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Which of the following sets is not linearly independent?
- a){1, x, 1 + x + x2} in a vector space of all polynomials over the field of real numbers
- b){1, x, x2,.......} in a vector space of all polynomials over the field of real numbers
- c){ ( 1 . 1 , 0 . 0), (0, 1 , - 1 , 0), (0, 0, 0. 3)} in V4(R )
- d){ ( 1 ,2 ,1 ) , ( 3 ,1 ,5 ) , (3, -4 , 7)} in V3(R)}
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Which of the following sets is not linearly independent?a){1, x, 1 + x...
(A) Let a, b, c be scalars such that
bers.
(B) Let 
be any finite subset of S having n vectors .
be any finite subset of S having n vectors .Here, m1,m2 .....mn are some non-negative integers. Let a1,a2 .....an, be scalars such that
By definition of equality of two polynomials,
∴ S is linearly independent.
(C) Let a, b, c be scalars such that a (1 , 1 ,0 , 0) + b(0, 1, - 1 , 0) + c(0, 0, 0, 3) = ( 0 , 0 , 0 , 0 )
⇒ (a, a + b, - b , 3c) = (0, 0, 0, 0)
⇒ (a, a + b, - b , 3c) = (0, 0, 0, 0)
⇒ a = 0, b = 0, c = 0
Hence, given set of vectors in V4 (R) is linearly independent.
(D) Let a, b, c be scalars such that a(1, 2 , 1 ) + b ( 3 , 1 , 5 ) + c (3, - 4 , 7) = (0, 0, 0)
⇒ (a + 3b + 3c, 2a + b - 4c, a + 5b + 7c) = (0, 0, 0)
⇒ a + 3b + 3c = 0, 2a + b - 4c = 0, a + 5b + 7c = 0
⇒ |A| = 0
∴ Rank (A) < 3 =* given set of vectors is linearly dependent.
Most Upvoted Answer
Which of the following sets is not linearly independent?a){1, x, 1 + x...
(A) Let a, b, c be scalars such that
bers.
(B) Let 
be any finite subset of S having n vectors .
be any finite subset of S having n vectors .Here, m1,m2 .....mn are some non-negative integers. Let a1,a2 .....an, be scalars such that
By definition of equality of two polynomials,
∴ S is linearly independent.
(C) Let a, b, c be scalars such that a (1 , 1 ,0 , 0) + b(0, 1, - 1 , 0) + c(0, 0, 0, 3) = ( 0 , 0 , 0 , 0 )
⇒ (a, a + b, - b , 3c) = (0, 0, 0, 0)
⇒ (a, a + b, - b , 3c) = (0, 0, 0, 0)
⇒ a = 0, b = 0, c = 0
Hence, given set of vectors in V4 (R) is linearly independent.
(D) Let a, b, c be scalars such that a(1, 2 , 1 ) + b ( 3 , 1 , 5 ) + c (3, - 4 , 7) = (0, 0, 0)
⇒ (a + 3b + 3c, 2a + b - 4c, a + 5b + 7c) = (0, 0, 0)
⇒ a + 3b + 3c = 0, 2a + b - 4c = 0, a + 5b + 7c = 0
⇒ |A| = 0
∴ Rank (A) < 3 =* given set of vectors is linearly dependent.
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Community Answer
Which of the following sets is not linearly independent?a){1, x, 1 + x...
Explanation:
In order to determine which set is not linearly independent, we need to check if any vector in the set can be written as a linear combination of the other vectors in the set. If we find such a combination, then the set is not linearly independent.
Option A: {1, x, 1, x, x^2} in a vector space of all polynomials over the field of real numbers
To check if this set is linearly independent, we set up the following equation:
a(1) + b(x) + c(1) + d(x) + e(x^2) = 0
where a, b, c, d, and e are coefficients. For this equation to hold for all values of x, we must have a = c = 0 and b = d = e = 0.
Therefore, the set {1, x, 1, x, x^2} is linearly independent.
Option B: {1, x, x^2, ...} in a vector space of all polynomials over the field of real numbers
This set is not linearly independent because any polynomial in the form of a constant multiple of x^n, where n is a non-negative integer, can be written as a linear combination of the other polynomials in the set.
Option C: {(1, 1, 0, 0), (0, 1, -1, 0), (0, 0, 0, 3)} in V4(R)
To check if this set is linearly independent, we set up the following equation:
a(1, 1, 0, 0) + b(0, 1, -1, 0) + c(0, 0, 0, 3) = (0, 0, 0, 0)
This equation can only hold if a = b = c = 0.
Therefore, the set {(1, 1, 0, 0), (0, 1, -1, 0), (0, 0, 0, 3)} is linearly independent.
Option D: {(1, 2, 1), (3, 1, 5), (3, -4, 7)} in V3(R)
To check if this set is linearly independent, we set up the following equation:
a(1, 2, 1) + b(3, 1, 5) + c(3, -4, 7) = (0, 0, 0)
Expanding this equation gives:
(a + 3b + 3c, 2a + b - 4c, a + 5b + 7c) = (0, 0, 0)
This equation can only hold if a = b = c = 0.
Therefore, the set {(1, 2, 1), (3, 1, 5), (3, -4, 7)} is linearly independent.
Conclusion:
The set that is not linearly independent is option D: {(1, 2, 1), (3, 1, 5), (3, -4,
In order to determine which set is not linearly independent, we need to check if any vector in the set can be written as a linear combination of the other vectors in the set. If we find such a combination, then the set is not linearly independent.
Option A: {1, x, 1, x, x^2} in a vector space of all polynomials over the field of real numbers
To check if this set is linearly independent, we set up the following equation:
a(1) + b(x) + c(1) + d(x) + e(x^2) = 0
where a, b, c, d, and e are coefficients. For this equation to hold for all values of x, we must have a = c = 0 and b = d = e = 0.
Therefore, the set {1, x, 1, x, x^2} is linearly independent.
Option B: {1, x, x^2, ...} in a vector space of all polynomials over the field of real numbers
This set is not linearly independent because any polynomial in the form of a constant multiple of x^n, where n is a non-negative integer, can be written as a linear combination of the other polynomials in the set.
Option C: {(1, 1, 0, 0), (0, 1, -1, 0), (0, 0, 0, 3)} in V4(R)
To check if this set is linearly independent, we set up the following equation:
a(1, 1, 0, 0) + b(0, 1, -1, 0) + c(0, 0, 0, 3) = (0, 0, 0, 0)
This equation can only hold if a = b = c = 0.
Therefore, the set {(1, 1, 0, 0), (0, 1, -1, 0), (0, 0, 0, 3)} is linearly independent.
Option D: {(1, 2, 1), (3, 1, 5), (3, -4, 7)} in V3(R)
To check if this set is linearly independent, we set up the following equation:
a(1, 2, 1) + b(3, 1, 5) + c(3, -4, 7) = (0, 0, 0)
Expanding this equation gives:
(a + 3b + 3c, 2a + b - 4c, a + 5b + 7c) = (0, 0, 0)
This equation can only hold if a = b = c = 0.
Therefore, the set {(1, 2, 1), (3, 1, 5), (3, -4, 7)} is linearly independent.
Conclusion:
The set that is not linearly independent is option D: {(1, 2, 1), (3, 1, 5), (3, -4,
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