9.5. In each cell of a 10 by 10 table, a minus sign is written. In one operation, it is allowed to simultaneously change the signs to their opposites in all cells of some column and some row (plus to minus and vice versa). What is the minimum number of operations required to make all the signs in the table pluses?
Answer. In 100 operations.

Solution. There are 19 cells in total in the row and column passing through a given cell, so if we perform operations on all pairs of rows and columns of the table (a total of $10 \times 10=100$ operations), each sign in the table will change 19 times, turning from minus to plus, so 100 operations are sufficient.

We will call the operation of changing the signs in all cells of a certain column and a certain row an operation relative to the cell at the intersection of these row and column. Cells relative to which we performed operations will be called red, and the rest - blue. Rows and columns containing an even number of red cells will be called even, and those containing an odd number of red cells - odd. Suppose it is possible to change all the signs in the table in fewer than 100 operations, then consider some blue cell A in row X and column Y. For the sign in A to change, row X and column Y together must contain an odd number of red cells, we can assume row X is even, and column Y is odd. Note that at the intersection of rows and columns of the same parity, there should be a red cell, and at the intersection of rows and columns of different parity - a blue cell, otherwise the sign in this cell will not change after all operations. Therefore, the number of red cells in each even row is equal to the number of even columns, and the number of blue cells is equal to the number of odd columns in the table. There is at least one even row X, so there is an even number of odd columns in the table. But the number of red cells in each odd row (odd!) is equal to the number of odd columns, which is an even number - a contradiction with the fact that there is at least one odd column. Therefore, it is impossible to manage with fewer than 100 operations.

Instructions. Correct example for 100 operations with explanations: 2 points. Proof of the minimality of $100: 5$ points.