Answer:
Step-by-step explanation:
consider B in matrix form
We have
[tex]\left[\begin{array}{ccc}1&1&1\\1&1&0\\0&1&1\end{array}\right][/tex]
Reduce this to row echelon form
by R1= R1-R3
we get
[tex]\left[\begin{array}{ccc}1&0&0\\1&1&0\\0&1&1\end{array}\right][/tex]
Now R2-R1 gives Identity matrix in row echelon form. So rank =3 hence this is a basis for R cube
to find (1,2,3) as linear combination of B
Let a, b, and c be the scalars such that
a(1,1,1)+b(1,1,0)+c(0,1,1) = (1,2,3)
Equate corresponding terms as
a+b= 1: a+b+c =2: a+c =3
Solving b = -1, c = 1 and a = 2
(1,2,3) = 2(1,1,1)-1(1,1,0)+1(0,1,1)
