Answer with Step-by-step explanation:
We are given that
[tex]x_1,x_2[/tex] and [tex]x_3[/tex] are linearly independent.
By definition of linear independent there exits three scalar [tex]a_1,a_2[/tex] and [tex]a_3[/tex] such that
[tex]a_1x_1+a_2x_2+a_3x_3=0[/tex]
Where [tex]a_1=a_2=a_3=0[/tex]
[tex]y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1[/tex]
We have to prove that [tex]y_1,y_2[/tex] and [tex]y_3[/tex] are linearly independent.
Let [tex]b_1,b_2[/tex] and [tex]b_3[/tex] such that
[tex]b_1y_1+b_2y_2+b_3y_3=0[/tex]
[tex]b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0[/tex]
[tex]b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0[/tex]
[tex](b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0[/tex]
[tex]b_1+b_3=0[/tex]
[tex]b_1=-b_3[/tex]...(1)
[tex]b_1+b_2=0[/tex]
[tex]b_1=-b_2[/tex]..(2)
[tex]b_2+b_3=0[/tex]
[tex]b_2=-b_3[/tex]..(3)
Because [tex]x_1,x_2[/tex] and [tex]x_3[/tex] are linearly independent.
From equation (1) and (3)
[tex]b_1=b_2[/tex]...(4)
Adding equation (2) and (4)
[tex]2b_1==0[/tex]
[tex]b_1=0[/tex]
From equation (1) and (2)
[tex]b_3=0,b_2=0,b_3=0[/tex]
Hence, [tex]y_1,y_2[/tex] and [tex]y_3[/tex] area linearly independent.
