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 (resit) / 4

Source gap / TODO: the original printed page 4, confirmed visually, omits the unknown vector. It literally equates a $3\times 3$ matrix with a $3\times 1$ vector, so it does not define a well-formed system of linear equations. The following answer is conditional on the usual intended interpretation $A\mathbf{x}=(1,k^2,k{)}^T$; the missing factor is not present in the source.

Under that interpretation, the rows satisfy $r_1-2r_2+r_3=0$. Applying this linear relation to the right-hand side gives the necessary consistency conditionIt is sufficient too: the upper-left $2\times 2$ minor has determinant $4-9=-5\ne 0$, so the first two rows are linearly independent. The third row is $2r_2-r_1$, and imposes precisely the condition already found. ThereforeFor completeness, setting $x_3=t$ givesand direct substitution satisfies all three linear equations exactly for those two values of $k$. An unconditional answer to the malformed printed equality awaits a source correction.

Apply flow conservation at the four junctions, using the directions in the original diagram:The total external inflow and outflow are both 250 cars per hour. Accordingly one of these linear equations is redundant. Put $F_2=t$. Successive substitution yieldsin cars per hour. The nonnegativity restriction expresses clockwise traffic on every arc. Conversely, every $t\ge 0$ gives nonnegative flows satisfying all four conservation equations, so this is the whole physically allowable family. Increasing $t$ adds a common network circulation around the roundabout without changing any external inflow or outflow.

Since $F_1=200+t$ and $t\ge 0$, its minimum is attained at $t=0$:Zero flow on the $B\to C$ arc is allowed; no strictly positive lower bound was specified.