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Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Then
- a)rank (T) = 0, nullity (T) = 3
- b)rank (T) = 2, nullity (T) = 1
- c)rank (T) = 3, nullity (T) = 0
- d)rank (T) = 1, nullity (T) = 2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let T : R3 → R3 be the linear transformation define by T(x, y, z)...
Let T : R2 → R3 be the L.T. defined as
T (x, y, z) = (x + y, y + z, z + x) ∀ (x, y, z) ∈ R3
N (T ) = {( x, y, z)|T ( x, y, z) = 0}
= {(x, y, z)|x + y = 0,y + z = 0, z + x = 0}
= {(0, 0, 0)| x, y, z ∈R}
⇒ dim N (T) = 0
Rank (N) + Nullity T = dim R3 = 3
Rank (T) + 0 = 3 ⇒ Rank of T = 3
T (x, y, z) = (x + y, y + z, z + x) ∀ (x, y, z) ∈ R3
N (T ) = {( x, y, z)|T ( x, y, z) = 0}
= {(x, y, z)|x + y = 0,y + z = 0, z + x = 0}
= {(0, 0, 0)| x, y, z ∈R}
⇒ dim N (T) = 0
Rank (N) + Nullity T = dim R3 = 3
Rank (T) + 0 = 3 ⇒ Rank of T = 3
Most Upvoted Answer
Let T : R3 → R3 be the linear transformation define by T(x, y, z)...
R2 be a linear transformation given by T(x, y, z) = (2x - y, 3y + z). In other words, T takes a vector in R3 and maps it to a vector in R2 by multiplying the first component by 2, subtracting the second component, and adding the third component to the second component.
To find the standard matrix of T, we need to find the images of the standard basis vectors in R3 under T. The standard basis vectors in R3 are e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1).
Applying T to each of these vectors, we have:
T(e1) = T(1, 0, 0) = (2(1) - 0, 3(0) + 0) = (2, 0)
T(e2) = T(0, 1, 0) = (2(0) - 1, 3(1) + 0) = (-1, 3)
T(e3) = T(0, 0, 1) = (2(0) - 0, 3(0) + 1) = (0, 1)
Therefore, the images of the standard basis vectors under T are (2, 0), (-1, 3), and (0, 1).
The standard matrix of T is formed by taking the images of the standard basis vectors as columns:
[2 -1 0]
[0 3 1]
So, the standard matrix of T is:
[2 -1 0]
[0 3 1]
To find the standard matrix of T, we need to find the images of the standard basis vectors in R3 under T. The standard basis vectors in R3 are e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1).
Applying T to each of these vectors, we have:
T(e1) = T(1, 0, 0) = (2(1) - 0, 3(0) + 0) = (2, 0)
T(e2) = T(0, 1, 0) = (2(0) - 1, 3(1) + 0) = (-1, 3)
T(e3) = T(0, 0, 1) = (2(0) - 0, 3(0) + 1) = (0, 1)
Therefore, the images of the standard basis vectors under T are (2, 0), (-1, 3), and (0, 1).
The standard matrix of T is formed by taking the images of the standard basis vectors as columns:
[2 -1 0]
[0 3 1]
So, the standard matrix of T is:
[2 -1 0]
[0 3 1]
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Let T : R3 → R3 be the linear transformation define by T(x, y, z)...
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Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer?
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Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer?.
Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T : R3 → R3 be the linear transformation define by T(x, y, z) = (x + y, y+ z, z + x) for all (x, y, z) ∈ 3. Thena)rank (T) = 0, nullity (T) = 3b)rank (T) = 2, nullity (T) = 1c)rank (T) = 3, nullity (T) = 0d)rank (T) = 1, nullity (T) = 2Correct answer is option 'C'. Can you explain this answer?.
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