Question

Using MATLAB solve:

The vectors v1=(1,-1,1), v2=(0,1,2), v3=(3,0,1) span R3. Express w=(x,y,z) as linear combination of v1,v2,v3.

Answer 1

Given, v₁ = (1,-1,1), v₂ = (0,1,2), v₃ = (3,0,1).

Let (x,y,z) ne any arbitrary vector in R³.

Let us consider a relation av₁+bv₂+cv₃ = (x,y,z), where a,b,c are real numbers.

Then, a(1,-1,1)+b(0,1,2)+c(3,0,1) = (x,y,z)

i.e., a+3c = x.........(i)

-a+b = y............(ii)

a+2b+c = z.........(iii)

Now, adding (i) and (ii) we get, b+3c = y+x.........(iv)

Adding (ii) and (iii) we get, 3b+c = y+z...............(v)

Adding (iv) and (v) we get, 4b+4c = x+2y+z

i.e., b+c = (x+2y+z)/4.......(vi)

Subtracting (vi) from (iv) we get, 2c = (y+x)-(x+2y+z)/4

i.e., c = (3x+2y-z)/8

Subtracting (vi) from (v) we get, 2b = (y+z)-(x+2y+z)/4

i.e., b = (3z+2y-x)/8

Putting this value in (ii) we get, a = [(3z+2y-x)/8]-y

i.e., a = (3z-6y-x)/8

Therefore, a = (3z-6y-x)/8, b = (3z+2y-x)/8, c = (3x+2y-z)/8.

Hence, the vectors v₁ = (1,-1,1), v₂ = (0,1,2), v₃ = (3,0,1) span R³.

And, W = [(3z-6y-x)/8]v₁+[(3z+2y-x)/8]v₂+[(3x+2y-z)/8]v₃.