Mathematics Exam > Mathematics Questions > Let T be a linear operator on R3defined byT(1...
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Let T be a linear operator on R3 defined by
T(1, 0,0) = (1,2,1)
T(0, 1,0) = (3, 1,5)
T(0,0,1) = (3,-4, 7)
Then
T(1, 0,0) = (1,2,1)
T(0, 1,0) = (3, 1,5)
T(0,0,1) = (3,-4, 7)
Then
- a)T is invertible and T-1(x, y, z)
- b)T is not invertible
- c)T-1(x,y,z) =
- d)T is invertible and T-1(x,y,z) =
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0)...
Let T be a linear operator on R3 defined by
T(1,0, 0) = (1, 2, 1)
T(0, 1, 0) = (3, 1, 5)
T(0,0, 1) = ( 3,- 4 , 7)
We need to find the image of (x, y, z) under T.
Let there exist scalars α, β and γ such that
(x,y, z) = α(l, 0, 0) + β(0, 1, 0) + γ(0, 0, 1)
or equivalently (x, y, z) =(α, β, y) Implies a = x, p = y andy = z Therefore,(x, y, z) = x ( l , 0, 0) + γ( 0, 1, 0) + z (0, 0, 1)
Taking the image under linear transformation T, we get
T(x, y, z) =xT’( l , 0, 0) + y T ( 0, 1 , 0 ) + zT(0, 0, 1) = x ( l , 2 , 1 ) + y ( 3 , 1, 5) + z(3, - 4 , 7)
=(x + 3y + 3z, 2x + y — 4z, x + 5y + Iz)
Now let (x, y, z) ∈ ker T
Then T(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get
(x + 3y + 3z, 2x + y - 4z, x + 5y + 7z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
x + 3y + 3z = 0
2x + y - 4z = 0
x + 5y + 7z = 0
Now solving for x, y, z, we get
Therefore, ker T =
Hence,dim (ker T) = 1
Therefore, T is not one-one and hence T is not invertible.
T(1,0, 0) = (1, 2, 1)
T(0, 1, 0) = (3, 1, 5)
T(0,0, 1) = ( 3,- 4 , 7)
We need to find the image of (x, y, z) under T.
Let there exist scalars α, β and γ such that
(x,y, z) = α(l, 0, 0) + β(0, 1, 0) + γ(0, 0, 1)
or equivalently (x, y, z) =(α, β, y) Implies a = x, p = y andy = z Therefore,(x, y, z) = x ( l , 0, 0) + γ( 0, 1, 0) + z (0, 0, 1)
Taking the image under linear transformation T, we get
T(x, y, z) =xT’( l , 0, 0) + y T ( 0, 1 , 0 ) + zT(0, 0, 1) = x ( l , 2 , 1 ) + y ( 3 , 1, 5) + z(3, - 4 , 7)
=(x + 3y + 3z, 2x + y — 4z, x + 5y + Iz)
Now let (x, y, z) ∈ ker T
Then T(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get
(x + 3y + 3z, 2x + y - 4z, x + 5y + 7z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
x + 3y + 3z = 0
2x + y - 4z = 0
x + 5y + 7z = 0
Now solving for x, y, z, we get
Therefore, ker T =
Hence,dim (ker T) = 1
Therefore, T is not one-one and hence T is not invertible.
Most Upvoted Answer
Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0)...
Let T be a linear operator on R3 defined by
T(1,0, 0) = (1, 2, 1)
T(0, 1, 0) = (3, 1, 5)
T(0,0, 1) = ( 3,- 4 , 7)
We need to find the image of (x, y, z) under T.
Let there exist scalars α, β and γ such that
(x,y, z) = α(l, 0, 0) + β(0, 1, 0) + γ(0, 0, 1)
or equivalently (x, y, z) =(α, β, y) Implies a = x, p = y andy = z Therefore,(x, y, z) = x ( l , 0, 0) + γ( 0, 1, 0) + z (0, 0, 1)
Taking the image under linear transformation T, we get
T(x, y, z) =xT’( l , 0, 0) + y T ( 0, 1 , 0 ) + zT(0, 0, 1) = x ( l , 2 , 1 ) + y ( 3 , 1, 5) + z(3, - 4 , 7)
=(x + 3y + 3z, 2x + y — 4z, x + 5y + Iz)
Now let (x, y, z) ∈ ker T
Then T(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get
(x + 3y + 3z, 2x + y - 4z, x + 5y + 7z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
x + 3y + 3z = 0
2x + y - 4z = 0
x + 5y + 7z = 0
Now solving for x, y, z, we get
Therefore, ker T =
Hence,dim (ker T) = 1
Therefore, T is not one-one and hence T is not invertible.
T(1,0, 0) = (1, 2, 1)
T(0, 1, 0) = (3, 1, 5)
T(0,0, 1) = ( 3,- 4 , 7)
We need to find the image of (x, y, z) under T.
Let there exist scalars α, β and γ such that
(x,y, z) = α(l, 0, 0) + β(0, 1, 0) + γ(0, 0, 1)
or equivalently (x, y, z) =(α, β, y) Implies a = x, p = y andy = z Therefore,(x, y, z) = x ( l , 0, 0) + γ( 0, 1, 0) + z (0, 0, 1)
Taking the image under linear transformation T, we get
T(x, y, z) =xT’( l , 0, 0) + y T ( 0, 1 , 0 ) + zT(0, 0, 1) = x ( l , 2 , 1 ) + y ( 3 , 1, 5) + z(3, - 4 , 7)
=(x + 3y + 3z, 2x + y — 4z, x + 5y + Iz)
Now let (x, y, z) ∈ ker T
Then T(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get
(x + 3y + 3z, 2x + y - 4z, x + 5y + 7z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
x + 3y + 3z = 0
2x + y - 4z = 0
x + 5y + 7z = 0
Now solving for x, y, z, we get
Therefore, ker T =
Hence,dim (ker T) = 1
Therefore, T is not one-one and hence T is not invertible.
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Community Answer
Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0)...
Determinant of basis formed by its vectors is 0
so it is not invertible
so it is not invertible
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Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. Can you explain this answer?
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Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. 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 be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. 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 be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. Can you explain this answer?.
Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. 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 be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. 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 be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice Let T be a linear operator on R3defined byT(1, 0,0) = (1,2,1)T(0, 1,0) = (3, 1,5)T(0,0,1) = (3,-4, 7)Thena)T is invertible and T-1(x, y, z)b)T is not invertiblec)T-1(x,y,z) =d)T is invertible andT-1(x,y,z) =Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Mathematics tests.
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