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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'.
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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?
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