 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/digital-systems-linear-algebra-introduction-to-formal-proof/2/d/solution

x_{1} = - 2 x_{2} - \frac{17}{5} x_{4}, x_{3} = \frac{3}{5} x_{4} .

x_{2} = s

and

x_{4} = 5 t

ker A = ⎩ ⎨ ⎧ s - 2 100 + t - 17 035 : s, t \in R ⎭ ⎬ ⎫ .

The two displayed vectors are linearly independent, as their second and fourth coordinates show, and each satisfies

A x = 0

. Thus

a null-space basis is {(- 2, 1, 0, 0)^{T}, (- 17, 0, 3, 5)^{T}}, dim ker A = 2.

This also agrees with the rank-nullity theorem:

4 - 2 = 2

 New to topics? Read the docs here!