Question

Alguém consegue me ajudar a calcular o valor das matrizes ?


*Eu até consegui calcular mas não consigo chegar na resposta do gabarito de jeito nenhum :c

Answer 1

Explicação passo-a-passo:

a) 2A - B + 3C

   $2.\left[\begin{array}{ccc}0&4&-2\\6&2&8\\\end{array}\right]-\left[\begin{array}{ccc}-3&6&9\\12&-6&0\\\end{array}\right]+3.\left[\begin{array}{ccc}0&-1&0\\1&-1&2\\\end{array}\right]$

   $\left[\begin{array}{ccc}2.0&2.4&2.(-2)\\2.6&2.2&2.8\\\end{array}\right]-\left[\begin{array}{ccc}-3&6&9\\12&-6&0\\\end{array}\right]+\left[\begin{array}{ccc}3.0&3.(-1)&3.0\\3.1&3.(-1)&3.2\\\end{array}\right]$

   $\left[\begin{array}{ccc}0&8&-4\\12&4&16\\\end{array}\right]-\left[\begin{array}{ccc}-3&6&9\\12&-6&0\\\end{array}\right]+\left[\begin{array}{ccc}0&-3&0\\3&-3&6\\\end{array}\right]$

   $\left[\begin{array}{ccc}0-(-3)&8-6&-4-9\\12-12&4-(-6)&16-0\\\end{array}\right]+\left[\begin{array}{ccc}0&-3&0\\3&-3&6\\\end{array}\right]$

   $\left[\begin{array}{ccc}3&2&-13\\0&10&16\\\end{array}\right]+\left[\begin{array}{ccc}0&-3&0\\3&-3&6\\\end{array}\right]$

   $\left[\begin{array}{ccc}3+0&2+(-3)&-13+0\\0+3&10+(-3)&16+6\\\end{array}\right]=\left[\begin{array}{ccc}3&-1&-13\\3&7&22\\\end{array}\right]$

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

b) $\frac{1}{2}A-(\frac{1}{3}B+C)$

   $\frac{1}{2}.\left[\begin{array}{ccc}0&4&-2\\6&2&8\\\end{array}\right]-(\frac{1}{3}.\left[\begin{array}{ccc}-3&6&9\\12&-6&0\\\end{array}\right]+\left[\begin{array}{ccc}0&-1&0\\1&-1&2\\\end{array}\right])$

   $\left[\begin{array}{ccc}\frac{1}{2}.0&\frac{1}{2}.4&\frac{1}{2}.(-2)\\\frac{1}{2}.6 &\frac{1}{2}.2&\frac{1}{2}.8\\\end{array}\right]-(\left[\begin{array}{ccc}\frac{1}{3}.(-3) &\frac{1}{3}.6&\frac{1}{3}.9\\\frac{1}{3}.12&\frac{1}{3}.(-6)&\frac{1}{3}.0\\\end{array}\right]+\left[\begin{array}{ccc}0&-1&0\\1&-1&2\\\end{array}\right]$

   $\left[\begin{array}{ccc}0&2&-1\\3&1&4\\\end{array}\right]-(\left[\begin{array}{ccc}-1&2&3\\4&-2&0\\\end{array}\right]+\left[\begin{array}{ccc}0&-1&0\\1&-1&2\\\end{array}\right])$

   $\left[\begin{array}{ccc}0&2&-1\\3&1&4\\\end{array}\right]-(\left[\begin{array}{ccc}-1+0&2+(-1)&3+0\\4+1&-2+(-1)&0+2\\\end{array}\right])$

   $\left[\begin{array}{ccc}0&2&-1\\3&1&4\\\end{array}\right]-\left[\begin{array}{ccc}-1&1&3\\5&-3&2\\\end{array}\right]$

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