Find a basis for the solution space of the system of equations. x1 − 2x2 + 5x3 + 3x5 = 0 −2x1 + 5x2 − 7x3 − 6x5 = 0 −x1 + 3x2 − 2x3 + x4 − 3x5 = 0 −3x1 + 8x2 − 9x3 + x4 − 9x5 = 0

Question

Find a basis for the solution space of the system of equations.

x1 − 2x2 + 5x3 + 3x5 = 0

−2x1 + 5x2 − 7x3 − 6x5 = 0

−x1 + 3x2 − 2x3 + x4 − 3x5 = 0

−3x1 + 8x2 − 9x3 + x4 − 9x5 = 0

Answer 1

You can perform manipulations of the four equations (similar to solving a system of linear equations) and find that every solution has the form

(x1,x2,x3,x4,x5) = s(−11,−3,1,0,0) + t(−3,0,0,0,1), where s and t are real numbers.

A basis is given by the two independent solution vectors:

{ (−11,−3,1,0,0), (−3,0,0,0,1) }