IGNOU MTE-02 Solved Assignment 2023

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IGNOU MTE-02 Solved Assignment 2023 Sample Solution
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  1. Which of the following statements are true and which are false? Justify your answer with a short proof or a counterexample.
i) The function f : R → R f : R → R f:RrarrRf: \mathbf{R} \rightarrow \mathbf{R}f:R→R defined by f ( x ) = cos ⁡ x f ( x ) = cos ⁡ x f(x)=cos xf(x)=\cos xf(x)=cos⁡x is 1 − 1 1 − 1 1-11-11−1.
Answer:
To determine whether the statement “The function f : R → R f : R → R f:RrarrRf: \mathbf{R} \rightarrow \mathbf{R}f:R→R defined by f ( x ) = cos ⁡ x f ( x ) = cos ⁡ x f(x)=cos xf(x) = \cos xf(x)=cos⁡x is 1-1″ is true or false, we need to understand what a 1-1 (one-to-one) function is and then apply this definition to the cosine function.

Definition of a 1-1 Function:

A function f : A → B f : A → B f:A rarr Bf: A \rightarrow Bf:A→B is called one-to-one (or injective) if for every x 1 , x 2 ∈ A x 1 , x 2 ∈ A x_(1),x_(2)in Ax_1, x_2 \in Ax1,x2∈A, whenever f ( x 1 ) = f ( x 2 ) f ( x 1 ) = f ( x 2 ) f(x_(1))=f(x_(2))f(x_1) = f(x_2)f(x1)=f(x2), it must be the case that x 1 = x 2 x 1 = x 2 x_(1)=x_(2)x_1 = x_2x1=x2. In simpler terms, no two different inputs in the domain of the function should map to the same output in the codomain.

Applying the Definition to f ( x ) = cos ⁡ x f ( x ) = cos ⁡ x f(x)=cos xf(x) = \cos xf(x)=cos⁡x:

To test if f ( x ) = cos ⁡ x f ( x ) = cos ⁡ x f(x)=cos xf(x) = \cos xf(x)=cos⁡x is 1-1, we need to see if there are any distinct values x 1 x 1 x_(1)x_1x1 and x 2 x 2 x_(2)x_2x2 in the domain R R R\mathbf{R}R (the set of all real numbers) such that cos ⁡ x 1 = cos ⁡ x 2 cos ⁡ x 1 = cos ⁡ x 2 cos x_(1)=cos x_(2)\cos x_1 = \cos x_2cos⁡x1=cos⁡x2 but x 1 ≠ x 2 x 1 ≠ x 2 x_(1)!=x_(2)x_1 \neq x_2x1≠x2.

Counterexample:

Consider the values x 1 = 0 x 1 = 0 x_(1)=0x_1 = 0x1=0 and x 2 = 2 π x 2 = 2 π x_(2)=2pix_2 = 2\pix2=2π. These are distinct values in R R R\mathbf{R}R, but:
  • cos ⁡ 0 = 1 cos ⁡ 0 = 1 cos 0=1\cos 0 = 1cos⁡0=1
  • cos ⁡ 2 π = 1 cos ⁡ 2 π = 1 cos 2pi=1\cos 2\pi = 1cos⁡2π=1
Here, cos ⁡ x 1 = cos ⁡ x 2 cos ⁡ x 1 = cos ⁡ x 2 cos x_(1)=cos x_(2)\cos x_1 = \cos x_2cos⁡x1=cos⁡x2 even though x 1 ≠ x 2 x 1 ≠ x 2 x_(1)!=x_(2)x_1 \neq x_2x1≠x2. This shows that the function f ( x ) = cos ⁡ x f ( x ) = cos ⁡ x f(x)=cos xf(x) = \cos xf(x)=cos⁡x is not 1-1, as it violates the definition of a one-to-one function.

