Let A = [ajj] be an n x n matrix with real entries such that the sum o...
Solution:
To determine whether the given matrix A is non-singular or singular and whether 0 is an eigenvalue of A, we can use the properties of the matrix and its row sums.
Properties of A:
1. A is an n x n matrix with real entries.
2. The sum of all the entries in each row is zero.
Statement (I): A is non-singular
To determine whether A is non-singular, we need to check if the determinant of A is nonzero. If the determinant is nonzero, then A is non-singular. Otherwise, A is singular.
Statement (II): A is singular
To determine whether A is singular, we need to check if the determinant of A is zero. If the determinant is zero, then A is singular. Otherwise, A is non-singular.
Statement (III): 0 is an eigenvalue of A
To determine whether 0 is an eigenvalue of A, we need to check if the equation Av = 0 has a nontrivial solution. If the equation has a nontrivial solution, then 0 is an eigenvalue of A. Otherwise, 0 is not an eigenvalue of A.
Explanation:
Since the sum of all the entries in each row of A is zero, we can say that the rows of A are linearly dependent. This implies that the rank of A is less than n.
Since the rank of A is less than n, the determinant of A is zero. Therefore, A is singular. This satisfies statement (II).
Since A is singular, the equation Av = 0 has a nontrivial solution. Therefore, 0 is an eigenvalue of A. This satisfies statement (III).
However, we cannot conclude that A is non-singular based on the given information. It is possible for a matrix to be singular even if the sum of all the entries in each row is zero. Therefore, statement (I) is not true.
Hence, the correct answer is option (c): (II) and (III) are true.
Let A = [ajj] be an n x n matrix with real entries such that the sum o...
Option b