Description

This mock test of Test: Linear Algebra - 9 for Mathematics helps you for every Mathematics entrance exam.
This contains 20 Multiple Choice Questions for Mathematics Test: Linear Algebra - 9 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Linear Algebra - 9 quiz give you a good mix of easy questions and tough questions. Mathematics
students definitely take this Test: Linear Algebra - 9 exercise for a better result in the exam. You can find other Test: Linear Algebra - 9 extra questions,
long questions & short questions for Mathematics on EduRev as well by searching above.

QUESTION: 1

Let T be the linear operator on R^{3} defined by T(x_{1}, x_{2}, x_{3}) = ( 2x_{1}, x_{1} - x_{2}, 5x_{1} + 4x_{2} + x_{3}). Then T^{-1} is

Solution:

Let T be the linear operator on R^{3} defined by

T(x_{1} x_{2}, x_{3}) = (2x, x_{1} - x_{2}, 5x_{1} + 4x_{2} + x_{3})

Let (x_{1}, x_{2}, x_{3}) ∈ ker T.

Then we need to find T^{-1}

Now, T(x_{1,} x_{2}, x_{3}) - (0, 0, 0)

Using the definition of T, we get

(2x, x_{l} - x_{2}, 5x_{1} + 4x_{2} + x_{3}) = (0, 0, 0)

Comparing the components of the coordinates, we get

2x_{1} = 0, x_{1} - x_{2 }= 0, 5x_{1} + 4 x 2 + x 3 = 0

Solving for x_{1} x_{2} and x_{3}, we get x_{1} = 0, x_{2} = 0, x_{3} = 0

Hence,ker T= {(0, 0, 0)}

and hence, T is one-one.

Since R^{3} is a 3-dimensional vector space and T : R^{3} —> R^{3} be a one-one linear transformation and hence, T is onto. Therefore, T^{-1} exist.

Let (a, b, c) be the image of (x_{1}, x_{2}, x_{3}) under T^{-1}.

Then T^{-1}(x_{1,}x_{2}, x_{3}) = (a, b, c) ImpliesT(a, b, c) = (x_{1,}x_{2}, x_{3})

Using the definition of T, we get

(2a a - b 5a + 4b + c) = (x_{1,}x_{2}, x_{3}) Comparing the components, we get 2a = x_{1}, a - b = x_{2}, 5a l 4b I c = x_{3} Solving for a, b, c, we get

Therefore, T^{-1}(x_{1} x_{2}, x_{3}) =

QUESTION: 2

The minimum polynomial m(λ) of M =

Solution:

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,

QUESTION: 3

The eigen values of a 3 x 3 real matrix A are 1,2 and - 3. Then

Solution:

We are given that the eigen values of a 3 x 3 real matrix A are 1,2 and -3 say λ_{1} λ_{2} and λ_{3}.

The characteristic equation of the given matrix is,

By Cayley Hamilton’s theorem, we can write this equation in the matrix form as,

QUESTION: 4

The minimal polynomial of is

Solution:

We need to find the minimal polynomial of the matrix

The characteristic polynomial of the given matrix is,

ch_{A}(x) = (x - 1)^{2} (x - 2) (x - 2)

and the minimal polynomial of the given matrix

= l.c.m. of (x - 1)^{2}, (x - 2) and (x-2)

= (x-1)^{2} (x-2)

QUESTION: 5

Let A be 3 x 3 matrix whose characteristic roots are 3 ,2 ,-1 .

If B = A^{2} - A, then |B| is

Solution:

We are given that A be a 3 x 3 matrix whose ' characteristic roots are 3, 2, -1 say λ_{1}, λ_{2} and λ_{3} and B = A^{2} - A

λ_{1} = 3,

λ_{1 }= (3)^{2} - 3 = 6

λ^{2} = 2,

λ_{2} = (2)^{2} - 2 = 2

λ_{3} = -1

λ_{3} = (-1)^{2} - (-1) = 2

The determinant of B is, | B | = product of its eigen values

= λ_{1}'λ_{2}'λ_{3}'

= 6 x 2 x 2 = 24

QUESTION: 6

Let

Then, the minimal polynomial of the matrix is

Solution:

