Question

C= (cij) 3×4, tal que cij= (1+i)2 +j​

Answer 1

Explicação passo-a-passo:

C = c11 c12 c13 c14

c21 c22 c23 c24

c31 c32 c33 c34

cij = (1+i)² + j

c11 = (1+1)² + 1= 2² +1= 4+1=5

c12= (1+1)² + 2 = 2² +2= 4+2= 6

c13= (1+1)² +3 = 2²+3=4 +3= 7

c14= (1+1)² +4= 2²+4= 4+4= 8

c21= (1+2)²+1 = 3²+1= 9+1= 10

c22= (1+2)²+2= 3² +2= 9+2= 11

c23= (1+2)² +3= 9+3= 12

c24= 9 +4= 13

c31 = (1+3)²+1= 4²+1= 16 +1 = 17

c32= 16 +2= 18

c33= 16 +3= 19

c34= 16 +4= 20

C = 5 6 7 8

10 11 12 13

17 18 19 20