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xn = xn-1 + xn - 2 for 3 < n < 10) of the vector space 
the dimension of W is
If the linear transformation T(v) = Av rotates the vectors v1 = (-1, 0) and v2 = (0, 1) clockwise π/2 radians, the resulting vectors are:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
If the linear transformation T(v) = Av rotates the vectors v1 = (-1, 0) and v2 = (0, 1) clockwise π/2 radians, the resulting vectors are:
a)
b)
c)
d)
|
Veda Institute answered |
We are given that the linear transformation T(v) = Av rotates the vector v1 = (-1, 0) and v2 = (0,1) clockwise π/2radians.
We know that if a vector (a, b) is rotated through an angle a clockwise direction under linear transformation T.Then
Here, v1 = (-1 , 0) and a = π/2 radians
Therefore, T(v1) = T(-1, 0)
=
= (0,1)
Therefore, T(-1, 0) = (0,1)
Similarly, for v2 = (0, 1) and α = π/2,
and T(v2) = T(0 ,1)
= (1,0)
Therefore, T(0,1) = (1 ,0 ).
We know that if a vector (a, b) is rotated through an angle a clockwise direction under linear transformation T.Then
Here, v1 = (-1 , 0) and a = π/2 radians
Therefore, T(v1) = T(-1, 0)
=
= (0,1)Therefore, T(-1, 0) = (0,1)
Similarly, for v2 = (0, 1) and α = π/2,
and T(v2) = T(0 ,1)
= (1,0)
Therefore, T(0,1) = (1 ,0 ).
Let A = 
be an n x n matrix such that aij = 3,
and j. Then the nullity of A is- a)n - 1
- b)n - 3
- c)n
- d)0
Correct answer is option 'A'. Can you explain this answer?
Let A = be an n x n matrix such that aij = 3,and j. Then the nullity of A is
be an n x n matrix such that aij = 3,and j. Then the nullity of A isa)
n - 1
b)
n - 3
c)
n
d)
0
|
Chirag Verma answered |
We are given that the n x n matrix A = [aij] such that aij = 3
and j.
Then , we need to find the nullity o f the given matrix.
Let the matrix,
Reduce the matrix to echelon form using the operations "R2 --> R2 - R1 ", "R3 ---> R3 - R1 " .......
"Rn --> Rn - R1".
These operations yield--
Here, rank (A) =1. By Sylwester’s law,
dim (A) =rank (A) + nullity (A)
or n = 1 + nullity (A)
or nullity(A) = n - 1
and j.Then , we need to find the nullity o f the given matrix.
Let the matrix,
Reduce the matrix to echelon form using the operations "R2 --> R2 - R1 ", "R3 ---> R3 - R1 " .......
"Rn --> Rn - R1".
These operations yield--
Here, rank (A) =1. By Sylwester’s law,
dim (A) =rank (A) + nullity (A)
or n = 1 + nullity (A)
or nullity(A) = n - 1
If 3x + 2 + z = 0
x + 4y + z = 0
2x + y + 4z = 0
be a system of equations, then- a)It is inconsistent
- b)it has only the trivial solution x = 0,y = 0, z = 0
- c)It can be reduced to a single equation and so a solution does not exis
- d)the determinant of the matrix of coefficient is zero
Correct answer is option 'B'. Can you explain this answer?
If 3x + 2 + z = 0
x + 4y + z = 0
2x + y + 4z = 0
be a system of equations, then
x + 4y + z = 0
2x + y + 4z = 0
be a system of equations, then
a)
It is inconsistent
b)
it has only the trivial solution x = 0,y = 0, z = 0
c)
It can be reduced to a single equation and so a solution does not exis
d)
the determinant of the matrix of coefficient is zero
|
|
Chirag Verma answered |
We are given that the system of equations,
3x + 2y + z = 0
x + 4y + z = 0
2x + y + 4z = 0
It can be written in the matrix form as,
Reduce this system to echelon form using the operation
“R2 -> 3R2 - R1 ” and " R3 --> 3R3 - 2R2".
These operations yield-
and also "R3 —> 10R3 + R2"
which gives, x = 0, y = 0 and z - 0, Hence, th is system of equations has only the trivial solution.
3x + 2y + z = 0
x + 4y + z = 0
2x + y + 4z = 0
It can be written in the matrix form as,
Reduce this system to echelon form using the operation
“R2 -> 3R2 - R1 ” and " R3 --> 3R3 - 2R2".
These operations yield-
and also "R3 —> 10R3 + R2"
which gives, x = 0, y = 0 and z - 0, Hence, th is system of equations has only the trivial solution.
Consider the system of simultaneous equation
x + 2y + z = 6;
2x+y + 2z = 6;
x+y + z = 5
This system has- a)unique solution
- b)infinite number of solution
- c)no solution
- d)exactly two solution
Correct answer is option 'C'. Can you explain this answer?
Consider the system of simultaneous equation
x + 2y + z = 6;
2x+y + 2z = 6;
x+y + z = 5
This system has
x + 2y + z = 6;
2x+y + 2z = 6;
x+y + z = 5
This system has
a)
unique solution
b)
infinite number of solution
c)
no solution
d)
exactly two solution
|
|
Chirag Verma answered |
We are given that the system of equation
This system of equation can be written in augmented matrix as,
Applying the operations "R2 → R2 - 2R1" and "R3 → R3 - R1" these operations yields:
also applying R3 → 3R3 - R2, which yields -
Since the rank of coefficient matrix ≠ the rank of augmented matrix.
Therefore, the given system of equations has no solution.
This system of equation can be written in augmented matrix as,
Applying the operations "R2 → R2 - 2R1" and "R3 → R3 - R1" these operations yields:
also applying R3 → 3R3 - R2, which yields -
Since the rank of coefficient matrix ≠ the rank of augmented matrix.
Therefore, the given system of equations has no solution.
Suppose T1 : V —> U and T2 : U ---> W be a linear transformations then- a)Rank T20T1 ≥ Rank T2
- b)Rank T20T1 ≤ Rank T2
- c)RankT20T1 ≥ Rank T1
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Suppose T1 : V —> U and T2 : U ---> W be a linear transformations then
a)
Rank T20T1 ≥ Rank T2
b)
Rank T20T1 ≤ Rank T2
c)
RankT20T1 ≥ Rank T
1
d)
None of these
|
|
Chirag Verma answered |
Since T1(V) 
we also have T2(T1(V)) 
T2(U) and
so, dim (T2(T1(V))) - dim (T2(U)) then Rank T2oT2 = dim ((T2oT1) (v)) = dim [T2 (T1(V))]
≤ dim T2(U)
= Rank T2
we also have T2(T1(V)) T2(U) andso, dim (T2(T1(V))) - dim (T2(U)) then Rank T2oT2 = dim ((T2oT1) (v)) = dim [T2 (T1(V))]
≤ dim T2(U)
= Rank T2
The system of equations
x + y + z = 0
3x + 6y + z = 0
αx + 2y + z = 0
has infinitely many solutions, then a is equal to - a)7
- b)7/5
- c)5/7
- d)4
Correct answer is option 'B'. Can you explain this answer?
