Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Digital systems, linear algebra, introduction to formal proof / 4

The third left-hand side is three times the first minus the second. Consistency of this linear system therefore requires $2=3a-a^2$, orThis is also sufficient: the first two rows are linearly independent, and on setting the free variable $x_3=t$ they giveSubstitution into the third equation gives $3a-a^2=2$ for either allowed $a$. Explicitly the solution families are $(1-t,-1-3t,t)$ and $(2-t,-3t,t)$. For every other $a$, the same row relation gives an immediate inconsistency.

Denote the columns by $c_1,\dots ,c_4$. The first two are linearly independent, because the top two coordinates have determinant $3\cdot 2-(-1)\cdot 1=7$. Direct calculation givesThus a column-space basis is $\{(3,1,9{)}^T,(-1,2,4{)}^T\}$, andFor the last description, the third row equals twice the first row plus three times the second, so every column lies in that plane. Since the two independent columns already span a two-dimensional vector subspace, they fill the plane.

The target vector satisfies $5=2(-2)+3(3)$, so it is in the column space. An especially simple linear combination isIn terms of the displayed basis, the same vector is $-c_1/7+11c_2/7$. The column space itself is unique; the freedom is in choosing its basis.

No. Take the one-by-one matrices $A=[1]$, $B=[-1]$. Both column spaces are $C=\mathbb{R}$, whereas $A+B=[0]$ has column space $\{0\}$. More generally, taking $B=-A$ cancels any nonzero column space. One always has $\mathrm{Col}(A+B)\subseteq C$, because each sum column lies in the vector subspace $C$, but equality can fail through cancellation.