Question

Alguém me ajuda pfvr ​

Answer 1

Resposta:

Alternativa d)

$D=\left[\begin{array}{ccc}-8&-9&16\\2&4&10\\10&5&5\end{array}\right]$

Explicação passo a passo:

$A=\left[\begin{array}{ccc}1&2&3\\-4&5&6\\4&6&8\end{array}\right] \\\\\\B=\left[\begin{array}{ccc}-7&-8&9\\12&6&5\\8&7&4\end{array}\right] \\\\\\C=\left[\begin{array}{ccc}2&3&-4\\6&7&1\\2&8&7\end{array}\right]$

Para fazer a operacao A+B-C basta somar ou subtrair cada numero de cada matriz com o numero na posicao correspodente da outra matriz (essa parte ficou confusa, mas se faz da seguinte forma):

$A+B=\left[\begin{array}{ccc}(1+(-7))&(2+(-8))&(3+9)\\(-4+12)&(5+6)&(6+5)\\(4+8)&(6+7)&(8+4)\end{array}\right] = \left[\begin{array}{ccc}-6&-6&12\\8&11&11\\12&13&12\end{array}\right]$

Agora fazemos a subtração (A+B)-C:

$A+B-C=D=\left[\begin{array}{ccc}-6&-6&12\\8&11&11\\12&13&12\end{array}\right]-\left[\begin{array}{ccc}2&3&-4\\6&7&1\\2&8&7\end{array}\right]$

$D=\left[\begin{array}{ccc}(-6-2)&(-6-3)&(12-(-4))\\(8-6)&(11-7)&(11-1)\\(12-2)&(13-8)&(12-7)\end{array}\right]$

$D=\left[\begin{array}{ccc}-8&-9&16\\2&4&10\\10&5&5\end{array}\right]$

Podíamos, ainda, ter feito a conta direto, formando a seguinte matriz:

$D=\left[\begin{array}{ccc}(1+(-7)-2)&(2+(-8)-3)&(3+9-(-4)\\(-4+12-6)&(5+6-7)&(6+5-1)\\(4+8-2)&6+7-8)&(8+4-7)\end{array}\right]$

$D=\left[\begin{array}{ccc}-8&-9&16\\2&4&10\\10&5&5\end{array}\right]$

Espero ter ajudado :)