The system of equations
x + y + z = 0
3x + 6y + z = 0
αx + 2y + z = 0
has infinitely many solutions, then a is equal to
x + y + z = 0
3x + 6y + z = 0
αx + 2y + z = 0
has infinitely many solutions, then a is equal to
a)
7
b)
7/5
c)
5/7
d)
4
|
|
Chirag Verma answered |
We are given that the system of equations,
x + y+z= 0
3x + 6y + z = 0
αx + 2y + z = 0
has infinitely many solutions. We need to find the value of a. The given system of equation may be written in matrix form as,
Reduce this system of equation to echelon form using the operations
‘R2 --> R2 - R1' and R3 ---> R3 - R1, these operations yields -
Also applying , R3 —> 5R3 - R2 we get -
Since, this system of equations has infinitely man solution. It is possible when 5α - 7 = 0
or α = 7/5
x + y+z= 0
3x + 6y + z = 0
αx + 2y + z = 0
has infinitely many solutions. We need to find the value of a. The given system of equation may be written in matrix form as,
Reduce this system of equation to echelon form using the operations
‘R2 --> R2 - R1' and R3 ---> R3 - R1, these operations yields -
Also applying , R3 —> 5R3 - R2 we get -
Since, this system of equations has infinitely man solution. It is possible when 5α - 7 = 0
or α = 7/5
The characteristic polynomial of 3 x 3 matrix A |λI - A| = λ3 + 3λ2 + 4λ - 3
Let x - trace (A) and y = |A|, the determinant of A.
Then,- a)x/y = 3/4
- b)x/y = 4/3
- c)x = y = -3
- d)x = 3 and y = -3
Correct answer is option 'C'. Can you explain this answer?
The characteristic polynomial of 3 x 3 matrix A |λI - A| = λ3 + 3λ2 + 4λ - 3
Let x - trace (A) and y = |A|, the determinant of A.
Then,
Let x - trace (A) and y = |A|, the determinant of A.
Then,
a)
x/y = 3/4
b)
x/y = 4/3
c)
x = y = -3
d)
x = 3 and y = -3
|
|
Chirag Verma answered |
We are given that the characteristic polynomial o f 3 x 3 matrix, A is
λ3 + 3λ2 + 4λ - 3 = 0
and x = trace (A)
y = det(A)
The general characteristic polynomial of 3 x 3 matrix is given by,
λ3 - trace (A) λ2 + λ + |A| = 0
Comparing this equation by the given equation, we get
trace = -3
or x = -3
and |A| = -3
or y = -3
Hence, x = y = -3
λ3 + 3λ2 + 4λ - 3 = 0
and x = trace (A)
y = det(A)
The general characteristic polynomial of 3 x 3 matrix is given by,
λ3 - trace (A) λ2 + λ + |A| = 0
Comparing this equation by the given equation, we get
trace = -3
or x = -3
and |A| = -3
or y = -3
Hence, x = y = -3
The minimal polynomial of the 3 x 3 real matrix 
is- a)(x-a)(x-b)
- b)(x-a)2 (x-b)
- c)(x-a)2 (x-b)2
- d)(x-a)(x-b)2
Correct answer is option 'A'. Can you explain this answer?
The minimal polynomial of the 3 x 3 real matrix is
isa)
(x-a)(x-b)
b)
(x-a)2 (x-b)
c)
(x-a)2 (x-b)2
d)
(x-a)(x-b)2
|
|
Chirag Verma answered |
We need to find the minimal polynomial of the 3 x 3 real matrix
A =
First we need to find the characteristic polynomial of the given matrix, that is
chA(x) = |xI - A| = (x - a) (x - a) (x - b)
Further, the minimal polynomial of the given matrix
= l.c.m. of (x - a), (x - a), (x - b)
or mA (x) =(x - a) (x - b)
A =
First we need to find the characteristic polynomial of the given matrix, that is
chA(x) = |xI - A| = (x - a) (x - a) (x - b)
Further, the minimal polynomial of the given matrix
= l.c.m. of (x - a), (x - a), (x - b)
or mA (x) =(x - a) (x - b)
Let T : Cn --> Cn be a linear operator having n distinct eigen values. Then,- a)T is invertible
- b)T is invertible as well as diagonalisable
- c)T is not diagonalisable
- d)T is diagonalisable
Correct answer is option 'D'. Can you explain this answer?
Let T : Cn --> Cn be a linear operator having n distinct eigen values. Then,
a)
T is invertible
b)
T is invertible as well as diagonalisable
c)
T is not diagonalisable
d)
T is diagonalisable
|
|
Chirag Verma answered |
We are given that a linear transformation T: Cn --> Cn having n distinct eigen values. Thus T is diagonalisable.
A linear operator defined on an n dimensional vector space V having n distinct eigen values is diagonalisable.
A linear operator defined on an n dimensional vector space V having n distinct eigen values is diagonalisable.
Which one of the following is a linear transformation?- a)T : R2 ---> R2 defined by T(x, y) = (xy, x)
- b)T : R2 ---> R3 defined by T(x, y) = (x + 3, 2y, x+y)
- c)T : R3 ---> R2 defined by T(x, y, z) = (Ix| , y + z)
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Which one of the following is a linear transformation?
a)
T : R2 ---> R2 defined by T(x, y) = (xy, x)
b)
T : R2 ---> R3 defined by T(x, y) = (x + 3, 2y, x+y)
c)
T : R3 ---> R2 defined by T(x, y, z) = (Ix| , y + z)
d)
None of these
|
|
Veda Institute answered |
Let (1,2) and (3,4) be two vectors of R2.
Then T[(1, 2) + (3,4)] = T(4, 6)]
= (4 x 6, 4) = (24, 6)
and T(l, 2) = (1 x 2,1) = (2, 1)
T(3, 4) = (3 x 4,.3) = (12, 3) Thus,T(l, 2) + T(3, 4) = (2, 1) + (12, 3) = (14, 4)
Therefore, T[(1, 2) + (3 ,4)] ≠ T(1 ,2) + T(3,4)
Hence, T is non linear.
To show that a given map T is a linear transformation we must show that
T(αx + βy) = αT(x) + βT(y)
where x,y are vectors and a, P are scalars while to show that the given map is non-linear conveniently we try to find a related contrast example.
Then T[(1, 2) + (3,4)] = T(4, 6)]
= (4 x 6, 4) = (24, 6)
and T(l, 2) = (1 x 2,1) = (2, 1)
T(3, 4) = (3 x 4,.3) = (12, 3) Thus,T(l, 2) + T(3, 4) = (2, 1) + (12, 3) = (14, 4)
Therefore, T[(1, 2) + (3 ,4)] ≠ T(1 ,2) + T(3,4)
Hence, T is non linear.
To show that a given map T is a linear transformation we must show that
T(αx + βy) = αT(x) + βT(y)
where x,y are vectors and a, P are scalars while to show that the given map is non-linear conveniently we try to find a related contrast example.
The system of equation 2x + y = 5, x - 3y = -1
3x + 4y = k is consistent, when k is- a)1
- b)2
- c)5
- d)10
Correct answer is option 'D'. Can you explain this answer?
The system of equation 2x + y = 5, x - 3y = -1
3x + 4y = k is consistent, when k is
3x + 4y = k is consistent, when k is
a)
1
b)
2
c)
5
d)
10
|
|
Chirag Verma answered |
We are given that the system of equations,
2x + y = 5
x - 3y = - 1
3x + 4y = k is consistent.
We need to find the value of k. The given system of equation may be written as,
Since, this system of equation is consistent. Therefore
or 2(-3k + 4) -1(k +3) + 5(4 + 9) = 0
-7k + 70 = 0
k =10
2x + y = 5
x - 3y = - 1
3x + 4y = k is consistent.
We need to find the value of k. The given system of equation may be written as,
Since, this system of equation is consistent. Therefore
or 2(-3k + 4) -1(k +3) + 5(4 + 9) = 0
-7k + 70 = 0
k =10
Let A be 3 x 3 matrix whose characteristic roots are 3 ,2 ,-1 .
If B = A2 - A, then |B| is- a)24
- b)-2
- c)12
- d)-12
Correct answer is option 'A'. Can you explain this answer?
