Find the number of $3 \times 3$ non-singular matrices with four entries as $1$ and all others as $0$.
Hello,
I have some trouble solving the above problem.
(A similar question has been asked before, it has zero answers and also requires symmetric matrices.)
We want to ensure that no column and no row are entirely $0$.
Now, since there are $4$ ones, we can iterate on the row that has $2$ ones.
Suppose row $1$ has $2$ ones. There are $3$ ways to place them. The rest of the ones can be placed in the other two rows in $3 \cdot 3$ ways but, we wish to subtract case where one column is unattended. These are $4$ ways. So $5$ ways net.
Thus, add $3 \cdot 5 = 15$.
One for each row (where $2$ ones are there), so total = $15 \cdot 3 = \boxed{45}$
Is this correct?
One possible flaw I can see in my solution is that it assumes that $\det = 0$ only if at least one row or column is entirely $0$. Is this statement true for this problem?
EDIT: Thank you to @DatBoi for mentioning flaw in my solution. I forgot the case where any two row / column are equal. Does anybody know correct solution?
Thanks
Best Answer
Here is one way to count. Since no column can be entirely zero, the entries $1$ must be distributed with two columns containing a single such entry, and the remaining column two of them. Moreover the first two columns mentioned must have their entries $1$ in distinct rows to avoid linear dependency; given this, the only restriction on the final column is that it should not have both its entries $1$ in a same row as one of the first two columns (that would create a zero row and also make the final column the sum of the first two).
Let us count first the possibilities with the three mentioned columns in that order, and then account for possible permutations. For the first part we have $3$ possibilities for the first column, $2$ for the second, and $2$ for the third, for $3\times2\times2=12$ possibilities in all. Then we permute the final column to the first, second and third place while keeping the relative order of the first two columns, which multiplies the number of solutions by $3$. With some reflection one sees that every valid solution is accounted for in exactly one way, so the number is $12\times3=36$.