Conclusion:

The statement “The function f : R → R f : R → R f:RrarrRf: \mathbf{R} \rightarrow \mathbf{R}f:R→R defined by f ( x ) = cos ⁡ x f ( x ) = cos ⁡ x f(x)=cos xf(x) = \cos xf(x)=cos⁡x is 1-1″ is false. The counterexample of x 1 = 0 x 1 = 0 x_(1)=0x_1 = 0x1=0 and x 2 = 2 π x 2 = 2 π x_(2)=2pix_2 = 2\pix2=2π demonstrates that two different inputs can yield the same output, which contradicts the definition of a one-to-one function.
ii) The operation * defined by x ∗ y = log ⁡ ( x y ) x ∗ y = log ⁡ ( x y ) x**y=log(xy)x * y=\log (x y)x∗y=log⁡(xy) is a binary operation on S S SSS, where S S SSS is the set { x ∈ R ∣ x > 0 } { x ∈ R ∣ x > 0 } {x inR∣x > 0}\{x \in \mathbf{R} \mid x>0\}{x∈R∣x>0}.
Answer:
To determine whether the statement “The operation ∗ ∗ ***∗ defined by x ∗ y = log ⁡ ( x y ) x ∗ y = log ⁡ ( x y ) x**y=log(xy)x * y = \log(xy)x∗y=log⁡(xy) is a binary operation on S S SSS, where S S SSS is the set { x ∈ R ∣ x > 0 } { x ∈ R ∣ x > 0 } {x inR∣x > 0}\{x \in \mathbf{R} \mid x > 0\}{x∈R∣x>0}” is true or false, we need to understand what a binary operation is and then apply this definition to the given operation.

Definition of a Binary Operation:

A binary operation on a set S S SSS is a rule that assigns to each ordered pair of elements in S S SSS a unique element in S S SSS. In other words, if ∗ ∗ ***∗ is a binary operation on S S SSS, then for every x , y ∈ S x , y ∈ S x,y in Sx, y \in Sx,y∈S, it must be the case that x ∗ y ∈ S x ∗ y ∈ S x**y in Sx * y \in Sx∗y∈S.

Applying the Definition to x ∗ y = log ⁡ ( x y ) x ∗ y = log ⁡ ( x y ) x**y=log(xy)x * y = \log(xy)x∗y=log⁡(xy):

We need to check if for every x , y ∈ S x , y ∈ S x,y in Sx, y \in Sx,y∈S, where S = { x ∈ R ∣ x > 0 } S = { x ∈ R ∣ x > 0 } S={x inR∣x > 0}S = \{x \in \mathbf{R} \mid x > 0\}S={x∈R∣x>0}, the result of x ∗ y = log ⁡ ( x y ) x ∗ y = log ⁡ ( x y ) x**y=log(xy)x * y = \log(xy)x∗y=log⁡(xy) is also in S S SSS.
  1. Closure Property: For x , y ∈ S x , y ∈ S x,y in Sx, y \in Sx,y∈S, we have x > 0 x > 0 x > 0x > 0x>0 and y > 0 y > 0 y > 0y > 0y>0. The product x y x y xyxyxy is also greater than 0 since the product of two positive numbers is positive. The logarithm function, log log log\loglog, is defined for positive real numbers. Therefore, log ⁡ ( x y ) log ⁡ ( x y ) log(xy)\log(xy)log⁡(xy) is a real number. Since log ⁡ ( x y ) log ⁡ ( x y ) log(xy)\log(xy)log⁡(xy) is defined and results in a real number for all x , y > 0 x , y > 0 x,y > 0x, y > 0x,y>0, the operation ∗ ∗ ***∗ is closed in S S SSS.
  2. Result in S S SSS: We need to ensure that log ⁡ ( x y ) log ⁡ ( x y ) log(xy)\log(xy)log⁡(xy) is also greater than 0 to be in S S SSS. However, this is not necessarily the case. For example, if x = y = 1 2 x = y = 1 2 x=y=(1)/(2)x = y = \frac{1}{2}x=y=12, then x y = 1 4 x y = 1 4 xy=(1)/(4)xy = \frac{1}{4}xy=14, and log ⁡ ( 1 4 ) log ⁡ 1 4 log((1)/(4))\log\left(\frac{1}{4}\right)log⁡(14) is negative, which is not in S S SSS.

