Question

Alguem pfv pode me ajudar nessas duas questões pra agora​

Answer 1

Resposta:

Explicação passo-a-passo:

questão 5: Letra A, soma de matrizes

$\left[\begin{array}{ccc}4&-2\\3&5\\\end{array}\right]  +\left[\begin{array}{ccc}6&2\\0&-3\\\end{array}\right] =  \left[\begin{array}{ccc} 4+6&-2+2\\3+0&5-3\\\end{array}\right] = \left[\begin{array}{ccc}10&0\\3&2\\\end{array}\right]$

Letra B, Subtração de matrizes

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

Letra D, produto de matrizes

$\left[\begin{array}{ccc}4&-2\\3&6\\\end{array}\right] * \left[\begin{array}{ccc}6&2\\0&-3\\\end{array}\right] = \left[\begin{array}{ccc}4*6 + -2*0&4*2 + -2*-3\\3*6 + 6*0&3*2+6*-3\\\end{array}\right] = \left[\begin{array}{ccc}24&14\\18&-12\\\end{array}\right]$

Questão 6, letra A, determinante

$\left|\begin{array}{ccc}3&-4\\2&9\\\end{array}\right| = 3*9 - (-4*2) = 27 +8 = 35$

Letra B, determinante

$\left|\begin{array}{ccc}1&-3&0\\5&4&1\\-1&7&8\end{arra}\right| \left\begin{array}{ccc}1&2&\\4&5&\\7&8&\end{array}\right = (1*4*8) + (-3*1*7)+(0*4*8) -(2*4*8)-(1*1*7)-(0*4*-1) = 32 -21 -64-7=-60$