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?
$A$ be a $10\times 10$ matrix in which each row has exactly one entry equal to 1. find the possible value of the determinant
-
1You 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
1 Answers
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.)
-
0The 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