Question

46. Resolva os sistemas.
Me ajudem eu sei a resposta, mas quero saber como montar a conta ​

Answer 1

Seja X e Y as matrizes:

$X~=~\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\\\\\\Y~=~\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]$

Temos:

$\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]~+~\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]~=~\left[\begin{array}{ccc}2&12\\8&-6\end{array}\right]\\\\\\\left[\begin{array}{ccc}a+e&b+f\\c+g&d+h\end{array}\right]~=~\left[\begin{array}{ccc}2&12\\8&-6\end{array}\right]\\\\\\\left\{\begin{array}{cccc}a+e&=&2&~~~~~~~~~[Eq.1]\\b+f&=&12&~~~~~~~~~[Eq.2]\\c+g&=&8&~~~~~~~~~[Eq.3]\\d+h&=&-2&~~~~~~~~~[Eq.4]\end{array}\right$

$\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]~-~\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]~=~\left[\begin{array}{ccc}5&-7\\4&3\end{array}\right]\\\\\\\left[\begin{array}{ccc}-a+e&-b+f\\-c+g&-d+h\end{array}\right]~=~\left[\begin{array}{ccc}5&-7\\4&3\end{array}\right]\\\\\\\left\{\begin{array}{cccc}-a+e&=&5&~~~~~~~~~[Eq.5]\\-b+f&=&-7&~~~~~~~~~[Eq.6]\\-c+g&=&4&~~~~~~~~~[Eq.7]\\-d+h&=&3&~~~~~~~~~[Eq.8]\end{array}\right$

Utilizando o método da adição, vamos somar:

--> Eq.1  + Eq.5

--> Eq.2 + Eq.6

--> Eq.3 + Eq.7

--> Eq.4 + Eq.8

Chegamos aos valores:

$\rightarrow~~e~=~\frac{7}{2}~~ou~~3,5\\\\\rightarrow~~f~=~\frac{5}{2}~~ou~~2,5\\\\\rightarrow~~g~=~6\\\\\rightarrow~~h~=~-\frac{3}{2}~~ou~~-1,5$

Substituindo os valores nas equações chegaremos aos valores:

$\rightarrow~~a~=~-\frac{3}{2}~~ou~~-1,5\\\\\rightarrow~~b~=~\frac{19}{2}~~ou~~9,5\\\\\rightarrow~~c~=~2\\\\\rightarrow~~d~=~-\frac{15}{2}~~ou~~-7,5$

b)

Seja X e Y as matrizes:

$X~=~\left[\begin{array}{ccc}a&b&c\end{array}\right] \\\\\\Y~=~\left[\begin{array}{ccc}e&f&g\end{array}\right]$

Temos:

$\left[\begin{array}{ccc}a&b&c\end{array}\right]~+~\left[\begin{array}{ccc}e&f&g\end{array}\right]~=~\left[\begin{array}{ccc}3&5&2\end{array}\right]\\\\\\\left[\begin{array}{ccc}a+e&b+f&c+g\end{array}\right]~=~\left[\begin{array}{ccc}3&5&2\end{array}\right]\\\\\\\left\{\begin{array}{cccc}a+e&=&3&~~~~~~~~~[Eq.1]\\b+f&=&5&~~~~~~~~~[Eq.2]\\c+g&=&2&~~~~~~~~~[Eq.3]\end{array}\right$

$\left[\begin{array}{ccc}a&b&c\end{array}\right]~-2~\left[\begin{array}{ccc}e&f&g\end{array}\right]~=~\left[\begin{array}{ccc}-3&0&8\end{array}\right]\\\\\\\left[\begin{array}{ccc}a-2e&b-2f&c-2g\end{array}\right]~=~\left[\begin{array}{ccc}-3&0&8\end{array}\right]\\\\\\\left\{\begin{array}{cccc}a-2e&=&-3&~~~~~~~~~[Eq.4]\\b-2f&=&0&~~~~~~~~~[Eq.5]\\c-2g&=&8&~~~~~~~~~[Eq.6]\end{array}\right$

Utilizando o método da adição, vamos somar:

--> Eq.1  - Eq.4

--> Eq.2 - Eq.5

--> Eq.3 - Eq.6

Chegamos aos valores:

$\rightarrow~~e~=~0\\\\\rightarrow~~f~=~\frac{5}{3}\\\\\rightarrow~~g~=~\frac{10}{3}$

Substituindo os valores nas equações chegaremos aos valores:

$\rightarrow~~a~=~3\\\\\rightarrow~~b~=~\frac{10}{3}\\\\\rightarrow~~c~=~-\frac{4}{3}$