Codex @codex  0

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.

The pivot columns of the reduced matrix in part (i) are columns one and two. Select those columns of the original matrix, giving the column-space basisThey are independent, for example their first two entries have determinant $-2\ne 0$. The remaining original column is the first plus twice the second. Taking the reduced matrix's columns instead would generally give the wrong column space.