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If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, then
- a)(x - k)r is a factor of |A|
- b)(x - k)r - 1 is a factor of |A|
- c)(x - k)r + 1 is a factor of |A|
- d)(x - k)r is a factor of A
Correct answer is option 'A'. Can you explain this answer?
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If A is a square matrix of order n such that its elements are polynomi...
Explanation:
Let A be a square matrix of order n such that its elements are polynomial in x. Let the r-rows of A become identical for x = k. Then we can write:
A = [a1(x), a2(x), ..., ar-1(x), ar(x), ..., an(x)]
where ar(x) = ar-1(x) for x = k. Let B be the matrix obtained from A by subtracting ar-1(x) from ar(x) in the r-th row. Then we have:
B = [a1(x), a2(x), ..., ar-1(x), 0, ..., an(x)] - [0, 0, ..., 0, ar-1(x) - ar-1(x), ..., 0]
= [a1(x), a2(x), ..., ar-1(x), 0, ..., an(x)] - [0, 0, ..., 0, 0, ..., 0]
= [a1(x), a2(x), ..., ar-1(x), ar-1(x) - ar(x), ..., an(x)]
= [a1(x), a2(x), ..., ar-1(x), (x - k)ar-1(x), ..., an(x)]
Therefore, the r-th row of B is:
[(x - k)ar-1(x), 0, ..., 0]
Now let C be the matrix obtained from B by dividing the r-th row by (x - k)r. Then we have:
C = [a1(x), a2(x), ..., ar-1(x), ar-1(x) / (x - k)r, ..., an(x)]
Therefore, the determinant of C is:
|C| = (ar-1(x) / (x - k)r) times the determinant of the matrix obtained from C by deleting the r-th row.
Since ar-1(x) = ar(x) for x = k, we have:
|C| = (ar-1(k) / (k - k)r) times the determinant of the matrix obtained from C by deleting the r-th row.
= 0 times the determinant of the matrix obtained from C by deleting the r-th row.
Therefore, (x - k)r is a factor of |C|. Since A is obtained from C by multiplying the r-th row by (x - k)r, we have:
|A| = (x - k)r times |C|
Therefore, (x - k)r is a factor of |A|. Hence, option (A) is correct.
Let A be a square matrix of order n such that its elements are polynomial in x. Let the r-rows of A become identical for x = k. Then we can write:
A = [a1(x), a2(x), ..., ar-1(x), ar(x), ..., an(x)]
where ar(x) = ar-1(x) for x = k. Let B be the matrix obtained from A by subtracting ar-1(x) from ar(x) in the r-th row. Then we have:
B = [a1(x), a2(x), ..., ar-1(x), 0, ..., an(x)] - [0, 0, ..., 0, ar-1(x) - ar-1(x), ..., 0]
= [a1(x), a2(x), ..., ar-1(x), 0, ..., an(x)] - [0, 0, ..., 0, 0, ..., 0]
= [a1(x), a2(x), ..., ar-1(x), ar-1(x) - ar(x), ..., an(x)]
= [a1(x), a2(x), ..., ar-1(x), (x - k)ar-1(x), ..., an(x)]
Therefore, the r-th row of B is:
[(x - k)ar-1(x), 0, ..., 0]
Now let C be the matrix obtained from B by dividing the r-th row by (x - k)r. Then we have:
C = [a1(x), a2(x), ..., ar-1(x), ar-1(x) / (x - k)r, ..., an(x)]
Therefore, the determinant of C is:
|C| = (ar-1(x) / (x - k)r) times the determinant of the matrix obtained from C by deleting the r-th row.
Since ar-1(x) = ar(x) for x = k, we have:
|C| = (ar-1(k) / (k - k)r) times the determinant of the matrix obtained from C by deleting the r-th row.
= 0 times the determinant of the matrix obtained from C by deleting the r-th row.
Therefore, (x - k)r is a factor of |C|. Since A is obtained from C by multiplying the r-th row by (x - k)r, we have:
|A| = (x - k)r times |C|
Therefore, (x - k)r is a factor of |A|. Hence, option (A) is correct.
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If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer?
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If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer?.
If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If A is a square matrix of order n such that its elements are polynomial in x and its r-rows become identical for x = k, thena)(x - k)r is a factor of |A|b)(x - k)r - 1 is a factor of |A|c)(x - k)r + 1 is a factor of |A|d)(x - k)r is a factor of ACorrect answer is option 'A'. Can you explain this answer?.
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