 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/e/solution

11 - 1 22 - 2 - 1 46 41 - 7 273 \sim 100200 - 1 50 4 - 3 0 250 .

The last row establishes consistency; equivalently the third original row is the second minus twice the first, and the right-hand entries obey

3 = 7 - 2 \cdot 2

. With

x_{2} = s

and

x_{4} = 5 t

x_{3} = 1 + 3 t

and the first gives

x_{1} = 3 - 2 s - 17 t

. All real solutions are therefore

x = 3010 + s - 2 100 + t - 17 035, s, t \in R .

This affine solution space is one particular solution plus the null space. The two free variables show that there are no additional restrictions.

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