Question

Sejam as matrizes A = (aij)2x2, com aij­ = 2i – j2 e B = (bij)2x2, com bij­ = aij-1 Calcule:

a) A - B

b) B - A

c) (A+B)t

d) At-Bt

Answer 1

A = (aij) 2×2, aij = 2i - 2j
B = (bij) 2×2, bij = aij -1 => bij = 2i - 2j -1

A = [a11 a12] => A = [0 2] . [a21 a22] [-2 0]
a11 = 2i - 2j = 2×1 – 2×1 = 2 - 2 => 0
a12 = 2i - 2j = 2×1 - 2×2 = 2 – 4 => - 2
a21 = 2i - 2j = 2×2 - 2×1 = 4 – 2 => 2
a22 = 2i - 2j = 2×2 - 2×2 = 4 – 4 => 0

B = [b11 b12 ] => B = [-1 -3] . [b21 b22 ] [ 1 -1]
b11 =2i - 2j -1 = 2×1 - 2×1 -1 = 2 - 2 - 1 = 0 - 1 => - 1
b12 =2i - 2j -1 = 2×1 - 2×2 - 1 = 2- 4 -1 = 2 - 5 => - 3
b21 =2i - 2j -1 =2×2 - 2×1 -1 = 4 – 2 - 1 = 4 - 3 => 1
b22 =2i - 2j -1 =2×2 - 2×2 -1 = 4 - 4 -1 = 0 - 1 => - 1

a)A–B:[0-(-1) 2-(-3)] = [+1 2+3] => [1 5] . [-2-1 0-(-1)] [-3 +1] [-3 1]

b)B–A: [-1-0 -3-2] = [-1 -5] => [-1 -5] . [1-(-2) -1-0] [1+2 -1] [+3 -1]

c)A+B: [0-1 2-3] => [-1 -1] . [-2+1 0-1] [-1 -1]
(A+B)^t = [-1 -1]
[-1 -1]
d) A^t = [0 -2] B^t = [-1 1] . [2 0] [-3 -1]
A^t–B^t = [0-(-1) -2-1] => [1 -3] . [2-(-3) 0-(-1)] [5 1]

Espero ter ajudado;