Let A be 3 x 3 matrix whose characteristic roots are 3 ,2 ,-1 .
If B = A2 - A, then |B| is
If B = A2 - A, then |B| is
a)
24
b)
-2
c)
12
d)
-12
|
|
Chirag Verma answered |
We are given that A be a 3 x 3 matrix whose ' characteristic roots are 3, 2, -1 say λ1, λ2 and λ3 and B = A2 - A
λ1 = 3,
λ1 = (3)2 - 3 = 6
λ2 = 2,
λ2 = (2)2 - 2 = 2
λ3 = -1
λ3 = (-1)2 - (-1) = 2
The determinant of B is, | B | = product of its eigen values
= λ1'λ2'λ3'
= 6 x 2 x 2 = 24
λ1 = 3,
λ1 = (3)2 - 3 = 6
λ2 = 2,
λ2 = (2)2 - 2 = 2
λ3 = -1
λ3 = (-1)2 - (-1) = 2
The determinant of B is, | B | = product of its eigen values
= λ1'λ2'λ3'
= 6 x 2 x 2 = 24
If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be - a)3
- b)0
- c)2
- d)1
Correct answer is option 'A'. Can you explain this answer?
If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be
a)
3
b)
0
c)
2
d)
1
|
|
Veda Institute answered |
Here A & B a re 3 × 2 real matrices such that rank (AB) = 1
So, |AB| = 0 ⇒ |A| |B| = 0 (∴ |AB| = |A| |B|)
⇒ either |A| or |B| should be zero
So, |BA| = |B||A| = 0
⇒ BA is singular
Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)
So, |AB| = 0 ⇒ |A| |B| = 0 (∴ |AB| = |A| |B|)
⇒ either |A| or |B| should be zero
So, |BA| = |B||A| = 0
⇒ BA is singular
Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)
The matrix
is a- a)Hermitain matrix
- b)skew-Hermitian matrix
- c)symmetric matrix
- d)skew-symmetric matrix
- e)
Correct answer is option 'B'. Can you explain this answer?
The matrix is a
is aa)
Hermitain matrix
b)
skew-Hermitian matrix
c)
symmetric matrix
d)
skew-symmetric matrix
e)
|
|
Chirag Verma answered |
Correct Answer :- b
Explanation : Given a matrix is equal to its conjugate transpose hence it is hermitian. also, it is skew also because the transpose of a given matrix is equal to negative of the given matrix.
Hence it is a skew-Hermitian matrix
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- 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?
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) =
|
|
Chirag Verma answered |
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.
Let F be a field and T be a linear operator on F2 defined by T(x1, x2) = (x1 + x2, x1). Then T-1(x1, x2) is- a)(x2, x1 - x2)
- b)(x1, x1 - x2)
- c)(x1 - x2, x2)
- d)(x1 - x2, x1)
Correct answer is option 'A'. Can you explain this answer?
Let F be a field and T be a linear operator on F2 defined by T(x1, x2) = (x1 + x2, x1). Then T-1(x1, x2) is
a)
(x2, x1 - x2)
b)
(x1, x1 - x2)
c)
(x1 - x2, x2)
d)
(x1 - x2, x1)
|
Ira Malhotra answered |
Explanation:
Given linear operator:
T(x1, x2) = (x1 + x2, x1)
Finding the inverse of T:
To find the inverse of T, we need to solve for T-1(x1, x2) such that T(T-1(x1, x2)) = (x1, x2) and T-1(T(x1, x2)) = (x1, x2).
Solving for T-1(x1, x2):
Let T-1(x1, x2) = (a, b), then we have:
T(a, b) = (a + b, a) = (x1, x2)
Equating components:
a + b = x1
a = x2
Solving for a and b:
From the equations above, we can solve for a and b:
a = x2
b = x1 - x2
Therefore, T-1(x1, x2) = (x2, x1 - x2), which corresponds to option 'a'.
Given linear operator:
T(x1, x2) = (x1 + x2, x1)
Finding the inverse of T:
To find the inverse of T, we need to solve for T-1(x1, x2) such that T(T-1(x1, x2)) = (x1, x2) and T-1(T(x1, x2)) = (x1, x2).
Solving for T-1(x1, x2):
Let T-1(x1, x2) = (a, b), then we have:
T(a, b) = (a + b, a) = (x1, x2)
Equating components:
a + b = x1
a = x2
Solving for a and b:
From the equations above, we can solve for a and b:
a = x2
b = x1 - x2
Therefore, T-1(x1, x2) = (x2, x1 - x2), which corresponds to option 'a'.
Which one of the following is true?- a) dim Horn (M2,3, R4) = 24
- b) dim Horn (M2,3, R4) = 9
- c) dim Horn (M2,3, R4) = 6
- d) dim Horn (M2,3, R4) = 12
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is true?
a)
dim Horn (M2
,
3, R4) = 24b)
dim Horn (M2,3, R4) = 9
c)
dim Horn (M2,3, R4) = 6
d)
dim Horn (M2,3, R4) = 12
|
|
Veda Institute answered |
We need to find the dimension of Horn (M2,3, R4).
Since dim M2
and dim R4 = 4
Therefore,dim Horn (M2,3, R4)
= 6 x 4 = 24
Let Mn,m be the vector space of all n x m matrices over R. Then
dim Mn,m = n x m
Since dim M2
,
3 = 2 x 3 = 6and dim R4 = 4
Therefore,dim Horn (M2,3, R4)
= 6 x 4 = 24
Let Mn,m be the vector space of all n x m matrices over R. Then
dim Mn,m = n x m
The minimum polynomial m(λ) of M = 
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The minimum polynomial m(λ) of M =
a)
b)
c)
d)
|
|
Chirag Verma answered |
We are given that the matrix
M =
We need to find the minimal polynomial of the given matrix. The characteristic polynomial of the given matrix is given by,
Hence, the minimal polynomial of the given matrix M is,
M =
We need to find the minimal polynomial of the given matrix. The characteristic polynomial of the given matrix is given by,
Hence, the minimal polynomial of the given matrix M is,If the linear transformation T(v) = Av rotates the vectors (-1, 0) and (0, 1) clockwise π/2 radians then:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If the linear transformation T(v) = Av rotates the vectors (-1, 0) and (0, 1) clockwise π/2 radians then:
a)
b)
c)
d)
|
|
Veda Institute answered |
We are given that the linear transformation T(v) = Av rotates the vectors (-1, 0) and (0, 1) clockwise π/2 radians. We need to find the matrix A.
We know that if a vector (a, b) is rotated through an angle a clockwise under T. Then
Here, (a, b) = (-1, 0) and a = π/2
Therefore, T (-1, 0)
= (0, 1) = 0(-1, 0) + 1(0, 1)
and (a, b) = (0,1), α = π/2
= (1, 0) = - 1(-1, 0) + 0(0, 1)
Therefore, the matrix of T is A =
We know that if a vector (a, b) is rotated through an angle a clockwise under T. Then
Here, (a, b) = (-1, 0) and a = π/2
Therefore, T (-1, 0)
= (0, 1) = 0(-1, 0) + 1(0, 1)
and (a, b) = (0,1), α = π/2
= (1, 0) = - 1(-1, 0) + 0(0, 1)
Therefore, the matrix of T is A =
Let T be the linear operator on R3 defined by T(x1, x2, x3) = ( 2x1, x1 - x2, 5x1 + 4x2 + x3). Then T-1 is- a)
- b)
- c)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Let T be the linear operator on R3 defined by T(x1, x2, x3) = ( 2x1, x1 - x2, 5x1 + 4x2 + x3). Then T-1 is
a)
b)
c)
d)
None of these
|
|
Veda Institute answered |
Let T be the linear operator on R3 defined by
T(x1 x2, x3) = (2x, x1 - x2, 5x1 + 4x2 + x3)
Let (x1, x2, x3) ∈ ker T.
Then we need to find T-1
Now, T(x1, x2, x3) - (0, 0, 0)
Using the definition of T, we get
(2x, xl - x2, 5x1 + 4x2 + x3) = (0, 0, 0)
Comparing the components of the coordinates, we get
2x1 = 0, x1 - x2 = 0, 5x1 + 4 x 2 + x 3 = 0
Solving for x1 x2 and x3, we get x1 = 0, x2 = 0, x3 = 0
Hence,ker T= {(0, 0, 0)}
and hence, T is one-one.
Since R3 is a 3-dimensional vector space and T : R3 —> R3 be a one-one linear transformation and hence, T is onto. Therefore, T-1 exist.
Let (a, b, c) be the image of (x1, x2, x3) under T-1.
Then T-1(x1,x2, x3) = (a, b, c) ImpliesT(a, b, c) = (x1,x2, x3)
Using the definition of T, we get
(2a a - b 5a + 4b + c) = (x1,x2, x3) Comparing the components, we get 2a = x1, a - b = x2, 5a l 4b I c = x3 Solving for a, b, c, we get
Therefore, T-1(x1 x2, x3) =
T(x1 x2, x3) = (2x, x1 - x2, 5x1 + 4x2 + x3)
Let (x1, x2, x3) ∈ ker T.
Then we need to find T-1
Now, T(x1, x2, x3) - (0, 0, 0)
Using the definition of T, we get
(2x, xl - x2, 5x1 + 4x2 + x3) = (0, 0, 0)
Comparing the components of the coordinates, we get
2x1 = 0, x1 - x2 = 0, 5x1 + 4 x 2 + x 3 = 0
Solving for x1 x2 and x3, we get x1 = 0, x2 = 0, x3 = 0
Hence,ker T= {(0, 0, 0)}
and hence, T is one-one.
Since R3 is a 3-dimensional vector space and T : R3 —> R3 be a one-one linear transformation and hence, T is onto. Therefore, T-1 exist.
Let (a, b, c) be the image of (x1, x2, x3) under T-1.
Then T-1(x1,x2, x3) = (a, b, c) ImpliesT(a, b, c) = (x1,x2, x3)
Using the definition of T, we get
(2a a - b 5a + 4b + c) = (x1,x2, x3) Comparing the components, we get 2a = x1, a - b = x2, 5a l 4b I c = x3 Solving for a, b, c, we get
Therefore, T-1(x1 x2, x3) =
If T : R2 --> R3 is a linear transformation T(1, 0) = (2, 3, 1) and T(1,1) = (3,0,2), then which one of the following is correct?- a)T(x, y) = (x + y, 2x + y , 3 x - 3y),
- b)T(x, y) = (2x + y, 3x - 3y , x + y ) ,
- c)7(x, y) = (2 x -y ,3 x + 3y , x - y ) ,
- d)T(x, y) = ( x ^ y , 2x - y , 3 x - 3y) ,
Correct answer is option 'B'. Can you explain this answer?
If T : R2 --> R3 is a linear transformation T(1, 0) = (2, 3, 1) and T(1,1) = (3,0,2), then which one of the following is correct?
a)
T(x, y) = (x + y, 2x + y , 3 x - 3y),
b)
T(x, y) = (2x + y, 3x - 3y , x + y ) ,
c)
7(x, y) = (2 x -y ,3 x + 3y , x - y ) ,
d)
T(x, y) = ( x ^ y , 2x - y , 3 x - 3y) ,
|
|
Veda Institute answered |
We are given that a linear transformation T: R2--> R3
show that
T(1,0) = (2,3,1)
and T(1 , 1) = (3, 0, 2)
We need to determine the image of (x, y) under the linear transformation T.
Let there exist scalars α, β,
such that (x,y) = a(l,0) + β(l, 1) or equivalently
(x,y) =(α + β, β)
Comparing the components of the co-ordinates we get,
α + β = x, β = y
Solving for α and β, we get
α = x-y, β = y
therefore, (x, y) = (x - y) (1,0) + y( 1,1) taking the image under linear transformation T, we get
Using the linearity condition, we get T(x,y) = (x-y)T(l,0)+yT(l,l)
Substituting the value of T(1,0) and T(1, 1) we get
T(x, y) = (x -y) (2, 3, 1) + y(3, 0, 2)
= (2x -2y + 3y, 3x - 3y, x - y + 2y)
= (2x + y, 3x - 3y, x + y)
Therefore , the image of (x, y) under T that is
show that
T(1,0) = (2,3,1)
and T(1 , 1) = (3, 0, 2)
We need to determine the image of (x, y) under the linear transformation T.
Let there exist scalars α, β,
such that (x,y) = a(l,0) + β(l, 1) or equivalently
(x,y) =(α + β, β)
Comparing the components of the co-ordinates we get,
α + β = x, β = y
Solving for α and β, we get
α = x-y, β = y
therefore, (x, y) = (x - y) (1,0) + y( 1,1) taking the image under linear transformation T, we get
Using the linearity condition, we get T(x,y) = (x-y)T(l,0)+yT(l,l)
Substituting the value of T(1,0) and T(1, 1) we get
T(x, y) = (x -y) (2, 3, 1) + y(3, 0, 2)
= (2x -2y + 3y, 3x - 3y, x - y + 2y)
= (2x + y, 3x - 3y, x + y)
Therefore , the image of (x, y) under T that is
The number of onto linear transformation from R3 to R4 is- a)0
- b)1
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
The number of onto linear transformation from R3 to R4 is
a)
0
b)
1
c)
2
d)
3
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|
Veda Institute answered |
We need to find the number of onto linear transformation from R3 to R4.
Since dim R4 = 4 > 3 = dim R3. Therefore, there is no onto linear transformation from R3 to R4.
Suppose T : U ---> V be an onto linear transformation.
Then dim V ≤ dim U
Since dim R4 = 4 > 3 = dim R3. Therefore, there is no onto linear transformation from R3 to R4.
Suppose T : U ---> V be an onto linear transformation.
Then dim V ≤ dim U
Consider the system x + y + z = 0; x - y - z = 0, then the system of equations have- a)no solution
- b)infinite solution
- c)unique solution
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Consider the system x + y + z = 0; x - y - z = 0, then the system of equations have
a)
no solution
b)
infinite solution
c)
unique solution
d)
None of the above
|
|
Chirag Verma answered |
We are given that the system of equation,
x + y + z = 0
x - y - z = 0
This system of equations has rank 2 and 3 unknowns,
that is, rank < no. of unknown
Hence, this system of equation has infinitly many solution.
x + y + z = 0
x - y - z = 0
This system of equations has rank 2 and 3 unknowns,
that is, rank < no. of unknown
Hence, this system of equation has infinitly many solution.
A is any matrix which satisfy A3 - A2 + A - I = 0 and A3x3, then A4 is- a)A = 0
- b) A = I
- c)There is no such matrix
- d)A3 + A2 - A
Correct answer is option 'D'. Can you explain this answer?
A is any matrix which satisfy A3 - A2 + A - I = 0 and A3x3, then A4 is
a)
A = 0
b)
A = I
c)
There is no such matrix
d)
A3 + A2 - A
|
|
Chirag Verma answered |
We are given that, A be any 3 x 3 matrix which satisfy
A3- A2 + A - I = 0
and we need to find A4.
