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Let the linear transformation S and T : 
be defined by
S(x, y, z) = (2x, 4x-y, 2x + 3y-z)
and T(x, y, z) = (x cos θ - y sin θ, x sin θ + y cos θ, z)
where 0 < θ < π/2. then,
be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)
and T(x, y, z) = (x cos θ - y sin θ, x sin θ + y cos θ, z)
where 0 < θ < π/2. then,
- a)S is one to one but not T
- b)T is one to one but not S
- c)both S and T are one to one
- d)neither S nor T is one-to-one
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Let the linear transformation S and T :be defined byS(x, y, z) = (2x, ...
We are given that two linear transformations S and T :
R3 --> R3 defined by
S(x, y, z) = (2x, 4x - y, 2x + 3y - z) and T(x, y, z) = (x cos θ - y sin θ, x sin θ - y cos θ, z)
Let (x,y,z) ∈ ker.S
then S(x, y, z) =(0, 0, 0)
Using the definition o f linear transformation, we get
(2x, 4x - y, 2x + 3y - z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
2x = 0
4x - y = 0
2x + 3y - z = 0
On solving these equation for x, y, z, we get
x = 0, y = 0 and z = 0
Hence,ker S = {0, 0, 0}
Thus, S is one-one.
Now, since A is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore S is onto.
Next Let (x, y, z) ∈ ker T
then T(x, y, z) = (0, 0,0)
Using the definition of linear transformation, we get (x cosθ - y sinθ, x sinθ + y cosθ, z) = (0, 0, 0)
Comparing the components of the co-ordinates, we get
x cosθ - y sinθ = 0
x sinθ + y cosθ = 0
z = 0
Now, solving for x, y and z, we get
x = 0 , y = 0 and z = 0
hence, ker T = {0 , 0, 0 }
Therefore, T is one-one
Now, since T is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore T is onto.
R3 --> R3 defined by
S(x, y, z) = (2x, 4x - y, 2x + 3y - z) and T(x, y, z) = (x cos θ - y sin θ, x sin θ - y cos θ, z)
Let (x,y,z) ∈ ker.S
then S(x, y, z) =(0, 0, 0)
Using the definition o f linear transformation, we get
(2x, 4x - y, 2x + 3y - z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
2x = 0
4x - y = 0
2x + 3y - z = 0
On solving these equation for x, y, z, we get
x = 0, y = 0 and z = 0
Hence,ker S = {0, 0, 0}
Thus, S is one-one.
Now, since A is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore S is onto.
Next Let (x, y, z) ∈ ker T
then T(x, y, z) = (0, 0,0)
Using the definition of linear transformation, we get (x cosθ - y sinθ, x sinθ + y cosθ, z) = (0, 0, 0)
Comparing the components of the co-ordinates, we get
x cosθ - y sinθ = 0
x sinθ + y cosθ = 0
z = 0
Now, solving for x, y and z, we get
x = 0 , y = 0 and z = 0
hence, ker T = {0 , 0, 0 }
Therefore, T is one-one
Now, since T is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore T is onto.
Most Upvoted Answer
Let the linear transformation S and T :be defined byS(x, y, z) = (2x, ...
We are given that two linear transformations S and T :
R3 --> R3 defined by
S(x, y, z) = (2x, 4x - y, 2x + 3y - z) and T(x, y, z) = (x cos θ - y sin θ, x sin θ - y cos θ, z)
Let (x,y,z) ∈ ker.S
then S(x, y, z) =(0, 0, 0)
Using the definition o f linear transformation, we get
(2x, 4x - y, 2x + 3y - z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
2x = 0
4x - y = 0
2x + 3y - z = 0
On solving these equation for x, y, z, we get
x = 0, y = 0 and z = 0
Hence,ker S = {0, 0, 0}
Thus, S is one-one.
Now, since A is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore S is onto.
Next Let (x, y, z) ∈ ker T
then T(x, y, z) = (0, 0,0)
Using the definition of linear transformation, we get (x cosθ - y sinθ, x sinθ + y cosθ, z) = (0, 0, 0)
Comparing the components of the co-ordinates, we get
x cosθ - y sinθ = 0
x sinθ + y cosθ = 0
z = 0
Now, solving for x, y and z, we get
x = 0 , y = 0 and z = 0
hence, ker T = {0 , 0, 0 }
Therefore, T is one-one
Now, since T is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore T is onto.
R3 --> R3 defined by
S(x, y, z) = (2x, 4x - y, 2x + 3y - z) and T(x, y, z) = (x cos θ - y sin θ, x sin θ - y cos θ, z)
Let (x,y,z) ∈ ker.S
then S(x, y, z) =(0, 0, 0)
Using the definition o f linear transformation, we get
(2x, 4x - y, 2x + 3y - z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
2x = 0
4x - y = 0
2x + 3y - z = 0
On solving these equation for x, y, z, we get
x = 0, y = 0 and z = 0
Hence,ker S = {0, 0, 0}
Thus, S is one-one.
Now, since A is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore S is onto.
Next Let (x, y, z) ∈ ker T
then T(x, y, z) = (0, 0,0)
Using the definition of linear transformation, we get (x cosθ - y sinθ, x sinθ + y cosθ, z) = (0, 0, 0)
Comparing the components of the co-ordinates, we get
x cosθ - y sinθ = 0
x sinθ + y cosθ = 0
z = 0
Now, solving for x, y and z, we get
x = 0 , y = 0 and z = 0
hence, ker T = {0 , 0, 0 }
Therefore, T is one-one
Now, since T is a one-one linear transformation from R3 to R3, which is a finite dimensional vector space therefore T is onto.
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Let the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect answer is option 'C'. Can you explain this answer?
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Let the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect 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 the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect 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 the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect answer is option 'C'. Can you explain this answer?.
Let the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect 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 the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect 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 the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Let the linear transformation S and T :be defined byS(x, y, z) = (2x, 4x-y, 2x + 3y-z)and T(x, y, z) = (x cos θ- y sin θ, x sin θ + y cos θ, z)where 0 < θ < π/2. then,a)S is one to one but not Tb)T is one to one but not Sc)both S and Tare one to oned)neither S nor Tis one-to-oneCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
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