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If every minor of order r of a matrix A is zero, then rank of A is
- a)greater than r
- b)equal to r
- c)less than or equal to r
- d)less than r
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
If every minor of order r of a matrix A is zero, then rank of A isa)gr...
Explanation:
Given, every minor of order r of a matrix A is zero.
Let's first understand what is meant by minor of a matrix:
Minor of order r of a matrix A is the determinant of any square submatrix of A of size r x r.
Now, let's proceed to prove the given statement.
Proof:
Assume that rank(A) is greater than or equal to r.
Then, there exists a submatrix of A of order r that has a non-zero determinant.
This implies that the minor of order r of A is non-zero, which contradicts the given statement that every minor of order r of A is zero.
Hence, our assumption that rank(A) is greater than or equal to r is false.
Now, assume that rank(A) is less than r.
Then, there exists a linearly dependent set of r columns of A.
Without loss of generality, let these columns be the first r columns of A.
Then, any submatrix of A of order r formed by selecting any r rows and any r columns from the first r columns of A will have a determinant equal to zero, since the columns are linearly dependent.
This implies that every minor of order r of A is zero, which satisfies the given statement.
Hence, our assumption that rank(A) is less than r is true.
Therefore, the rank of A is less than r.
Hence, the correct answer is option 'D'.
Given, every minor of order r of a matrix A is zero.
Let's first understand what is meant by minor of a matrix:
Minor of order r of a matrix A is the determinant of any square submatrix of A of size r x r.
Now, let's proceed to prove the given statement.
Proof:
Assume that rank(A) is greater than or equal to r.
Then, there exists a submatrix of A of order r that has a non-zero determinant.
This implies that the minor of order r of A is non-zero, which contradicts the given statement that every minor of order r of A is zero.
Hence, our assumption that rank(A) is greater than or equal to r is false.
Now, assume that rank(A) is less than r.
Then, there exists a linearly dependent set of r columns of A.
Without loss of generality, let these columns be the first r columns of A.
Then, any submatrix of A of order r formed by selecting any r rows and any r columns from the first r columns of A will have a determinant equal to zero, since the columns are linearly dependent.
This implies that every minor of order r of A is zero, which satisfies the given statement.
Hence, our assumption that rank(A) is less than r is true.
Therefore, the rank of A is less than r.
Hence, the correct answer is option 'D'.
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If every minor of order r of a matrix A is zero, then rank of A isa)greater than rb)equal to rc)less than or equal to rd)less than rCorrect answer is option 'D'. Can you explain this answer? for IIT JAM 2025 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If every minor of order r of a matrix A is zero, then rank of A isa)greater than rb)equal to rc)less than or equal to rd)less than rCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If every minor of order r of a matrix A is zero, then rank of A isa)greater than rb)equal to rc)less than or equal to rd)less than rCorrect answer is option 'D'. Can you explain this answer?.
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