Question

sejam as matrizes:
olhem a imagem por favor
a)A.B
b)B.A
c)A.C
d)Bt.C
e)B.At​

Answer 1

Explicação passo-a-passo:

Sendo $$\begin{lgathered}A = \left[\begin{array}{ccc}1&3\\2&0\\-1&4\end{array}\right]\end{lgathered}$$ e $$\begin{lgathered}B = \left[\begin{array}{ccc}2&1\\3&1\end{array}\right]\end{lgathered}$$ , vamos determinar as matrizes transpostas de A e B.

Para isso, o que era linha vira coluna e o que era coluna vira linha.

Sendo assim:

$$\begin{lgathered}A^T = \left[\begin{array}{ccc}1&2&-1\\3&0&4\end{array}\right]\end{lgathered}$$

$$\begin{lgathered}B^T = \left[\begin{array}{ccc}2&3\\1&1\end{array}\right]\end{lgathered}$$

Logo:

a) A.B

$$\begin{lgathered}A.B =\left[\begin{array}{ccc}1&3\\2&0\\-1&4\end{array}\right].\left[\begin{array}{ccc}2&1\\3&1\end{array}\right]\end{lgathered}$$

$$\begin{lgathered}A.B= \left[\begin{array}{ccc}11&4\\4&2\\10&3\end{array}\right]\end{lgathered}$$

b) B.A

Perceba que a matriz B é (2x2) e a matriz A é (3x2).

Como o número de colunas de B é diferente do número de linhas de A, então não é possível realizar a multiplicação.

c) A.C

Sendo $$\begin{lgathered}C = \left[\begin{array}{cc}4\\-1\end{array}\right]\end{lgathered}$$ , temos que:

$$\begin{lgathered}A.C = \left[\begin{array}{ccc}1&3\\2&0\\-1&4\end{array}\right].\left[\begin{array}{cc}4\\-1\end{array}\right]\end{lgathered}$$

$$\begin{lgathered}A.C= \left[\begin{array}{ccc}1\\8\\-8\end{array}\right]\end{lgathered}$$

d) $$B^T.C$$

$$\begin{lgathered}B^T.C = \left[\begin{array}{ccc}2&3\\1&1\end{array}\right]. \left[\begin{array}{cc}4\\-1\end{array}\right]\end{lgathered}$$

$$\begin{lgathered}B^T.C= \left[\begin{array}{cc}5\\3\end{array}\right]\end{lgathered}$$

e) $$B.A^T$$

$$\begin{lgathered}B.A^T= \left[\begin{array}{ccc}2&1\\3&1\end{array}\right].\left[\begin{array}{ccc}1&2&-1\\3&0&4\end{array}\right]\end{lgathered}$$

$$\begin{lgathered}B.A^T= \left[\begin{array}{ccc}5&4&2\\6&6&1\end{array}\right]\end