The Hungarian Method reduces the cost matrix by rows and then columns, checks whether the resulting zeros can already be covered with $n$ lines, and adjusts the matrix if not, until an optimal one-to-one assignment of independent zeros appears. …

Step 1 — Row reduction (subtract row minimums $8,7,9$):

Step 2 — Column reduction (subtract column minimums $2,0,0$):

Step 3/4 — Cover zeros and test: row $W_2$ covers $(W_2J_1,W_2J_2)$; column $J_3$ covers $(W_1J_3,W_3J_3)$ — all zeros covered with 2 lines, which is less than $n=3$, so the solution is not yet optimal.

Step 5 — Adjust: smallest uncovered entry is $W_1J_1=1$. Subtract 1 from every uncovered cell and add 1 to the intersection cell $W_2J_3$:

Step 3/4 (repeat): column $J_1$, column $J_3$, and row $W_2$ together cover all five zeros using 3 lines $=n$ → optimal.

Step 6 — Assignment: $W_3$'s only zero is at $J_3$ → assign $W_3\to J_3$; remaining $2\times2$ block gives $W_1$'s only zero at $J_1$ → assign $W_1\to J_1$; this leaves $W_2\to J_2$. …

Independent verification: recompute the total cost by summing the assigned cells directly in the ORIGINAL cost matrix (11+7+9=27), separate from the reduction-based derivation, and also check a handful of alternative permutations by hand …

Reading off the final cost from the REDUCED matrix's zero cells (which are cost differences, not real costs) instead of the original matrix; and miscounting the minimum number of cov …