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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?

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    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?2012-05-09
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    Third possibility may be zero.But i am not sure about that.2012-05-09
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    The third possibility is indeed $0$. The [Leibniz formula](http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants) for the determinant easily implies that $1,-1$, and $0$ are the only possibilities, but you may not be familiar with it.2012-05-09
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    Unfortunately i am not familiar with it.2012-05-09
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    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.2012-05-09

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Adding or subtracting a row of a matrix from another does not change its determinant, so we may assume each column of the matrix has at most one entry that is 1.

Swapping rows of a matrix changes the sign of the determinant only; so if we perform row swaps so that the resulting matrix is diagonal, we'll have determined the determinant up to a sign.

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.)

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    The opening sentence is not clear, since subtracting a row from another may break the requirement that every row has _exactly_ on nonzero entry (which is equal to $1$), so you are not reducing to another instance of the same problem. On the other hand you may generalise the statement to require only that each row has _at most_ one nonzero entry, and then the above proof works. The only doubt remains is whether the possibilty of getting $0$ might be due _only_ to weaking the condition, but a simple example shows it is not.2014-06-30