Conclusion:

The statement “The operation ∗ ∗ ***∗ defined by x ∗ y = log ⁡ ( x y ) x ∗ y = log ⁡ ( x y ) x**y=log(xy)x * y = \log(xy)x∗y=log⁡(xy) is a binary operation on S S SSS, where S S SSS is the set { x ∈ R ∣ x > 0 } { x ∈ R ∣ x > 0 } {x inR∣x > 0}\{x \in \mathbf{R} \mid x > 0\}{x∈R∣x>0}” is false. The operation ∗ ∗ ***∗ is not closed in S S SSS because there exist x , y ∈ S x , y ∈ S x,y in Sx, y \in Sx,y∈S such that x ∗ y = log ⁡ ( x y ) x ∗ y = log ⁡ ( x y ) x**y=log(xy)x * y = \log(xy)x∗y=log⁡(xy) is not in S S SSS. The example x = y = 1 2 x = y = 1 2 x=y=(1)/(2)x = y = \frac{1}{2}x=y=12 demonstrates that the result of the operation can be outside of S S SSS, violating the definition of a binary operation.
iii) The set { ( x 1 ⋅ x 2 , … , x n ) ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 } x 1 ⋅ x 2 , … , x n ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 {(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}\left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1=2 x_2+3\right\}{(x1⋅x2,…,xn)∣x1,x2,…,xn∈R,x1=2x2+3} is a subspace of R n R n R^(n)\mathbf{R}^nRn.
Answer:
To determine whether the statement “The set { ( x 1 ⋅ x 2 , … , x n ) ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 } x 1 ⋅ x 2 , … , x n ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 {(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}\left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1 = 2x_2 + 3\right\}{(x1⋅x2,…,xn)∣x1,x2,…,xn∈R,x1=2x2+3} is a subspace of R n R n R^(n)\mathbf{R}^nRn” is true or false, we need to understand what a subspace is and then apply this definition to the given set.

Definition of a Subspace:

A subset W W WWW of a vector space V V VVV is a subspace of V V VVV if and only if it satisfies three conditions:
  1. Zero Vector: The zero vector of V V VVV is in W W WWW.
  2. Closed under Addition: For every u , v ∈ W u , v ∈ W u,v in Wu, v \in Wu,v∈W, the sum u + v u + v u+vu + vu+v is in W W WWW.
  3. Closed under Scalar Multiplication: For every u ∈ W u ∈ W u in Wu \in Wu∈W and every scalar c c ccc, the product c u c u cucucu is in W W WWW.

Applying the Definition to the Given Set:

Let’s denote the set as S = { ( x 1 ⋅ x 2 , … , x n ) ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 } S = x 1 ⋅ x 2 , … , x n ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 S={(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}S = \left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1 = 2x_2 + 3\right\}S={(x1⋅x2,…,xn)∣x1,x2,…,xn∈R,x1=2x2+3} and check if it satisfies the subspace criteria in R n R n R^(n)\mathbf{R}^nRn.
  1. Zero Vector: The zero vector in R n R n R^(n)\mathbf{R}^nRn is ( 0 , 0 , … , 0 ) ( 0 , 0 , … , 0 ) (0,0,dots,0)(0, 0, \ldots, 0)(0,0,…,0). For this vector to be in S S SSS, we need x 1 = 2 x 2 + 3 = 0 x 1 = 2 x 2 + 3 = 0 x_(1)=2x_(2)+3=0x_1 = 2x_2 + 3 = 0x1=2x2+3=0, which is not possible since 2 x 2 + 3 2 x 2 + 3 2x_(2)+32x_2 + 32x2+3 is never zero for any real value of x 2 x 2 x_(2)x_2x2. Therefore, the zero vector is not in S S SSS.
Since the zero vector is not in S S SSS, S S SSS cannot be a subspace of R n R n R^(n)\mathbf{R}^nRn. This alone is sufficient to conclude that the statement is false. However, for completeness, let’s briefly consider the other two properties:
  1. Closed under Addition: Even if we were to consider this property, the set S S SSS is not necessarily closed under addition. For example, if we take two different vectors from S S SSS that satisfy x 1 = 2 x 2 + 3 x 1 = 2 x 2 + 3 x_(1)=2x_(2)+3x_1 = 2x_2 + 3x1=2x2+3, their sum may not satisfy this condition.
  2. Closed under Scalar Multiplication: Similarly, scalar multiplication of a vector in S S SSS may not result in a vector that still satisfies x 1 = 2 x 2 + 3 x 1 = 2 x 2 + 3 x_(1)=2x_(2)+3x_1 = 2x_2 + 3x1=2x2+3.

