Question

alguém me ajuda com a letra C e D, já tentei resolver e não consegui​

Answer 1

Explicação passo-a-passo:

c) $(5.A+B)^{t}+\frac{C}{2}$

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

   $(\left[\begin{array}{ccc}5.2&5.1\\5.3&5.2\\\end{array}\right]+\left[\begin{array}{ccc}1&5\\2&-2\\\end{array}\right])^{t}+\left[\begin{array}{ccc}2:2&0:2\\4:2&-6:2\\\end{array}\right]$

   $(\left[\begin{array}{ccc}10&5\\15&10\\\end{array}\right]+\left[\begin{array}{ccc}1&5\\2&-2\\\end{array}\right])^{t}+\left[\begin{array}{ccc}1&0\\2&-3\\\end{array}\right]$

   $(\left[\begin{array}{ccc}10+1&5+5\\15+2&10+(-2)\\\end{array}\right])^{t}+\left[\begin{array}{ccc}1&0\\2&-3\\\end{array}\right]$

   $\left[\begin{array}{ccc}11&10\\17&8\\\end{array}\right]^{t}+\left[\begin{array}{ccc}1&0\\2&-3\\\end{array}\right]$

   matriz transposta é uma matriz resultante da troca ordenadamente

   de linhas pelas colunas de outra matriz. Então:

   $\left[\begin{array}{ccc}11&17\\10&8\\\end{array}\right]+\left[\begin{array}{ccc}1&0\\2&-3\\\end{array}\right]$

   $\left[\begin{array}{ccc}11+1&17+0\\10+2&8+(-3)\\\end{array}\right]=\left[\begin{array}{ccc}12&17\\12&5\\\end{array}\right]$

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

d) $2.(A-C)+3B^{t}$

   

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

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

   $2.\left[\begin{array}{ccc}0&1\\-1&8\\\end{array}\right]+\left[\begin{array}{ccc}3.1&3.2\\3.5&3.(-2)\\\end{array}\right]$

   $\left[\begin{array}{ccc}2.0&2.1\\2.(-1)&2.8\\\end{array}\right]+\left[\begin{array}{ccc}3&6\\15&-6\\\end{array}\right]$

   $\left[\begin{array}{ccc}0&2\\-2&16\\\end{array}\right]+\left[\begin{array}{ccc}3&6\\15&-6\\\end{array}\right]$

   $\left[\begin{array}{ccc}0+3&2+6\\-2+15&16+(-6)\\\end{array}\right]=\left[\begin{array}{ccc}3&8\\13&10\\\end{array}\right]$