I was almost able to solve this question. I made it to the highlighted part in the solution below, but I was unable to f

[answerhappygod]

I was almost able to solve this question. I made it tothe highlighted part in the solution below, but I was unable tofigure out how they arrive at the solution for a, b, and c. Cananyone please show me how they get a = 1/2 * (x + y - z)from the SLE ?

I Was Almost Able To Solve This Question I Made It To The Highlighted Part In The Solution Below But I Was Unable To F 1 (3.65 KiB) Viewed 59 times

Solution:

I Was Almost Able To Solve This Question I Made It To The Highlighted Part In The Solution Below But I Was Unable To F 2 (36.83 KiB) Viewed 59 times

(ii) Show that R3=span([1,1,0],[0,1,1],[1,0,1]).
i) We show that an arbitrary vector [x,y,z] can be written as a linear combination of the vectors [1,1,0],[0,1,1], and [1,0,1]. So we need to find a,b,c∈R such that a+c=xa+b=yb+c=z.​ Row reducing gives ⎣⎡​110​011​101​xyz​⎦⎤​∼⎣⎡​100​010​1−12​xy−xz−(y−x)​⎦⎤​, and we see that the solution is a=21​(x+y−z),b=21​(z+y−x),c=21​(z+x−y). Hence for any vector [x,y,z]∈R3 we have [x,y,z]=21​(x+y−z)[1,1,0]+21​(z+y−x)[0,1,1]+21​(z+x−y)[1,0,1].