We are given that,

and

We need to find the minimal polynomial of the matrix

The characteristic polynomial of the given matrix M is,

ch_{m} (x) =(x - 3)^{2} (x - 3) (x + 3)^{2} (x - 0)^{2}

And the minimal polynomial of the given matrix M is

m_{x}(x) = l.c.m. o f (x - 3)^{2},

(x - 3),(x + 3)^{2}, (x - 0)^{2}

or m_{m}(x) = (x - 3)^{2} (x + 3)^{2}x^{2}

or m_{m} (x) = x^{2}(x - 3)^{2} (x + 3)^{2}

QUESTION: 7

Let T : C^{n }—> C^{n} be a linear operator of rank n - 2. Then,

Solution:

We are given that a linear operator T : C^{n}—> C^{n} of rank n - 2.

Since T : C^{n} —> C^{n }is a linear operator of rank n - 2.

Therefore, T is singular operator, Hence 0 must be an eigen value of T.

QUESTION: 8

Choose the correct matching from A, B, C and D for the transformation T_{1}, T_{2 }and T_{3} (mappings from R^{2} to R^{3}) as defined in Group 1 with the statements given in Group 2

Group 1

P.T_{1}x( x, y) = (x , x, 0)

Q.T_{2}(x,y) = (x, x +y ,y)

R.T_{3}(x,y) = (x,x+1,y)

Group 2

1. Linear transformation of rank 2

2. Not a linear transformation

3. Linear transformation of rank 1

Solution:

Consider the linear transformation defined by

T_{1}(x,y) = (x,x, 0)

Let (x_{1},x_{1}) and (x_{2},x_{2}) be two vectors and α, β be scalars. Then

T_{1}[α(x_{1} y_{1}) + β(x_{2}, y_{2})]

= T_{1}[(αx_{1} + βx_{2}, αy_{1} + βy_{2})]

= (αx_{1} + βx_{2}, αx_{1} + βx_{2}, 0)

= α(x_{1},x_{1}, 0) + β(x_{2}, x_{2}, 0)

= αT_{1}(x_{1},y_{1}) + βT_{1}(x_{2},y_{2})

or equivalently

Therefore, T_{1} is linear transformation.

Let (x,y) ∈ ker T_{1}.

Then T(x,y) = (0, 0, 0)

Using the definition of linear transformation T_{1},we get

(x, x, 0) = (0, 0, 0)

Comparing the components of the co-ordinates, we get,

x = 0 and y is arbitrary

Therefore,ker T_{1}{ = {(0, y ) : y ∈ R}

Hence,dim ker T_{1} = 1.

Using Rank nullity theorem Rank T_{1}

= 2 - dim ker T_{1 }

= 2 - 1 = 1.

Therefore, 'T_{1 }is a linear transformation of rank 1.

Next, consider the linear transformation T_{2} defined by

T_{2}(x,y) = (x,x+y,y)

Let (x_{1},y_{1}) and (x_{2},y_{2}) be any two vectors and α, β are scalars.

Then Hence, T_{2} is a linear transformation.

Let (x, y) ∈ ker T.

Then T_{2}(x,y) = (0, 0, 0)

Using the definition of linear transformation T_{2}, we get

(.x, x+ y,y ) = (0, 0, 0)

Comparing the components of the coordinates, we get

x = 0, x + y = 0, y = 0

Solving for x and y, we get x = 0, y = 0

Hence,ker T_{2} = {(0,0)}.

Therefore,dim ker T_{2} = 0

Using rank nullity theorem,

Rank T_{2} = 2 - dim ker T_{2}

= 2 - 0 - 2 .

Hence, T_{2} is a linear transformation of rank 2.

Next, consider the linear transformation T3 defined by

T_{3}(x,y) = (x,x+1, y)

Then T_{3}(0, 0) =(0, 1, 0)

but (0,1,0) is not a zero of codomain space.

Hence, the image of (0, 0) under T3 is not a zero.

Therefore T3 is not a linear transformation.

QUESTION: 9

Let V, W and X be three finite dimensional vector spaces such that dim V = dim X, Suppose S : V --> W and T : W ---> X are two linear maps such ToS : V --> X is injective.Then,

Solution:

Let V, W and X be three finite dimensional vector spaces such that dim V = dim X. Suppose S : V --> W and T : W --> X are two linear maps such ToS : V --> X is injective.Then S is injective and T is surjective.

