Let s u, v be a linearly independent set. prove that the {u + v, u - v} is linearly independent

This means that there exist constant a,b such that if:

                                au+bv=0

                             then a=b=0

{u+v,u-v} is a linearly independent set.

Let us consider there exists constant c,d such that:

                            c(u+v)+d(u-v)=0

To show:   c=d=0

The expression could also be written as:

 cu+cv+du-dv=0

( Since, using the distributive property)

Now on combining the like terms that is the terms with same vectors.

cu+du+cv-dv=0

i.e.

(c+d)u+(c-d)v=0

Since, we are given that u and v are linearly independent vectors this means that:

c+d=0(1)

and c-d=0 i.e c=d(2)

and from equation (1) using equation (2) we have:

2c=0

i.e. c=0

and similarly by equation (2) we have:

         d=0

Hence, we are proved with the result.

We get that the vectors {u+v,u-v} is linearly independent.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer:   2 multiplied by x =24

step-by-step explanation: