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Let S and T be two linear operators on R3 defined by
S (x,y,z) = (x,x + y , x - y - z )
T(x , y , z) = (x + 2z ,y - z ,x + y + z)
S (x,y,z) = (x,x + y , x - y - z )
T(x , y , z) = (x + 2z ,y - z ,x + y + z)
- a)S is invertible but not T
- b)T is invertible but not S
- c)both S and T are invertible
- d)neither S nor T is invertible
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let S and T be two linear operators on R3 defined byS (x,y,z) =...
Let S and T be two linear operators on R3 defined by
S(x, y, z) = (x, x + y, x - y - z)
and T(x, y, z) = (x + 2z, y - z, x + y + z).
Let (x, y, z) ∈ ker S.
Then S(x,y,z)= (0,0,0)
Using the definition of linear transformation S, we get
(x , x + y , x - y - z ) = (0, 0, 0) Comparing the components of the coordinates, we get
x = 0 , x+ y = 0, x - y - z = 0
Solving for x, y and z, we get
x = 0, y = 0, z = 0
Hence,ker S = {(0, 0, 0)}
Therefore, S is one-one.
Since S is a one-one linear operator on R3. Therefore, S is onto and hence invertible that is non-singular.
Next, Let (x, y, z) ∈ ker T. Then
T(x,y,z)=(0,0,0)
Using the definition of linear transformation T, we get
(x + 2z ,y - z , x + y + z) = (0, 0, 0) Comparing the components of the coordinates, we get
x + 2 z = 0 , y - z = 0, x + y + z = 0 Solving for x, y and z, we get x = -2z, y = z Therefore, ker
Hence, dim ker T= 1. Therefore, T is not one-one and hence T is not invertible that is T is singular.
S(x, y, z) = (x, x + y, x - y - z)
and T(x, y, z) = (x + 2z, y - z, x + y + z).
Let (x, y, z) ∈ ker S.
Then S(x,y,z)= (0,0,0)
Using the definition of linear transformation S, we get
(x , x + y , x - y - z ) = (0, 0, 0) Comparing the components of the coordinates, we get
x = 0 , x+ y = 0, x - y - z = 0
Solving for x, y and z, we get
x = 0, y = 0, z = 0
Hence,ker S = {(0, 0, 0)}
Therefore, S is one-one.
Since S is a one-one linear operator on R3. Therefore, S is onto and hence invertible that is non-singular.
Next, Let (x, y, z) ∈ ker T. Then
T(x,y,z)=(0,0,0)
Using the definition of linear transformation T, we get
(x + 2z ,y - z , x + y + z) = (0, 0, 0) Comparing the components of the coordinates, we get
x + 2 z = 0 , y - z = 0, x + y + z = 0 Solving for x, y and z, we get x = -2z, y = z Therefore, ker
Hence, dim ker T= 1. Therefore, T is not one-one and hence T is not invertible that is T is singular.
Most Upvoted Answer
Let S and T be two linear operators on R3 defined byS (x,y,z) =...
Question Analysis:
We are given two linear operators S and T on R3. We need to determine whether S is invertible but not T.
Definitions:
1. A linear operator is invertible if it has an inverse. In other words, a linear operator S is invertible if there exists another linear operator S' such that S(S'(x)) = x for all x in the vector space.
2. A linear operator is invertible if and only if its null space contains only the zero vector.
Solution:
Operator S:
The linear operator S is defined as follows:
S (x,y,z) = (x,x+y,x-y-z)
Null Space of S:
To determine whether S is invertible, we need to find its null space. The null space of S is the set of all vectors x such that S(x) = 0.
Setting S(x) = 0, we have:
x = 0
x+y = 0
x-y-z = 0
Solving these equations, we get:
x = 0
y = 0
z = 0
Therefore, the null space of S contains only the zero vector. This implies that S is invertible.
Operator T:
The linear operator T is defined as follows:
T(x,y,z) = (x,2z,y-z,x+y+z)
Null Space of T:
To determine whether T is invertible, we need to find its null space. The null space of T is the set of all vectors x such that T(x) = 0.
Setting T(x) = 0, we have:
x = 0
2z = 0
y-z = 0
x+y+z = 0
From the second equation, we have z = 0. Substituting this into the third equation, we get y = 0. Substituting these values into the fourth equation, we have x = 0.
Therefore, the null space of T contains only the zero vector. This implies that T is invertible.
Conclusion:
From the analysis above, we can conclude that S is invertible but T is not invertible. Therefore, the correct answer is option 'A'.
We are given two linear operators S and T on R3. We need to determine whether S is invertible but not T.
Definitions:
1. A linear operator is invertible if it has an inverse. In other words, a linear operator S is invertible if there exists another linear operator S' such that S(S'(x)) = x for all x in the vector space.
2. A linear operator is invertible if and only if its null space contains only the zero vector.
Solution:
Operator S:
The linear operator S is defined as follows:
S (x,y,z) = (x,x+y,x-y-z)
Null Space of S:
To determine whether S is invertible, we need to find its null space. The null space of S is the set of all vectors x such that S(x) = 0.
Setting S(x) = 0, we have:
x = 0
x+y = 0
x-y-z = 0
Solving these equations, we get:
x = 0
y = 0
z = 0
Therefore, the null space of S contains only the zero vector. This implies that S is invertible.
Operator T:
The linear operator T is defined as follows:
T(x,y,z) = (x,2z,y-z,x+y+z)
Null Space of T:
To determine whether T is invertible, we need to find its null space. The null space of T is the set of all vectors x such that T(x) = 0.
Setting T(x) = 0, we have:
x = 0
2z = 0
y-z = 0
x+y+z = 0
From the second equation, we have z = 0. Substituting this into the third equation, we get y = 0. Substituting these values into the fourth equation, we have x = 0.
Therefore, the null space of T contains only the zero vector. This implies that T is invertible.
Conclusion:
From the analysis above, we can conclude that S is invertible but T is not invertible. Therefore, the correct answer is option 'A'.
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