Question

determine A= (aij)3x3, com aij= (i- 3j)³

Answer 1

a11=(1-3.1)^3=(1-3)^3=(-2)^3=-8
a12=(1-3.2)^3=(1-6)^3=(-5)^3=-125
a13=(1-3.3)^3=(1-9)^3=(-8)^3=-512
a21=(2-3.1)^3=(2-3)^3=(-1)^3=-1
a22=(2-3.2)^3=(2-6)^3=(-4)^3=-64
a23=(2-3.3)^3=(2-9)^3=(-7)^3=-343
a31=(3-3.1)^3=(3-3)^3=0^3=0
a32=(3-3.2)^3=(3-6)^3=(-3)^3=-27
a33=(3-3.3)^3=(3-9)^3=(-6)^3=-216

Answer 2

A=|a11 a12 a13|      
    |a21 a22 a23|                                                        
    |a31 a32 a33|                                                        

a11=(1-3x1)^3=(1-3)^3=(-2)^3=-8            a11=-8
a12=(1-3x2)^3=(1-6)^3=(-5)^3=-125        a12=-125
a13=(1-3x3)^3=(1-9)^3=(-8)^3=-512        a13=-512

a21=(2-3x1)^3=(2-3)^3=(-1)^3=-1          a21=-1
a22=(2-3x2)^3=(2-6)^3=(-4)^3=-64        a22=-64
a23=(2-3x3)^3=(2-9)^3=(-7)^3=-243      a23=-243

a31=(3-3x1)^3=(3-3)^3=0^3=0                a31=0
a32=(3-3x2)^3=(3-6)^3=(-3)^3=-27        a32=-27
a33=(3-3x3)^3=(3-9)^3=(-6)^3=-216      a33=-216
então temos:
    |a11 a12 a13|   |-8  -125  -512|
A=|a21 a22 a23|=|-1  -64    -243|
    |a31 a31 a33|  |0    -27    -216|