Question

por favor me ajudem, coml resolvo?

Answer 1

12)

a) Pela igualdade temos:

x + 1 = 10
x = 10 - 1  --->  x = 9

x - y = 2
9 - y = 2
-y = 2 - 9
-y = -7 . (-1)  --->  y = 7


b) Montando o sistema:

(I)  { 3x - 2y =  1 
(II) { x + 3y = 4 

Isolando x em (II):

x = 4 - 3y

Substituindo x em (I):

3.(4 - 3y) - 2y = 1
12 - 9y - 2y = 1
-9y - 2y = 1 - 12
-11y= -11 . (-1)
11y = 11
y = 11/11  --->  y = 1


Determinando o valor de x:

x = 4 - 3y
x= 4 - 3.(1)
x = 4 - 3  --->  x = 1


13)

a) A + B

$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ + $\left[\begin{array}{ccc}7&8\\-5&1\end{array}\right]$ =

$\left[\begin{array}{ccc}1 + 7&3 + 8\\4 + (-5)&-6 + 1\end{array}\right]$ =

$\left[\begin{array}{ccc}8&11\\4 - 5&-5\end{array}\right]$ =

$\left[\begin{array}{ccc}8&11\\-1&-5\end{array}\right]$


b) B - C

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

$\left[\begin{array}{ccc}7 - 4&8 - 2\\-5 - 0&1 - (-2)\end{array}\right]$ =

$\left[\begin{array}{ccc}3&6\\-5&1 + 2\end{array}\right]$ =

$\left[\begin{array}{ccc}3&6\\-5&3\end{array}\right]$


c) 2A + B

$2 \ . \ \left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ + $\left[\begin{array}{ccc}7&8\\-5&1\end{array}\right]$ =

$\left[\begin{array}{ccc}2 \ . \ 1&2 \ . \ 3\\2 \ . \ 4&2 \ . \ (-6)\end{array}\right]$ + $\left[\begin{array}{ccc}7&8\\-5&1\end{array}\right]$ =

$\left[\begin{array}{ccc}2&6\\8&-12\end{array}\right]$ + $\left[\begin{array}{ccc}7&8\\-5&1\end{array}\right]$ =

$\left[\begin{array}{ccc}2 + 7&6 + 8\\8 + (-5)&-12 + 1\end{array}\right]$ =

$\left[\begin{array}{ccc}9&14\\8 - 5&-11\end{array}\right]$ =

$\left[\begin{array}{ccc}9&14\\3&-11\end{array}\right]$


d) A - 3B

$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ - 3 . $\left[\begin{array}{ccc}7&8\\-5&1\end{array}\right]$ =

$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ - $\left[\begin{array}{ccc}3 \ . \ 7&3 \ . \ 8\\3 \ . \ (-5)&3 \ . \ 1\end{array}\right]$ =

$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ - $\left[\begin{array}{ccc}21&24\\-15&3\end{array}\right]$ =

$\left[\begin{array}{ccc}1 - 21&3 - 24\\4 - (-15)&-6 - 3\end{array}\right]$ =

$\left[\begin{array}{ccc}-20&-18\\4 + 15&-9\end{array}\right]$ =

$\left[\begin{array}{ccc}-20&-18\\19&-9\end{array}\right]$


e) A + 3B + 2C


$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ + 3 . $\left[\begin{array}{ccc}7&8\\-5&1\end{array}\right]$ + 2 . $\left[\begin{array}{ccc}4&2\\0&-2\end{array}\right]$ =

$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ + $\left[\begin{array}{ccc}3 \ . \ 7&3 \ . \ 8\\3 \ . \ (-5)&3 \ . \ 1\end{array}\right]$ + $\left[\begin{array}{ccc}2 \ . \ 4&2 \ . \ 2\\2 \ . \ 0&2 \ . \ (-2)\end{array}\right]$ =

$\left[\begin{array}{ccc}1&3\\4&-6\end{array}\right]$ + $\left[\begin{array}{ccc}21&24\\-15&3\end{array}\right]$ + $\left[\begin{array}{ccc}8&4\\0&-4\end{array}\right]$ =

$\left[\begin{array}{ccc}1 + 21 + 8&3 + 24 + 4\\4 + (-15) + 0&-6 + 3 + (-4)\end{array}\right]$ =

$\left[\begin{array}{ccc}30&31\\4 - 15&-6 + 3 - 4\end{array}\right]$ =

$\left[\begin{array}{ccc}30&31\\-11&-7\end{array}\right]$