Conclusion:

The statement “The set { ( x 1 ⋅ x 2 , … , x n ) ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 } x 1 ⋅ x 2 , … , x n ∣ x 1 , x 2 , … , x n ∈ R , x 1 = 2 x 2 + 3 {(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}\left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1 = 2x_2 + 3\right\}{(x1⋅x2,…,xn)∣x1,x2,…,xn∈R,x1=2x2+3} is a subspace of R n R n R^(n)\mathbf{R}^nRn” is false. The primary reason is that the zero vector of R n R n R^(n)\mathbf{R}^nRn is not in the set, violating the first and most fundamental condition for being a subspace.
iv) There is no 7 × 5 7 × 5 7xx57 \times 57×5 matrix of rank 6 .
Answer:
To evaluate the statement “There is no 7 × 5 7 × 5 7xx57 \times 57×5 matrix of rank 6,” we need to understand the concept of the rank of a matrix and the constraints imposed by the dimensions of the matrix.

Definition of Matrix Rank:

The rank of a matrix is defined as the maximum number of linearly independent column vectors in the matrix. It can also be equivalently defined as the maximum number of linearly independent row vectors in the matrix.

Analyzing the 7 × 5 7 × 5 7xx57 \times 57×5 Matrix:

  • A 7 × 5 7 × 5 7xx57 \times 57×5 matrix has 7 rows and 5 columns.
  • The rank of a matrix cannot exceed the number of rows or the number of columns. In other words, the rank of a matrix is limited by the smaller of these two dimensions.

Applying the Definition to a 7 × 5 7 × 5 7xx57 \times 57×5 Matrix:

  • Since the matrix in question is a 7 × 5 7 × 5 7xx57 \times 57×5 matrix, the maximum number of linearly independent columns it can have is 5 (the number of columns).
  • Similarly, the maximum number of linearly independent rows it can have is also 5 (since there are only 5 columns to form linearly independent combinations).

Conclusion:

The statement “There is no 7 × 5 7 × 5 7xx57 \times 57×5 matrix of rank 6″ is true. The rank of a 7 × 5 7 × 5 7xx57 \times 57×5 matrix cannot exceed the number of its columns, which is 5. Therefore, it is impossible for such a matrix to have a rank of 6.
v) If V V VVV and V ′ V ′ V^(‘)V^{\prime}V′ are vector spaces and T : V → V ′ T : V → V ′ T:V rarrV^(‘)T: V \rightarrow V^{\prime}T:V→V′ is a linear transformation, then whenever u 1 , u 2 , … , u k u 1 , u 2 , … , u k u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_ku1,u2,…,uk are linearly independent, T u 1 , T u 2 , … , T u k T u 1 , T u 2 , … , T u k Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_kTu1,Tu2,…,Tuk are also linearly independent.
Answer:
The statement “If V V VVV and V ′ V ′ V^(‘)V’V′ are vector spaces and T : V → V ′ T : V → V ′ T:V rarrV^(‘)T: V \rightarrow V’T:V→V′ is a linear transformation, then whenever u 1 , u 2 , … , u k u 1 , u 2 , … , u k u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_ku1,u2,…,uk are linearly independent, T u 1 , T u 2 , … , T u k T u 1 , T u 2 , … , T u k Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_kTu1,Tu2,…,Tuk are also linearly independent” needs to be evaluated for its truthfulness.

