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QUESTION: 1

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.

Solution:

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

"R_{2} → R_{2} - R_{1}" and R_{3} → R_{3 }- R_{1}.

These operations yield -

and also, R_{3} → R_{3} - 3R_{2} gives,

also this system of equation has not unique solution if,

k-7 = 0

k = 7

QUESTION: 2

Solution for the system defined by the set of equations 4y + 3z = 8; 2x - z = 2 and 3x + 2y = 5 is

Solution:

We need to determine the solution for the system of equation

4y+3z= 8

2x - z = 2

3x + 2y= 5

This system of equation may be written in matrix form as,

Reduce this system of equation to echelon form using the operations

"R_{2} ⇔ R_{1}". This operations yields

and also applying "R_{3 }→ 2R_{3} - 3R_{1}" which yields

again, "R_{1}→ R_{3} - R_{2} " we get

Here the coefficient matrix

and augmented matrix is given by

Since, the rank of coefficient matrix ≠ the rank of augmented matrix.

Therefore, the solution does not exist.

QUESTION: 3

Consider the system of simultaneous equation

x + 2y + z = 6;

2x+y + 2z = 6;

x+y + z = 5

This system has

Solution:

We are given that the system of equation

This system of equation can be written in augmented matrix as,

Applying the operations "R_{2} → R_{2} - 2R_{1}" and "R_{3} → R_{3} - R_{1}" these operations yields:

also applying R_{3} → 3R_{3} - R_{2}, which yields -

Since the rank of coefficient matrix ≠ the rank of augmented matrix.

Therefore, the given system of equations has no solution.

QUESTION: 4

A_{2x2} matrix which satisfy A^{2} - A = 0, then

Solution:

We are given that A be any 2 x 2 matrix which satisfy A^{2} - A = 0

The characteristic equation of the given matrix is

There exists distinct eigen values.

Hence, A must be diagonalizable.

QUESTION: 5

A is any n x n matrix with entries equal to 1 then

Solution:

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).

QUESTION: 6

Consider the system x + y + z = 0; x - y - z = 0, then the system of equations have

Solution:

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.

QUESTION: 7

If the linear transformation T : R^{2} → R^{3} is such that T(1, 0) = (2,3,1) and T(1,1) = (3, 0,2), then

Solution:

We are given that a linear transformation T : R^{3}→ R^{3} 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).

QUESTION: 8

The unique linear transformation T : R^{2 }→ R^{2} such that T(1,2) = (2,3) and T(0, 1) = (1,4). Then, the rule for T is.

Solution:

We need to find the linear transformation T : R^{2 }→ R^{2} such that

T(1, 2) = (2,3) and T(0, 1) = (1, 4).

Let (x, y) ∈ R^{2} 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 y^{2} = 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)

QUESTION: 9

For a linear transformation T : R^{10} → R^{6}, the kernal has dimension 5. Then, the dimension of the range of T is

Solution:

We are given a linear transformation T : R^{10} → R^{6} and ker T has dimension 5. We need to find the dimension of range T.

using rank nullity theorem, we get dim range T = dim R^{10 }- dim ker T.

= 10 - 5 = 5

Therefore, the dim of range T is 5.

QUESTION: 10

Transformation

(x, y, z) → (x + y, y + z) : R^{3} → R^{2} is

Solution:

We are given that a linear transformation T : R^{3} → R^{2} defined by T(x, y, z) = (x + y, y + z)

Here, we need to check its linearity and also need to find ker T.

Let (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) be any two vector of R^{3}, Then

T[(x_{1}, y_{1}, z_{1}) + (x_{2}, y_{2}, z_{2})]

Again let α be a scalar number then,

T[α(x,y, z)]

Hence, T is a linear.

Let (x, y, z) ∈ ker T, then T(x, y, z) = (0,0)

using the definition of ker T, we get

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

Implies x + y - 0 and

y + z = 0

Implies x = - y and y = - z

Therefore, ker T = {(x, -x, x ) : x e R}

Therefore, dim ker T = 1

Hence, ker T is a proper subspace of R^{3}

QUESTION: 11

Let P_{2}[x] be the vector space of all polynomials over R of degree less than or equal to 2. Let D be the differential operator on P_{2}[x]. Then, matrix of D relative td the basis [x^{2}, 1, x] is equal to

Solution:

Let P_{2}(x) be the vector space of all polynomials over R of degree less than or equal to 2 and D be the differential operator defined on P_{2}[x].

We need to find the matrix of D related to the basis {x^{3}, 1, x} Now

Therefore, the matrix of D related to the basis {x^{2}, 1, x} is

QUESTION: 12

If T : V^{2}(R) → V_{3}(R) defined as T(a, b) = (a + b , a-b ,b) is a linear transformation, Then nullity of T is

Solution:

We are given that a linear transformation T: V_{2}(R) → V_{3}(R) defined by

T(a, b) = (a + b , a - b, b).

We need to find the nullity of linear transformation T.

Let (a, b) ∈ ker T, Then

T(a,b)= (0,0,0)

Now using the definition of linear transformation, we get

(a + b, a - b , b) = (0, 0, 0)

Comparing the components of the co-ordinates, we get

a + b = 0, a - b = 0 , b = 0

Solving for a, b, we get

a = 0,6 = 0

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

Thus, dim (ker T) = 0, that is nullity of linear transformation T is zero.

QUESTION: 13

The transition matrix P from the standard ordered basis to the ordered basis {(1, 1), (-1,0)} is

Solution:

We need to find the transition matrix P from the standard ordered basis to the ordered basis {(1, 1), (-1, 0)}.