A3- A2+ A - I = 0
or A3 = A 2 - A + I
or A4 = A3 - A2 + A
A3- A2 + A - I = 0
and we need to find A4.
A3- A2+ A - I = 0
or A3 = A 2 - A + I
or A4 = A3 - A2 + A
The determinant of the matrix 
is- a)0
- b)-9
- c)-27
- d)1
Correct answer is option 'C'. Can you explain this answer?
The determinant of the matrix is
isa)
0
b)
-9
c)
-27
d)
1
|
|
Chirag Verma answered |
We need to find the determinant of the matrix,
The determinant of the given matrix,
= (-3) (1) (1 - 4) + (6) (2) (1 - 4)
= 9 - 36 = -27
The determinant of the given matrix,
= (-3) (1) (1 - 4) + (6) (2) (1 - 4)
= 9 - 36 = -27
Let T : P3[0 ,1] --> P2[0 , 1] be defined by (Tp) (x) = P"(x) + P'(x). Then the matrix representation of T with respect to the basis {1, x, x2, x3} and {1, x, x2} of P3[0, 1] and P2[0, 1], respectively is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Let T : P3[0 ,1] --> P2[0 , 1] be defined by (Tp) (x) = P"(x) + P'(x). Then the matrix representation of T with respect to the basis {1, x, x2, x3} and {1, x, x2} of P3[0, 1] and P2[0, 1], respectively is
a)
b)
c)
d)
|
|
Chirag Verma answered |
Let T : P3[0, 1] --> P2[0,1] be a linear transformation defined by (Tp) (x) = p"(x) + p'(x ). We need to find the matrix of T with respect to the basis { l , x , x2, x3} and {1, x, x2} of P3[0, 1] and P2[0, 1] respectively.
Now,
and 
Therefore, the matrix of T with respect to the basis {1, x, x2, x3} and { 1,x,x2} of P3[0, 1] and P2[0, 1] respectively is
Now,
and Therefore, the matrix of T with respect to the basis {1, x, x2, x3} and { 1,x,x2} of P3[0, 1] and P2[0, 1] respectively is
Let T : R2-->R 2 be a map defined by
T(x,y) = (x + y , x - y)
Which of the following statement is correct?- a)T is linear and its kernel has infinite number of elements of R2
- b)T is non linear
- c)The kernel of T consist of only two elements of R2
- d)Nullity of T is zero
Correct answer is option 'D'. Can you explain this answer?
Let T : R2-->R 2 be a map defined by
T(x,y) = (x + y , x - y)
Which of the following statement is correct?
T(x,y) = (x + y , x - y)
Which of the following statement is correct?
a)
T is linear and its kernel has infinite number of elements of R2
b)
T is non linear
c)
The kernel of T consist of only two elements of R2
d)
Nullity of T is zero
|
|
Chirag Verma answered |
We are given that a linear transformation T : R2—> R2 is defined by
T(x,y) = (x + y , x - y)
= αT(x1
Hence, T is a linear transformation.
Let (x, y) ∈ ker T Then T(x, y) = (0, 0) Implies(x + y , x —y) = (0, 0)
Implies x +y = 0 and x - y = 0
Solving for x and y, we get
x = 0, y = 0
Hence ker T= {0, 0}
Therefore,
Nullity of T= dim ker T = 0
T(x,y) = (x + y , x - y)
= αT(x1
,
y
1) + βT(x2, y2)Hence, T is a linear transformation.
Let (x, y) ∈ ker T Then T(x, y) = (0, 0) Implies(x + y , x —y) = (0, 0)
Implies x +y = 0 and x - y = 0
Solving for x and y, we get
x = 0, y = 0
Hence ker T= {0, 0}
Therefore,
Nullity of T= dim ker T = 0
For a positive integer n, let Pn denote the space of all polynomials P(x) with coefficients in R such that degree P(x) ≤ n and let Bn denote the standard basis of Pn given by Bn= 
If T : P3 ---> P4 is the linear transformation defined by 
and A = (aij) is the 5 x 4 matrix of T with respect to the standard bases B3 and B4 then- a)a32 = 3/2 and a33 = 7/3
- b)a32 = 3/2 and a33 = 0
- c)a32 = 0 and a33 = 7/3
- d)a32 = 0 and a33 = 0
Correct answer is option 'B'. Can you explain this answer?
For a positive integer n, let Pn denote the space of all polynomials P(x) with coefficients in R such that degree P(x) ≤ n and let Bn denote the standard basis of Pn given by Bn=
If T : P3 ---> P4 is the linear transformation defined by and A = (aij) is the 5 x 4 matrix of T with respect to the standard bases B3 and B4 then
If T : P3 ---> P4 is the linear transformation defined by
and A = (aij) is the 5 x 4 matrix of T with respect to the standard bases B3 and B4 thena)
a32 = 3/2 and a33 = 7/3
b)
a32 = 3/2 and a33 = 0
c)
a32 = 0 and a33 = 7/3
d)
a32 = 0 and a33 = 0
|
|
Veda Institute answered |
For a positive integer n, let Pn denote the space of all polynomials P(x) with coefficients in M such that degree P(x) ≤ n and let Bn denote the standard basis of Pn given by Bn = {1, x, x2,....., xn}
If T : P3---->P4 is a linear transformation defined by
T(P(x)) = x2P'(x) +
and A = (aij) is the 5 x 4 matrix of T with respect to the standard bases B3 and B4
Hence, the matrix of T related to the basis
Hence a32 = 3/2 and a33 = 0
If T : P3---->P4 is a linear transformation defined by
T(P(x)) = x2P'(x) +
and A = (aij) is the 5 x 4 matrix of T with respect to the standard bases B3 and B4
Hence, the matrix of T related to the basis
Hence a32 = 3/2 and a33 = 0If M is a 7 x 5 matrix of rank 3 and N is a 5 x 7 matrix of rank 5, then rank (MN) is- a)5
- b)3
- c)2
- d)1
Correct answer is option 'B'. Can you explain this answer?
If M is a 7 x 5 matrix of rank 3 and N is a 5 x 7 matrix of rank 5, then rank (MN) is
a)
5
b)
3
c)
2
d)
1
|
|
Chirag Verma answered |
We are given that M is a 7 x 5 matrix of rank 3 and N is a 5 x 7 matrix of rank 5.
or
Since,
or
Since,
Let S = T : R3 —> R3 ; T is a linear transformation with T(1,0,1) = (1, 2, 3), T(1,2, 3) = (1,0,1)}. Then S is- a)A singleton set
- b)A finite set containing more than one element
- c)A countable infinite set
- d)An uncountable set
Correct answer is option 'D'. Can you explain this answer?
Let S = T : R3 —> R3 ; T is a linear transformation with T(1,0,1) = (1, 2, 3), T(1,2, 3) = (1,0,1)}. Then S is
a)
A singleton set
b)
A finite set containing more than one element
c)
A countable infinite set
d)
An uncountable set
|
|
Chirag Verma answered |
Let S = {T : R3---> R3 such that T is a linear transformation with
T(1, 0, 1) = (1, 2, 3), T(1, 2, 3) = (1,0,1)}.
Then we can define the linear transformation for the third independent element in any way. Therefore, we can get uncountably many linear transformation.
T(1, 0, 1) = (1, 2, 3), T(1, 2, 3) = (1,0,1)}.
Then we can define the linear transformation for the third independent element in any way. Therefore, we can get uncountably many linear transformation.
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?
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
|
Aanya Sharma answered |
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]
The rank of the matrix (m × n) where m<n cannot be more than?- a)n
- b)m
- c)m * n
- d)m - n
Correct answer is option 'B'. Can you explain this answer?
The rank of the matrix (m × n) where m<n cannot be more than?
a)
n
b)
m
c)
m * n
d)
m - n
|
|
Chirag Verma answered |
let us consider a 2 × 3 matrix
Where R1 ≠ R2 rank is 2
Another 2 × 3 matrix
Here, R1 = R2 rank is 1
And the rank of these two matrices is 1, 2
So rank cannot be more than m.
Where R1 ≠ R2 rank is 2
Another 2 × 3 matrix
Here, R1 = R2 rank is 1
And the rank of these two matrices is 1, 2
So rank cannot be more than m.
and I = 
. Which of the following is the zero matrix?
Consider the following linear transformation from the vector space R2 into the vector space R3.
T(x, y ) = (- x - y ,3x + 8y, 9x - 11y)
Then, the rank and nullity of T are respectively.- a)2 and 0
- b)1 and 0
- c)1 and 1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Consider the following linear transformation from the vector space R2 into the vector space R3.
T(x, y ) = (- x - y ,3x + 8y, 9x - 11y)
Then, the rank and nullity of T are respectively.
T(x, y ) = (- x - y ,3x + 8y, 9x - 11y)
Then, the rank and nullity of T are respectively.
a)
2 and 0
b)
1 and 0
c)
1 and 1
d)
None of these
|
|
Chirag Verma answered |
We are given that a linear transformation
T : R2 —> R3 defined by
T(x, y) = (-x -y, 3x + 8y, 9x - 11y)
We need to find the rank and nullity of T.
Let x, y ∈ ker T.
Then
T(x, y) = (0, 0, 0)
Using the definition of T, we get
Comparing the components on both sides, we get
Solving for * and y, we get x = 0, y = 0.
Hence,ker T= {(0, 0)},
Therefore,Nullity of T = dim ker T = 0.
Using Rank Nullity theorem, we get Rank of T = 2 - Nullity of T
= 2 - 0 = 2 .
Hence, Rank and Nullity of T are 2 and 0 respectively.
T : R2 —> R3 defined by
T(x, y) = (-x -y, 3x + 8y, 9x - 11y)
We need to find the rank and nullity of T.
Let x, y ∈ ker T.
Then
T(x, y) = (0, 0, 0)
Using the definition of T, we get
Comparing the components on both sides, we getSolving for * and y, we get x = 0, y = 0.Hence,ker T= {(0, 0)},
Therefore,Nullity of T = dim ker T = 0.
Using Rank Nullity theorem, we get Rank of T = 2 - Nullity of T
= 2 - 0 = 2 .
Hence, Rank and Nullity of T are 2 and 0 respectively.
Find the value of k for which the following simultaneous equations
x + y + z = 3; x + 2y + 3z = 4; x + 4y + kz = 6 will not have a unique solution.- a)0
- b)5
- c)6
- d)7
Correct answer is option 'D'. Can you explain this answer?
Find the value of k for which the following simultaneous equations
x + y + z = 3; x + 2y + 3z = 4; x + 4y + kz = 6 will not have a unique solution.
x + y + z = 3; x + 2y + 3z = 4; x + 4y + kz = 6 will not have a unique solution.
a)
0
b)
5
c)
6
d)
7
|
|
Chirag Verma answered |
We need to find the value of k for which the following simultaneous equations
x + y + z= 3
x + 2y + 3z = 4
x + 4y + kz = 6
will not have a unique solution. This system of equation may be written as,
Reduce this system of equation to echelon form using the operations
"R2 → R2 - R1" and R3 → R3 - R1.
These operations yield -
and also, R3 → R3 - 3R2 gives,
also this system of equation has not unique solution if,
k-7 = 0
k = 7
x + y + z= 3
x + 2y + 3z = 4
x + 4y + kz = 6
will not have a unique solution. This system of equation may be written as,
Reduce this system of equation to echelon form using the operations
"R2 → R2 - R1" and R3 → R3 - R1.
These operations yield -
and also, R3 → R3 - 3R2 gives,
also this system of equation has not unique solution if,
k-7 = 0
k = 7
Consider the system of linear equation
x + y + z = 3, x - y - z = 4, x - 5 y + kz = 6
Then, the value of k for which this system has an infinite number of solution is- a)k = - 5
- b)k = 0
- c)k = 1
- d)k = 3
Correct answer is option 'A'. Can you explain this answer?
Consider the system of linear equation
x + y + z = 3, x - y - z = 4, x - 5 y + kz = 6
Then, the value of k for which this system has an infinite number of solution is
x + y + z = 3, x - y - z = 4, x - 5 y + kz = 6
Then, the value of k for which this system has an infinite number of solution is
a)
k = - 5
b)
k = 0
c)
k = 1
d)
k = 3
|
|
Chirag Verma answered |
We are given that the system of equations,
x + y +z = 3
x - y - z = 4
x - 5y + kz = 6
has infinitely many solution and we need to find the value of k. The given system of equation may be written in the matrix form as,
Reduce this system of equations to echelon form using the operations "R2 —> R2 - R1", "R3 --> R3 - R1" these operations yield -
and also applying "R3 -> R3 - 3R2" we get.
Since, the system has infinite solution it is possible when,
k + 5 = 0 or k = - 5
x + y +z = 3
x - y - z = 4
x - 5y + kz = 6
has infinitely many solution and we need to find the value of k. The given system of equation may be written in the matrix form as,
Reduce this system of equations to echelon form using the operations "R2 —> R2 - R1", "R3 --> R3 - R1" these operations yield -
and also applying "R3 -> R3 - 3R2" we get.
Since, the system has infinite solution it is possible when,
k + 5 = 0 or k = - 5
If the linear transformation T : R2 → R3 is such that T(1, 0) = (2,3,1) and T(1,1) = (3, 0,2), then- a)T(x, y) = (x +y, 2x + y, 3x - 3y)
- b)T(x, y) = (2x + y, 3x - 3y, x +y)
- c)T(x, y) = (2x - y, 3x + 3y, x - y)
- d)T(x, y) = (x - y, 2x - y , 3x + 3y)
Correct answer is option 'B'. Can you explain this answer?
If the linear transformation T : R2 → R3 is such that T(1, 0) = (2,3,1) and T(1,1) = (3, 0,2), then
a)
T(x, y) = (x +y, 2x + y, 3x - 3y)
b)
T(x, y) = (2x + y, 3x - 3y, x +y)
c)
T(x, y) = (2x - y, 3x + 3y, x - y)
d)
T(x, y) = (x - y, 2x - y , 3x + 3y)
|
|
Veda Institute answered |
We are given that a linear transformation T : R3→ R3 defined by
T(1, 0) = (2, 3,1)
and T(l, 1) = (3, 0, 2)
We need to find the image of (x, y) under linear transformation T.
Let there exist scalars α and β such that
(x,y) = α(1,0) + β(1, 1)
or equivalently (x, y)= (α + β, β)
Comparing the components on both sides, we get
x = α + β and y = β
Solving for α and β, we get
α = x - y and β = y
Therefore,(x, y)= (x - y) (1, 0) + y(1, 1) Taking the image under linear transformation T. we get
implies T(x, y) = (2x + y, 3x - 3y, x+y).
T(1, 0) = (2, 3,1)
and T(l, 1) = (3, 0, 2)
We need to find the image of (x, y) under linear transformation T.
Let there exist scalars α and β such that
(x,y) = α(1,0) + β(1, 1)
or equivalently (x, y)= (α + β, β)
Comparing the components on both sides, we get
x = α + β and y = β
Solving for α and β, we get
α = x - y and β = y
Therefore,(x, y)= (x - y) (1, 0) + y(1, 1) Taking the image under linear transformation T. we get
implies T(x, y) = (2x + y, 3x - 3y, x+y).
The unique linear transformation T : R2 → R2 such that T(1,2) = (2,3) and T(0, 1) = (1,4). Then, the rule for T is.- a)T(x, y) = (y, - 5x + 4y)
- b)T(x,y) = (-5x + 4y,y)
- c)T(x, y) = (x, -5x + 4y)
- d)T(x, y) = (- 4x + 5y, y)
Correct answer is option 'A'. Can you explain this answer?
The unique linear transformation T : R2 → R2 such that T(1,2) = (2,3) and T(0, 1) = (1,4). Then, the rule for T is.
a)
T(x, y) = (y, - 5x + 4y)
b)
T(x,y) = (-5x + 4y,y)
c)
T(x, y) = (x, -5x + 4y)
d)
T(x, y) = (- 4x + 5y, y)
|
|
Veda Institute answered |
We need to find the linear transformation T : R2 → R2 such that
T(1, 2) = (2,3) and T(0, 1) = (1, 4).
Let (x, y) ∈ R2 and α, β are scalars such that
(x,y) = α(1 , 2) + β(0 ,1)
(x,y) = (α , 2α + β)
On comparing the Camponents of coordinates, we get
α = x and y2 = 2α + β
Solving for α and β, we get
α = x and β = y - 2x
Therefore,(x, y) = x (1, 2) + (y - 2x) (0, 1)
Taking the image under linear transformation T, we get
T(x, y) = T[x(1, 2 ) + (y - 2x) (0,1)]
Using the property of linearity, we get
T(x, y) = xT(1, 2) + (y - 2x) T(0, 1)
Substituting the value of T(1, 2) and T(0, 1), we get
T(x, y) = x(2, 3) + (y - 2x) (1, 4)
= (2x, 3x) + (y - 2x, 4y - 8x)
= (y, 4y-5x)
T(1, 2) = (2,3) and T(0, 1) = (1, 4).
Let (x, y) ∈ R2 and α, β are scalars such that
(x,y) = α(1 , 2) + β(0 ,1)
(x,y) = (α , 2α + β)
On comparing the Camponents of coordinates, we get
α = x and y2 = 2α + β
Solving for α and β, we get
α = x and β = y - 2x
Therefore,(x, y) = x (1, 2) + (y - 2x) (0, 1)
Taking the image under linear transformation T, we get
T(x, y) = T[x(1, 2 ) + (y - 2x) (0,1)]
Using the property of linearity, we get
T(x, y) = xT(1, 2) + (y - 2x) T(0, 1)
Substituting the value of T(1, 2) and T(0, 1), we get
T(x, y) = x(2, 3) + (y - 2x) (1, 4)
= (2x, 3x) + (y - 2x, 4y - 8x)
= (y, 4y-5x)
Define T(v) = Av, where A = 
. Then T(v):- a)reflects v about the x2-axis
- b)reflects v about the x1-axis
- c)rotates v clockwise π/2 radians about the origin
- d)rotates v counterclockwise x/2 radians about the origin
Correct answer is option 'A'. Can you explain this answer?
Define T(v) = Av, where A = . Then T(v):
. Then T(v):a)
reflects v about the x2-axis
b)
reflects v about the x1-axis
c)
rotates v clockwise π/2 radians about the origin
d)
rotates v counterclockwise x/2 radians about the origin
|
|
Veda Institute answered |
We are given that
T(v) = Av
where A =
We need to define T(v)
Let v =
, then T(v) = Av = 
Which reflects v about the x2-axis.
T(v) = Av
where A =
We need to define T(v)
Let v =
, then T(v) = Av = Which reflects v about the x2-axis.
Let A be a 4 x 4 matrix with real entries such that - 1, 1, 2, - 2 are its eigen values. If B = A4 - 5A2 + 5I, where I denotes the 4 x 4 identity matrix, then which of the following statements are correct?- a)trace of A - B is 0
- b)det (B) = 1
- c)det(A + B) = 0
- d)trace of A + B is 4
Correct answer is option 'C'. Can you explain this answer?
Let A be a 4 x 4 matrix with real entries such that - 1, 1, 2, - 2 are its eigen values. If B = A4 - 5A2 + 5I, where I denotes the 4 x 4 identity matrix, then which of the following statements are correct?
a)
trace of A - B is 0
b)
det (B) = 1
c)
det(A + B) = 0
d)
trace of A + B is 4
|
|
Chirag Verma answered |
Correct Answer :- C
Explanation : Det ( A) = 1*-1*2*-2 = 4
Det (B) = 14 - 52 + 5 = 1 (Hence product of its eigenvalues = 1) [take all 1,1,1,1]
Det(A+B) = -1+1 , 1+1, 2+1, -2+1 = 0*2*3*-1 = 0
If the system of equations
x - 2y - 3z = 1, (p + 2)z = 3, (2p + 1)y + z = 2 is inconsistent, then what is the value of P?- a)-2
- b)-1/2
- c)0
- d)2
Correct answer is option 'A'. Can you explain this answer?
If the system of equations
x - 2y - 3z = 1, (p + 2)z = 3, (2p + 1)y + z = 2 is inconsistent, then what is the value of P?
x - 2y - 3z = 1, (p + 2)z = 3, (2p + 1)y + z = 2 is inconsistent, then what is the value of P?
a)
-2
b)
-1/2
c)
0
d)
2
|
|
Chirag Verma answered |
We are given that the system of equations,
x - 2y - 3z = l, (p + 2)z =3, (2p + l) y + z = 2 is inconsistent. Then we need to find the value of p.
The augmented matrix of this system of equations is,
The solution of this matrix is inconsistent if
p + 2 = 0
or p = -2
x - 2y - 3z = l, (p + 2)z =3, (2p + l) y + z = 2 is inconsistent. Then we need to find the value of p.
The augmented matrix of this system of equations is,
The solution of this matrix is inconsistent if
p + 2 = 0
or p = -2
Let T : R3 --> R3 be a linear transformation defined by T(x, y, z) =(x + y - z, x + y + z, y - z)
Then, the matrix of the linear transformation T with respect to the ordered basis B = {(0,1,0), (0,0,1), (1,0, 0)| of R3 is- a)
- b)
- c)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Let T : R3 --> R3 be a linear transformation defined by T(x, y, z) =(x + y - z, x + y + z, y - z)
Then, the matrix of the linear transformation T with respect to the ordered basis B = {(0,1,0), (0,0,1), (1,0, 0)| of R3 is
Then, the matrix of the linear transformation T with respect to the ordered basis B = {(0,1,0), (0,0,1), (1,0, 0)| of R3 is
a)
b)
c)
d)
None of these
|
|
Chirag Verma answered |
Let T : R3 ---> R3 be a linear transformation defined by
T(x,y,z) = (x + y - z , x + y + z, y - z)
We need to find the matrix of T with respect to the ordered basis B = {(0, 1,0), (0,0, 1), (1, 0,0)} of R3.
Now
T(0,1,0) = (1,1,1)
= 1(0, 1, 0) + 1(0, 0, 1) + 1(0, 0, 0)
T(0, 0, 1) = (-1, 1, -1)
= 1(0, 1, 0 ) - 1(0, 0 , 1) - 1( 1, 0 , 0)
T(1, 0, 0) =(1,1,0)
= 1(0,1,0) + 0(0,0,1) + 1(1,0,0)
Therefore, the matrix of T with respect to the ordered basis
B = {(0, 1, 0), (0, 0, 1), (1, 0, 0)} is
T(x,y,z) = (x + y - z , x + y + z, y - z)
We need to find the matrix of T with respect to the ordered basis B = {(0, 1,0), (0,0, 1), (1, 0,0)} of R3.
