Question

dada as matrizes:

A=|2 5| B=|9 1| C=|3 -1|
|-4 8| |-7 0| |-9 5|

a) A+2B b) 2A-C c)A*B d)B(elevado a T) * C(elevado a T)

Answer 1

$A = \left[\begin{array}{ccc}2&5\\-4&8\end{array}\right] B=\left[\begin{array}{ccc}9&1\\-7&0\end{array}\right] C=\left[\begin{array}{ccc}3&-1\\-9&5\end{array}\right]$

Questão A)
$\left[\begin{array}{ccc}2&5\\-4&8\end{array}\right] + \left[\begin{array}{ccc}18&2\\-14&0\end{array}\right] = \left[\begin{array}{ccc}20&7\\-18&8\end{array}\right]$

Questão B)
$\left[\begin{array}{ccc}4&10\\-8&16\end{array}\right] - \left[\begin{array}{ccc}3&-1\\-9&5\end{array}\right] = \left[\begin{array}{ccc}1&9\\1&11\end{array}\right]$

Questão C)
$A \left[\begin{array}{ccc}2&5\\-4&8\end{array}\right] \times B \left[\begin{array}{ccc}9&1\\-7&0\end{array}\right] = \left[\begin{array}{ccc}-17&2\\-92&-4\end{array}\right]$

Questão D)
Ps. Matriz elevada a T, significa Matriz Transposta. Você só troca linha por coluna, e vice-versa.

$B^t=\left[\begin{array}{ccc}9&-7\\1&0\end{array}\right]\times C^t=\left[\begin{array}{ccc}3&-9\\-1&5\end{array}\right]=\left[\begin{array}{ccc}34&-116\\3&-9\end{array}\right]$