Question

7) Dadas as matrizes, calcule as matrizes:

Answer 1

Explicação passo-a-passo:

a)

$5A=5.\left[\begin{array}{ccc}2&5&0\\-1&6&-7\\\end{array}\right] \\\\5A=\left[\begin{array}{ccc}5.2&5.5&5.0\\5.(-1)&5.6&5.(-7)\\\end{array}\right] \\\\\\5A=\left[\begin{array}{ccc}10&25&0\\-5&30&-35\\\end{array}\right]$

b)

$A+B=\left[\begin{array}{ccc}2&5&0\\-1&6&-7\\\end{array}\right] +\left[\begin{array}{ccc}-4&11&3\\2&-2&-5\\\end{array}\right] \\\\A+B=\left[\begin{array}{ccc}2+(-4)&5+11&0+3\\-1+2&6+(-2)&-7+(-5)\\\end{array}\right]\\\\A+B=\left[\begin{array}{ccc}-2&16&3\\1&4&-12\\\end{array}\right]$

c)

$A-B=\left[\begin{array}{ccc}2&5&0\\-1&6&-7\\\end{array}\right] -\left[\begin{array}{ccc}-4&11&3\\2&-2&-5\\\end{array}\right] \\\\A-B=\left[\begin{array}{ccc}2-(-4)&5-11&0-3\\-1-2&6-(-2)&-7-(-5)\\\end{array}\right]\\\\A-B=\left[\begin{array}{ccc}6&-6&-3\\-3&8&-2\\\end{array}\right]$

d)

$3A=3.\left[\begin{array}{ccc}2&5&0\\-1&6&-7\\\end{array}\right] \\\\3A=\left[\begin{array}{ccc}3.2&3.5&3.0\\3.(-1)&3.6&3.(-7)\\\end{array}\right] \\\\\\3A=\left[\begin{array}{ccc}6&15&0\\-3&18&-21\\\end{array}\right]$

$4B=4.\left[\begin{array}{ccc}-4&11&3\\2&-2&-5\\\end{array}\right] \\\\4B=\left[\begin{array}{ccc}4.(-4)&4.11&4.3\\4.2&4.(-2)&4.(-5)\\\end{array}\right] \\\\\\4B=\left[\begin{array}{ccc}-16&44&12\\8&-8&-20\\\end{array}\right]$

$3A+4B=\left[\begin{array}{ccc}6&15&0\\-3&18&-21\\\end{array}\right] +\left[\begin{array}{ccc}-16&44&12\\8&-8&-20\\\end{array}\right]\\\\3A+4B=\left[\begin{array}{ccc}6-16&15+44&0+12\\-3+8&18-8&-21-20\\\end{array}\right] \\\\3A+4B=\left[\begin{array}{ccc}-10&59&12\\5&10&-41\\\end{array}\right]$