# $A$ be a $10\times 10$ matrix in which each row has exactly one entry equal to 1. find the possible value of the determinant

Let $A$ be a $10\times 10$ matrix in which each row has exactly one entry equal to $1$. And remaining nine entries of the row being $0$. Which of the following is not a possible value of the determinant? $0, 1 ,-1, 10$. I am able to identify for $2\times 2$ cross two matrices for which possible value of determinant is $1$ or $-1$. How to identify for such a big size matrix? Can we identify such matrix?

-
For $2 \times 2$ matrices, $1$ and $-1$ are not the only possibilities. What is the third possibility? Now try it for $3 \times 3$ matrices. What's the pattern? –  TonyK May 9 '12 at 22:32
Third possibility may be zero.But i am not sure about that. –  srijan May 9 '12 at 22:34
The third possibility is indeed $0$. The Leibniz formula for the determinant easily implies that $1,-1$, and $0$ are the only possibilities, but you may not be familiar with it. –  Brian M. Scott May 9 '12 at 22:39
Unfortunately i am not familiar with it. –  srijan May 9 '12 at 22:41
You can still prove the result by thinking about what happens when you row-reduce such a matrix, if you know how row-reduction affects the determinant. For that matter, you can see it if you think about the determinant in terms of cofactor expansions. –  Brian M. Scott May 9 '12 at 22:46
So now we have a diagonal matrix whose diagonal entries are either 1 or 0. The determinant of this matrix must be $0$ or $1$; and hence, the determinant of the original matrix must be $0$, $1$, or $-1$.
(The $-1$ possibility can arise: start with the identity matrix and interchange the last two rows. The 0 possibility can arise: start with a matrix whose first column is all $1$'s. And, of course, the identity matrix shows that $1$ is a possible value of the determinant.)