We know that two standard ordered basis for R^{2 }is {(1, 0), (0, 1)}.

Therefore, Let there exists scalars α and β such that

(1 ,0 )= α (1, 1) + β (-1,0 )

or equivalently (1,0) = (α - β, α )

Comparing the components of the coordinates, we get α - β = 1, α = 0 Solving for α and β, we get α = 0, β = 1 Therefore, (1, 0) = 0(1, 1) - 1(-1, 0) Again let there be scalars a and P such that

(0 ,1) = α(l, 1) + β ( - l , 0) equivalently(0, 1) = (α - β, α)

Comparing the components of the coordinates, we get α - β = 0, α = 1 Solving for α and β we get

α = 1, β = 1.

Therefore, (0, 1) = 1(1,1) + 1(-1, 0)

By using (1,0) = 0(1, 1) - 1(-1, 0)

and (0,1) = 1(1, 1) + 1 (- 1 ,0 )

The transition matrix is

QUESTION: 14

Let F be any field and let T be a linear operator on F^{2} defined by T(a, b) = (a + b, a), then T^{-1}(a, b) is equal to:

Solution:

Let F be any field and let T be a linear operator on F^{2} defined by

T(a, b) = (a + b, a)

We need to find T^{-1}(a, b).

Let (a, b) ∈ ker T.

Then T(a, b) = (0, 0)

Using the definition of linear transformation, we get

(a + b, a) = (0, 0)

Comparing the components of the coordinates, we get

a + b = 0, a = 0

Solving for α and β, we get a = 0, b = 0.

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

Hence, T is one-one. Since T is a linear transformation on finite dimensional vector space F^{2}. Therefore, T is onto. Hence, T is invertible linear transformation.

Let (x,y) be the image of (a,b) under T^{-1}.

Then (x, y) = T^{-1}(a, b) or equivalently T(x, y) = (a, b)

Using the definition of linear transformation,

we get (x+y,y)= (a,b)

Comparing the components of the coordinates, we get

x + y = a and y = b

Solving for x and y, we get x= a - b , y = b.

Therefore,T^{-1}(a, b) = (a - b, b)

QUESTION: 15

A linear transformation T : R^{2 }→ R^{2} such that T(3, 1) = (2, -4) and T(1, 1) = (0,2). Then, T(7, 8) is

Solution:

We are given that a linear transformation T : R^{2} → R^{2}

such that T(3, l ) = (2,- 4).

and T(1,1) = (0, 2)

We need to find T(7, 8).

Let there exist scalar α and β such that (7, 8)

= α (3,1) + β(1,1)

or equivalently (7,8) = (3α + β, α + β)

On comparing the components of coordinates, we get

3α + β = 7 and α + β = 8

Solving for α and β, we get

Therefore,

Now taking the image under linear transformation T, we get

Substituting the values of T(3, 1) and T(1, 1), we get

= (-1,2) + (0,17) = (-1,19)

Thus, the image of (7,8) under the linear transformation T is (-1, 19).

QUESTION: 16

A is any matrix which satisfy A^{3} - A^{2} + A - I = 0 and A_{3x3}, then A^{4 }is

Solution:

We are given that, A be any 3 x 3 matrix which satisfy

A^{3}- A^{2 }+ A - I = 0

and we need to find A^{4}.

A^{3}- A^{2}+ A - I = 0

or A^{3} = A^{ 2 }- A + I

or A^{4} = A^{3} - A^{2} + A

QUESTION: 17

The eigen values of the matrix are

Solution:

We need to find the eigen values of the matrix A =

The characteristics equation of the given matrix A,

|A - λI |= 0

or

or (4 - λ) (1 - λ) - 4 = 0

or λ^{2} - 5λ = 0

=> λ(λ-5) = 0

=> λ = 0,5

The eigen values of the given matrix is 0 and 5.

QUESTION: 18

Which one of the following is an eigen vector of the matrix

Solution:

We need to find the eigen vector of the matrix

The characteristic equation of the given matrix is,

Put λ = 5 in equation | A - λI | X = 0 that is,

or

5z =0

-3z +1 = 0

3z - 4f = 0

z = 0, t = 0

and also let x = k_{1}, y = k_{2}

Hence, the eigen vector corresponding to the eigen value λ= 5 is_{,
}

Let k_{1} = 1, and k_{2 }= -2, then

QUESTION: 19

The eigen vectors o f the matrix are written in the form and . What is a + b ?

Solution:

We are given that the eigen vectors o f the matrix are and . We need to find the value o f a + b. The characteristic equation o f the given matrix is, |A - λI| = 0

or (1 - λ)(2 - λ) = 0 or

λ = 1,2

Put λ = 1 in the equation [ A- λI ] X= 0

or a = 0 and also put X = 2 in the equation [A - λI] X= 0

or

or 1+2b = 0

or b = 1/2

Now, a + b = 0 + 1/2

a + b = 1/2

QUESTION: 20

An eigen vector o f A = is

Solution:

We need to find the eigen vector of the matrix A

=

The characteristic equation of the given matrix is, |A - λI| = 0

or = 0

or (1 - λ) (2 - λ,) (3 - λ) = 0

λ = 1,2,3

Suppose, X be the eigen vector corresponding to the eigen value λ = 3.

Hence, (A - λI) X = 0

or

or -2x + y = 0

-y + 2z = 0

Let x = k

Hence y = 2k and z = k

Now, the eigen vector is,

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