Find the basis for the solution space in R^5 of this linear system. Find this subspace dimension:

[1 2 3 4 5

2 -1 2 -1 2

0 5 4 9 8

1 7 7 13 13]

;; 1x+ 2y + 3z + 4s +5d = 0

;; 2x + -1y + 2z + -1s + 2d = 0

;; 0x + 5y + 4z + 9s + 8d = 0

;; 1x + 7y + 7z + 13s + 13d = 0

row reduction =

1 0 7/5 2/5 9/5

0 1 4/5 9/5 8/5

0 0 0 0 0

0 0 0 0 0

so x3 x4 x5 are free

x1 = -7/5 x3 - 2/5 x4 - 9/5 x5

x2 = -4/5 x3 - 9/5 x4 - 8/5 x5

therefore linearly independent and form a basis for their span of 3 dimensions

is this correct ?