Now
T(0,1,0) = (1,1,1)
= 1(0, 1, 0) + 1(0, 0, 1) + 1(0, 0, 0)
T(0, 0, 1) = (-1, 1, -1)
= 1(0, 1, 0 ) - 1(0, 0 , 1) - 1( 1, 0 , 0)
T(1, 0, 0) =(1,1,0)
= 1(0,1,0) + 0(0,0,1) + 1(1,0,0)
Therefore, the matrix of T with respect to the ordered basis
B = {(0, 1, 0), (0, 0, 1), (1, 0, 0)} is
Consider the vector space R3 and the maps f g : R3 —> R3 defined by f ( x , y, z) = (x, | y |, z) and g(x, y, z) = (x + 1, y - 1, z). Then,- a)both f and g are linear
- b)neither f nor g is linear
- c)g is linear but not f
- d)f is linear but not g
Correct answer is option 'B'. Can you explain this answer?
Consider the vector space R3 and the maps f g : R3 —> R3 defined by f ( x , y, z) = (x, | y |, z) and g(x, y, z) = (x + 1, y - 1, z). Then,
a)
both f and g are linear
b)
neither f nor g is linear
c)
g is linear but not f
d)
f is linear but not g
|
|
Chirag Verma answered |
We are given that two linear transformations , g : R3 --> R3 defined by
F(x ,y ,z ) = (x ,|y|,z)
and g(x, y, z) = (x + 1, y - 1, z)
Let (0 ,1,2) and (0, - 1 , 2) be two vectors of R3 then ( 0 ,1 , 2) = (0 ,1 ,2 )
and (0,-1 ,2 ) = (0 ,1 ,2 )
therefore f(0,1 ,2) + f(0 ,-1, 2) = (0,1,2) + (0,1,2) = (0, 2, 4)
but f [(0,1,2) + (0,-1,2)] = f(0 0,4) = (0 ,0 ,4)
Hence, f(0,1,2) + f(0, -1, 2) ≠ [(0 ,1, 2) + (0,- 1, 2)]
Thus , f is non linear
Next, g(0, 0, 0) = (1, -1, 0)
but(1,- 1, 0) ≠ (0, 0, 0)
hence, g is also non linear.
F(x ,y ,z ) = (x ,|y|,z)
and g(x, y, z) = (x + 1, y - 1, z)
Let (0 ,1,2) and (0, - 1 , 2) be two vectors of R3 then ( 0 ,1 , 2) = (0 ,1 ,2 )
and (0,-1 ,2 ) = (0 ,1 ,2 )
therefore f(0,1 ,2) + f(0 ,-1, 2) = (0,1,2) + (0,1,2) = (0, 2, 4)
but f [(0,1,2) + (0,-1,2)] = f(0 0,4) = (0 ,0 ,4)
Hence, f(0,1,2) + f(0, -1, 2) ≠ [(0 ,1, 2) + (0,- 1, 2)]
Thus , f is non linear
Next, g(0, 0, 0) = (1, -1, 0)
but(1,- 1, 0) ≠ (0, 0, 0)
hence, g is also non linear.
Let A be a square matrix and AT be its transpose matrix. Then A – AT is …. - a)Zero matrix
- b)Skew-symmetric matrix
- c)Identity matrix
- d)Symmetric matrix
Correct answer is option 'B'. Can you explain this answer?
Let A be a square matrix and AT be its transpose matrix. Then A – AT is ….
a)
Zero matrix
b)
Skew-symmetric matrix
c)
Identity matrix
d)
Symmetric matrix
|
|
Chirag Verma answered |
use the definition of transpose matrix, then we have A is skew-symmetric matrix.
A is any n x n matrix with entries equal to 1 then- a)multiplicity of 0 is n - 1
- b)multiplicity of 0 is 1
- c)multiplicity of 0 is n
- d)multiplicity of 0 is
Correct answer is option 'A'. Can you explain this answer?
A is any n x n matrix with entries equal to 1 then
a)
multiplicity of 0 is n - 1
b)
multiplicity of 0 is 1
c)
multiplicity of 0 is n
d)
multiplicity of 0 is
|
|
Chirag Verma answered |
We are given that A is any n x n with entries equal to 1. Suppose, if A is 2 x 2 matrix with entries equal to 1, that is,
multiplicity of 0 is 1.
For 3 x 3 matrix with entries equal to 1,
i.e.,
multiplicity of 0 is 2.
By continuing this process, for n x n matrix with entries equal to 1, that is,
A =
multiplicity of 0 is (n - 1).
multiplicity of 0 is 1.
For 3 x 3 matrix with entries equal to 1,
i.e.,
multiplicity of 0 is 2.
By continuing this process, for n x n matrix with entries equal to 1, that is,
A =
multiplicity of 0 is (n - 1).
The value of α for which the system of equations
x + y + z = 0
y + 2z = 0
αx + z = 0 has more than one solution is - a)-1
- b)0
- c)1/2
- d)1
Correct answer is option 'A'. Can you explain this answer?
The value of α for which the system of equations
x + y + z = 0
y + 2z = 0
αx + z = 0 has more than one solution is
x + y + z = 0
y + 2z = 0
αx + z = 0 has more than one solution is
a)
-1
b)
0
c)
1/2
d)
1
|
|
Chirag Verma answered |
We are given that the system of equation,
x + y + z =0
y + 2z = 0
αx + z= 0
has more than one solution. We need to find the value of α. The determinant of the coefficient matrix must be zero, i.e.,
x + y + z =0
y + 2z = 0
αx + z= 0
has more than one solution. We need to find the value of α. The determinant of the coefficient matrix must be zero, i.e.,
If A = 
, then calculate A9.- a) 511A + 510I
- b)309A +104I
- c)154A + 155I
- d)exp(9A)
Correct answer is option 'A'. Can you explain this answer?
If A = , then calculate A9.
, then calculate A9.a)
511A + 510I
b)
309A +104I
c)
154A + 155I
d)
exp(9A)
|
|
Veda Institute answered |
We are given the matrix A = 
The characteristic equation of the given matrix is |A-λI| = 0
or
or λ2 + 3λ + 2 = 0
By Cayley Hamilton theorem, this characteristic equation can be written in the matrix form as,
The characteristic equation of the given matrix is |A-λI| = 0
or
or λ2 + 3λ + 2 = 0
By Cayley Hamilton theorem, this characteristic equation can be written in the matrix form as,
The system of linear equations 4x + 2y =7, 2x + y = 6 has- a)a unique solution
- b)no solution
- c)an infinite number of solutions
- d)exactly two distinct solutions
Correct answer is option 'B'. Can you explain this answer?
The system of linear equations 4x + 2y =7, 2x + y = 6 has
a)
a unique solution
b)
no solution
c)
an infinite number of solutions
d)
exactly two distinct solutions
|
Pagal Insan answered |
Here, 4x + 2y = 7 and 2x + y = 6
[4 2 ] = [ 7]
[ 2 1 ] [ 6]
R1 : R1- 2R2
[ 0 0 ] = [ - 5 ]
[ 2 1] [ 6 ]
here In left side only one row and in right side is 2 row
so A < b="" is="" no="" solution="" />
# if A> B then a unique solution
# if A = B then is infinite number of solution.
[4 2 ] = [ 7]
[ 2 1 ] [ 6]
R1 : R1- 2R2
[ 0 0 ] = [ - 5 ]
[ 2 1] [ 6 ]
here In left side only one row and in right side is 2 row
so A < b="" is="" no="" solution="" />
# if A> B then a unique solution
# if A = B then is infinite number of solution.
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