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**iExcavate** Sorry this might be a bit messy! I apologize!

The subset S = {v1, v2, v3, v4, v5, v6, v7} of R^5 is given by:

v1 = (-2, -2, -1, 1, 2), v2 = (3, 3, 2, -1, -3), v3 = (5, 5, 4, -1, -5), v4 = (3, 2, 2, -2, -4), v5 = (4, 3, 3, -2, -5), v6 = (-2, -2, -1, 2, 3), v7 = (-3, -4, -1, 3, 4)

The reduced row echelon form of the matrix A is the matrix B where:

A = $\displaystyle \begin{bmatrix} -2 & 3 & 5 & 3 & 4 & -2 & -3 \\ -2 & 3 & 5 & 2 & 3 & -2 & -4 \\ 1 & 2 & 4 & 2 & 3 & -1 & -1 \\ 1 & -1 & -1 & -2 & -2 & 2 & 3 \\ 2 & -3 & -5 & -4 & -5 & 3 & 4 \end{bmatrix}$

AND

B = $\displaystyle \begin{bmatrix} 1 & 0 & 2 & 0 & 1 & 0 & 1 \\ 0 & 1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$

**(A) Write down a four element subset S1 of S so that every vector which is a linear combination of elements of S is already a linear combination of elements of S1.**

I have:

Columns associated with v1, v2, v4 and v6 have leading entries, thus s1 = v1=a, v2 = b, v4 = c, v6 = d

(where a, b, c, d are real values)

(B) Express each of the three elements of S which are not in S1 as a linear combination of elements of S1.

I currently have:

v3 = 2v1 + 3v2 (this is correct)

v5 = v1 + v2 + v3 (this is not correct)

v7 = v1 + v3 + 2v4 (this is not correct)

That took a while to write up!

Help, maybe from scratch?

Thanks!