Question

ALGUÉM BOM EM MATEMÁTICA PODERIA ME AJUDAR NESSAS QUESTÕES????​

Answer 1

Resposta:

Explicação passo-a-passo:

$3(\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right]) - (\left[\begin{array}{ccc}a\\b\\c\\\end{array}\right]) = 2(\left[\begin{array}{ccc}2\\3\\0\\\end{array}\right]) - (\left[\begin{array}{ccc}-1\\0\\2\\\end{array}\right])\\\\(\left[\begin{array}{ccc}3x\\3y\\3z\\\end{array}\right]) - (\left[\begin{array}{ccc}a\\b\\c\\\end{array}\right]) = (\left[\begin{array}{ccc}4\\6\\0\\\end{array}\right])-(\left[\begin{array}{ccc}-1\\0\\2\\\end{array}\right])$

$(\left[\begin{array}{ccc}3x\\3y\\3z\\\end{array}\right]) - (\left[\begin{array}{ccc}a\\b\\c\\\end{array}\right]) = (\left[\begin{array}{ccc}5\\6\\-2\\\end{array}\right])\\3x-a = 5\\3y - b = 6\\3z - c = -2$

x + y = A -B

$(\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right]) + (\left[\begin{array}{ccc}a\\b\\c\\\end{array}\right]) = (\left[\begin{array}{ccc}3\\3\\-2\\\end{array}\right])\\x+a = 3\\y+b = 3\\z + c = -2$

se x+a = 3

e 3x - a = 5

vamos utilizar a substituiçao a = 3-x entao

3x - (3-x) = 5

3x -3 + x = 5

4x = 5 + 3

x = 2    entao a = 1

se 3y -b = 6  e

y + b = 3

vamos utilizar a substituiçao b = 3 -y  entao

3y - (3 - y) = 6

3y -3 + y = 6

4y = 9

y = 9/ 4  e b = 3/4

Se z + c = -2 e

3z - c = -2  

vamos usar a substituiçao c = -2 - z

entao

3z - (-2-z) = -2

3z + 2 + z = -2

4z = -2 - 2

4z = - 4

z = -1    e c = -1

Agora temos tudo vamos formar as matrizes X e Y

$X = (\left[\begin{array}{ccc}2\\\frac{9}{4} \\-1\\\end{array}\right])$

$Y = (\left[\begin{array}{ccc}1\\\frac{3}{4} \\-1\\\end{array}\right])$

Espero ter ajudado! Imagina se eu tivesse feito todas, iria ficar gigante!!! ;3