2&0\\ \:\:-1&3&1\\ \:\:0&4&-2\end{pmatrix}-\begin{pmatrix}1&-1&0\\ \:\:\:2&3&4\\ \:\:\:0&1&-2\end{pmatrix}=\begin{pmatrix}1-1&2-\left(-1\right)&0-0\\ \left(-1\right)-2&3-3&1-4\\ 0-0&4-1&\left(-2\right)-\left(-2\right)\end{pmatrix}\\\\\\=\begin{pmatrix}0&3&0\\ -3&0&-3\\ 0&3&0\end{pmatrix}$