Linear algebra can aggregate scores from ballot data. Fix $n$ candidates and one column for each possible ranking. Let $S_{ij}$ be candidate $i$'s score in ranking $j$, and let $v_j$ count ballots of that ranking. Then the total score vector is $s=Sv$.
For three candidates with Borda scores $(2,1,0)$, two ballots $A>B>C$ and one ballot $B>C>A$ produce totals $(4,4,1)^T$. This aggregation is linear in ballot counts; converting scores into a ranking requires comparisons and tie handling.
Do not infer that every voting rule is linear. Ranked-pairs cycle avoidance, runoffs and many preference procedures are not fixed matrix multiplication. Pairwise comparisons can be stored in a matrix, but the rule applied to that matrix is a separate choice.
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