Definition of Linear Independence:

A set of vectors u 1 , u 2 , … , u k u 1 , u 2 , … , u k u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_ku1,u2,…,uk in a vector space is said to be linearly independent if the only solution to the linear equation c 1 u 1 + c 2 u 2 + … + c k u k = 0 c 1 u 1 + c 2 u 2 + … + c k u k = 0 c_(1)u_(1)+c_(2)u_(2)+dots+c_(k)u_(k)=0c_1 u_1 + c_2 u_2 + \ldots + c_k u_k = 0c1u1+c2u2+…+ckuk=0 (where 0 0 000 is the zero vector) is c 1 = c 2 = … = c k = 0 c 1 = c 2 = … = c k = 0 c_(1)=c_(2)=dots=c_(k)=0c_1 = c_2 = \ldots = c_k = 0c1=c2=…=ck=0.

Definition of a Linear Transformation:

A function T : V → V ′ T : V → V ′ T:V rarrV^(‘)T: V \rightarrow V’T:V→V′ is a linear transformation if for all u , v ∈ V u , v ∈ V u,v in Vu, v \in Vu,v∈V and scalars c c ccc, the following two properties hold:
  1. T ( u + v ) = T ( u ) + T ( v ) T ( u + v ) = T ( u ) + T ( v ) T(u+v)=T(u)+T(v)T(u + v) = T(u) + T(v)T(u+v)=T(u)+T(v)
  2. T ( c u ) = c T ( u ) T ( c u ) = c T ( u ) T(cu)=cT(u)T(cu) = cT(u)T(cu)=cT(u)

Analyzing the Statement:

The statement claims that if a set of vectors u 1 , u 2 , … , u k u 1 , u 2 , … , u k u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_ku1,u2,…,uk in V V VVV is linearly independent, then their images under T T TTT, namely T u 1 , T u 2 , … , T u k T u 1 , T u 2 , … , T u k Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_kTu1,Tu2,…,Tuk in V ′ V ′ V^(‘)V’V′, are also linearly independent.

Counterexample:

To show that the statement is false, we can provide a counterexample. Consider a linear transformation T T TTT that is not injective (one-to-one). For instance, let T T TTT be the zero transformation, which maps every vector in V V VVV to the zero vector in V ′ V ′ V^(‘)V’V′.
In this case, even if u 1 , u 2 , … , u k u 1 , u 2 , … , u k u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_ku1,u2,…,uk are linearly independent in V V VVV, their images under T T TTT will all be the zero vector in V ′ V ′ V^(‘)V’V′. The set { T u 1 , T u 2 , … , T u k } { T u 1 , T u 2 , … , T u k } {Tu_(1),Tu_(2),dots,Tu_(k)}\{T u_1, T u_2, \ldots, T u_k\}{Tu1,Tu2,…,Tuk} will then consist of only the zero vector, which is not a set of linearly independent vectors, as multiple non-zero scalar combinations of the zero vector will result in the zero vector.

Conclusion:

The statement “If V V VVV and V ′ V ′ V^(‘)V’V′ are vector spaces and T : V → V ′ T : V → V ′ T:V rarrV^(‘)T: V \rightarrow V’T:V→V′ is a linear transformation, then whenever u 1 , u 2 , … , u k u 1 , u 2 , … , u k u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_ku1,u2,…,uk are linearly independent, T u 1 , T u 2 , … , T u k T u 1 , T u 2 , … , T u k Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_kTu1,Tu2,…,Tuk are also linearly independent” is false. The existence of linear transformations that are not injective, such as the zero transformation, provides a counterexample where linearly independent vectors in V V VVV can map to vectors in V ′ V ′ V^(‘)V’V′ that are not linearly independent.
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