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
Posted: Wed Jul 13, 2022 5:08 am
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 ?
Solution:
(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 ⎣⎡110011101xyz⎦⎤∼⎣⎡1000101−12xy−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].
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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 ⎣⎡110011101xyz⎦⎤∼⎣⎡1000101−12xy−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].