This question was taken from MIT OCW and Introduction to Linear Algebra by Gilbert Strang
A $9$ by $9$ Sudoku matrix $S$ has the numbers $1,...,9$ in every row and column and in every $3$ by $3$ block. For the all-ones vector $\overrightarrow { x } = (1,...,1),$ what is $S\overrightarrow { x } $?
A better question is: Which row exchanges will produce another Sudoku matrix? Also, which exchanges of block rows give another Sudoko matrix?
$6$ orders for the first $3$ rows can be seen, all giving Sudoku matrices. Also, $6$ permutations of the next $3$ rows, and of the last $4$ rows. And $6$ block permutations of the block rows?
What I understand:
The first part is pretty simple for me to understand. Without even constructing matrix $S$, it's obvious that $S\overrightarrow { x } $ will be:
$\left[\begin{array}{r} 45 \\ 45 \\ 45 \\ 45 \\ 45 \\ 45 \\ 45 \\ 45 \\ 45 \\ \end{array}\right]$
since the numbers $(1,...,9)$ added up will always be $45$.
What I don't understand:
What does the author mean by "Which row exchanges will produce another Sudoku matrix?" I want an explanation that will help me solve the last parts of this question on my own?
Answer to your question is given in 18.06 problem set solutions (page 3). Excerpt from there
Solution: For a Sudoku matrix $S$, and $x = (1, 1, ..., 1)$, $Sx$ is a column vector with 9 elements, all equal to 45:
$$Sx = (45, 45, ...45) = 45 \cdot x$$
There are 6 permutations of three numbers: $(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2) and (3,2,1)$ as mentioned in Section 2.7. Group each three rows, i.e. row 1-3, 4-6 and 7-9, and we can do row permutations inside each group, which would still give us Sudoku matrix. This in total gives $6 \cdot 6 \cdot 6$ ways of creating new (Sudoku) matrices.
And exchange the order of the three row blocks will also give us Sudoku matrix. This gives another 6 ways of permutation. Combined with the row permutation inside each group, in total, we have $6^4 = 1296$ orders of 9 rows that stay Sudoku.