Let $A$ be a $3\times3$ matrix whose entries in the first column are all equal to $1$. If the other entries are $-1$, $0$ or $1$, then find how many distinct values can $\det A$ take.
I want to know how this can be solved in a clever way. I found a brute force way to do it: we can subtract the first row from the second and the third, thus getting a $2\times 2$ determinant. We may now solve the problem with enough patience, but I don't think that this is the best way to tackle the problem since it was asked in a contest and due to the time limit my approach would be rendered useless.
Best Answer
I don't know if this is what you consider "brute force", but here's how I would do it:
We know that $A = \begin{bmatrix}1 & a & b\\1 & c & d\\ 1 & e & f\end{bmatrix}$
and $\det(A) = ad + cf + eb - (af + cb + ed)$
Each of those terms is $-1, 0,$ or $ 1$.
Also, the first 3 cannot be 1 with the last 3 all being -1, since that would require the number of -1 entries to be both odd and even. Nor could one of the first 3 terms be 0 without one of the last 3 terms being 0.
Therefore, $|\det(A)|\le4$.
Now, it is easy to find an $A$ which has determinant zero. Also, for any $A$, we can switch the sign of the determinant by swapping the last two columns.
So it remains to be shown that we can find $A$ with determinants 1, 2, 3 and 4.
The first two seem rather easy:
$\left| \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix}\right| = 1$
$\left| \begin{bmatrix}1 & 0 & 0\\1 & 1 & -1\\ 1 & 1 & 1\end{bmatrix}\right| = 2$
Since the bottom two entries of the first two columns are the same, we see that
$\left| \begin{bmatrix}1 & -1 & 0\\1 & 1 & -1\\ 1 & 1 & 1\end{bmatrix}\right| = 4$
To make a determinant of 3, we need one of the 2x2 submatrices to have determinant 2, and the other to have determinant 1, which suggests adding a zero:
$\left| \begin{bmatrix}1 & -1 & 0\\1 & 1 & -1\\ 1 & 0 & 1\end{bmatrix}\right| = 3$
So there are nine possible values of the determinant.