QUESTION: 10

Let S and T be two linear operators on R^{3} defined by

S (x,y,z) = (x,x + y , x - y - z )

T(x , y , z) = (x + 2z ,y - z ,x + y + z)

Solution:

Let S and T be two linear operators on R^{3} 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 R^{3}. 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.

QUESTION: 11

Let A be the matrix of quadratic form (x_{1}- x_{2} + 2x_{3})^{2}.

Then, trace of A is

Solution:

We are given that A be a matrix of quardratic form (x_{1}-x_{2 }+ 2x_{3})^{2}

or x_{1}^{2} + x_{2}^{2} + 4x_{3}^{2} - 2x_{1}x_{2} - 4x_{2}x_{3} + 4x_{3}x_{1}

trace (A) = sum of coefficient of

x_{1}^{2}, x_{2}^{2} and x_{3}^{2}.

= 1 + 1 + 4

= 6

QUESTION: 12

The eigen values of a 3 x 3 real matrix P are 1, -2, 3. Then,

Solution:

We are given that the eigen values of a 3 x 3 real matrix P are 1, - 2 ,3 say λ_{1}, λ_{2}, λ_{3}.

The characteristic equation of the given matrix P is,

x^{2} - (trace P) x^{2} + (λ_{1}•λ2 + λ_{2}•λ_{3} + λ_{3}•λ_{1})x - |P| = 0

or x^{3 }- (1 - 2 + 3)x^{2} + (-2 - 6 + 3) x -(-6) = 0

or x^{3} — 2x^{2} — 5x + 6 = 0

By Cayley Hamilton theorem, we can write this equation in the matrix form.

p^{3} - 2P^{2} - 5p + 6I = 0

or 6I = 5P + 2P^{2} - P^{3}

or 6I = P(5I + 2P - P^{2})

or P^{-1} = 1/6(5I + 2P - P^{2})

QUESTION: 13

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

Solution:

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

‘R_{2} --> R_{2} - R_{1}' and R_{3} ---> R_{3} - R_{1}, these operations yields -

Also applying , R_{3} —> 5R_{3} - R_{2} we get -

Since, this system of equations has infinitely man solution. It is possible when 5α - 7 = 0

or α = 7/5

QUESTION: 14

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

Solution:

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 "R_{2} —> R_{2} - R_{1}", "R_{3 }--> R_{3} - R_{1}" these operations yield -

and also applying "R_{3} -> R_{3} - 3R_{2}" we get.

Since, the system has infinite solution it is possible when,

k + 5 = 0 or k = - 5

QUESTION: 15

Real matrices [A]_{3x1},[B]_{3x3}, [C]_{3 x 5} [D]_{5 x 3}, [E]_{5 x 5} and [F]_{5 x 1}, are given. Matrices [B] and [E] are symmetric. Following statements are made with respect to these matrices:

(i) Matrix product [F]^{T} [C]^{T} [B] [C] [F] is a scalar.

(ii) Matrix product [D]^{T} [F] [D] is always symmetric. With reference to above statements, which of the following applies?

Solution:

QUESTION: 16

For any two matrices A and B, which is true?

Solution:

QUESTION: 17

Matrix A = is

Solution:

QUESTION: 18

Matrix A = is

Solution:

QUESTION: 19

Match List I with List II and select the correct answer using the codes given below in the lists.

Codes

Solution:

QUESTION: 20

Given an orthogonal matrix

, then (AA')^{-1} is

Solution:

### Lec 9 | 18.06 Linear Algebra, Spring 2005

Video | 50:14 min

### Geometry of Linear Algebra | 18.06SC Linear Algebra, Fall 2011

Video | 16:36 min

### Dense-Linear-Algebra

Doc | 69 Pages

### Linear Algebra (Part - 2)

Doc | 2 Pages

- Test: Linear Algebra - 9
Test | 20 questions | 60 min

- Test: Linear Algebra - 10
Test | 20 questions | 60 min

- Test: Linear Algebra - 11
Test | 20 questions | 60 min

- Test: Linear Algebra - 7
Test | 20 questions | 60 min

- Linear Algebra MCQ - 2
Test | 30 questions | 90 min