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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? – 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
up vote 6 down vote accepted

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. – Marc van Leeuwen Jun 30 '14 at 12:01

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