Applying row reduction followed by column reduction to this cost matrix produces three independent zeros immediately, so the optimal assignment is reached without needing any adjustment step. …

Step 1 — Row reduction (subtract row minimums $4,5,4$):

Step 2 — Column reduction (column minimums: $J_1\to0$ unchanged, $J_2\to1$, $J_3\to0$ unchanged):

Step 3/4 — Cover zeros and test: the zeros are at $W_1J_1$, $W_2J_2$, $W_2J_3$, $W_3J_3$. Notice $W_1J_1$, $W_2J_2$, $W_3J_3$ already form a set of three independent zeros (no two share a row or column) — the minimum number of lines needed to cover all zeros is therefore 3 (row $W_2$ plus columns $J_1$ and $J_3$, or equivalently three lines through the independent zeros themselves), which equals $n=3$ → optimal immediately, no adjustment step needed.

Step 6 — Assignment: $W_1\to J_1$, $W_2\to J_2$, $W_3\to J_3$ — a valid one-to-one mapping, every worker and every job used exactly once. …

Assuming Step 5's adjustment is always required for every assignment problem — when the minimum line cover already equals $n$ right after row and column reduction, as here, the solution is optimal …