Question

sendo as matrizes a e b de ordem 2 e lei de formação a) A+B b) B-A c) 2(A+B)^T D)A x B

Answer 1

Resposta:

Veja a explicação abaixo.

Explicação passo-a-passo:

Temos que:

i → linha

j → coluna

Ele nos disse que a matriz é de ordem 2, isto é, a matriz é quadrada com 2 linhas e 2 colunas:

Uma matriz genérica (M) de ordem 2 é do tipo:

$M = \left[\begin{array}{cc}a_{11} &a_{12} \\a_{21} &a_{22} \\\end{array}\right]$

Agora, vamos estabelecer a matriz A:

A = ($a_{ij}$) = 3i + 2j

$a_{11} = 3(1) + 2(1) =3 + 2 = 5\\a_{12} = 3(1) + 2(2) = 3 + 4 = 7\\a_{21} = 3(2) + 2(1) = 6 + 2 = 8\\a_{22} = 3(2) + 2(2) = 6 + 4 = 10$

Sendo assim, temos que

$A=\left[\begin{array}{cc}5&7\\8&10\\\end{array}\right]$

Vamos então, estabelecer a matriz B:

B = ($b_{ij}$) = i + j

$b_{11} = 1 + 1 = 2 \\b_{12} = 1 + 2 = 3\\b_{21} = 2 + 1 = 3\\b_{22} = 2 + 2 = 4$

Logo,

$B=\left[\begin{array}{cc}2&3\\3&4\\\end{array}\right]$

Encontradas as matrizes A e B, podemos responder as questões:

a) A + B =

$\left[\begin{array}{cc}5&7\\8&10\\\end{array}\right]+\left[\begin{array}{cc}2&3\\3&4\\\end{array}\right]=\left[\begin{array}{cc}5+2&7+3\\8+3&10+4\\\end{array}\right]=\left[\begin{array}{cc}7&10\\11&14\\\end{array}\right]$

b) B - A

$\left[\begin{array}{cc}2&3\\3&4\\\end{array}\right] - \left[\begin{array}{cc}5&7\\8&10\\\end{array}\right]=\left[\begin{array}{cc}2-5&3-7\\3-8&4-10\\\end{array}\right]=\left[\begin{array}{cc}-3&-4\\-5&-6\\\end{array}\right]$

$c) 2.(A+B)^{t}$

da letra (a), sabemos que

$A+B=\left[\begin{array}{cc}7&10\\11&14\\\end{array}\right]$, então a transposta de A+B:

$(A+B)^{t} =\left[\begin{array}{cc}7&11\\10&14\\\end{array}\right]$

calculando $2.(A+B)^{t}$:

$2.\left[\begin{array}{cc}7&11\\10&14\\\end{array}\right]=\left[\begin{array}{cc}2.7&2.11\\2.10&2.14\\\end{array}\right]=\left[\begin{array}{cc}14&22\\20&28\\\end{array}\right]$

d) A.B =

$\left[\begin{array}{cc}5&7\\8&10\\\end{array}\right].\left[\begin{array}{cc}2&3\\3&4\\\end{array}\right]= \left[\begin{array}{cc}5\cdot \:2+7\cdot \:3&5\cdot \:3+7\cdot \:4\\8\cdot \:2+10\cdot \:3&8\cdot \:3+10\cdot \:4\\\end{array}\right]=\left[\begin{array}{cc}31&43\\46&64\\\end{array}\right]$