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IGNOU BMTE-141 Solved Assignment 2024 | B.Sc (G) CBCS
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IGNOU BMTE-141 Assignment Question Paper 2024
PART – A (30 Marks)
- i) Find the angle between the vectors
sqrt2i+2j+2k \sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k} andi+sqrt2j+sqrt2k \mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k} .
- a) Let
S S be any non-empty set and letV(S) V(S) be the set of all real valued functions onR \mathbb{R} . Define addition onV(s) V(s) by(f+g)(x)=f(x)+g(x) (f+g)(x)=f(x)+g(x) and scalar multiplication by(alpha*f)(x)=alpha f(x) (\alpha \cdot f)(x)=\alpha f(x) . Check that(V(S),+,*) (V(S),+, \cdot) is a vector space.
- a) Let
T:R^(3)rarrR^(3) T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 be a linear operator and suppose the matrix of the operator with respect to the ordered basis
- a) Find the eigenvalues and eigenvectors of the matrix
B=[[1,1,0],[-1,3,0],[1,-1,1]] B=\left[\begin{array}{rrr}1 & 1 & 0 \\ -1 & 3 & 0 \\ 1 & -1 & 1\end{array}\right] . Is the matrix diagonalisable? Justify your answer.
- a) Solve the folowing set of simultaneous equations using Cramer’s rule:
- a) Let
V V be the vector space of all real valued functions that are twice differentiable inR \mathbb{R} and
ii) Check whether
- a) Let
(x_(1),x_(2),x_(3)) \left(x_1, x_2, x_3\right) and(y_(1),y_(2),y_(3)) \left(y_1, y_2, y_3\right) represent the coordinates with respect to the basesB_(1)={(1,0,0),(1,1,0),(0,0,1)},B_(2)={(1,0,0),(0,1,1),(0,0,1)} B_1=\{(1,0,0),(1,1,0),(0,0,1)\}, B_2=\{(1,0,0),(0,1,1),(0,0,1)\} . If
- Which of the following statements are true and which are false? Justify your answer with a short proof or a counterexample.
i) IfW_(1) W_1 andW_(2) W_2 are proper subspaces of a non-zero, finite dimensional, vector spaceV V anddim(W_(1)) > (dim(V))/(2),dim(W_(2)) > (dim(V))/(2) \operatorname{dim}\left(W_1\right)>\frac{\operatorname{dim}(V)}{2}, \operatorname{dim}\left(W_2\right)>\frac{\operatorname{dim}(V)}{2} , theW_(1)nnW_(2)!={0} W_1 \cap W_2 \neq\{0\} .
BMTE-141 Sample Solution 2024
bmte-141-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d
- i) Find the angle between the vectors
sqrt2i+2j+2k \sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k} andi+sqrt2j+sqrt2k \mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k} .
-
Calculate the Dot Product
A*B \mathbf{A} \cdot \mathbf{B} :A*B=(sqrt2i+2j+2k)*(i+sqrt2j+sqrt2k) \mathbf{A} \cdot \mathbf{B} = (\sqrt{2}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{i} + \sqrt{2}\mathbf{j} + \sqrt{2}\mathbf{k}) =sqrt2*1+2*sqrt2+2*sqrt2 = \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2} -
Calculate the Magnitudes of
A \mathbf{A} andB \mathbf{B} :- Magnitude of
A \mathbf{A} ,|A| |\mathbf{A}| :|A|=sqrt((sqrt2)^(2)+2^(2)+2^(2)) |\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2} - Magnitude of
B \mathbf{B} ,|B| |\mathbf{B}| :|B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2)) |\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2}
- Magnitude of
-
Find the Angle
theta \theta :cos(theta)=(A*B)/(|A||B|) \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} theta=cos^(-1)((A*B)/(|A||B|)) \theta = \cos^{-1}\left(\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}\right) We will calculatetheta \theta after obtaining the dot product and magnitudes.
-
Dot Product
A*B \mathbf{A} \cdot \mathbf{B} :A*B=sqrt2*1+2*sqrt2+2*sqrt2=5sqrt2 \mathbf{A} \cdot \mathbf{B} = \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2} = 5\sqrt{2} -
Magnitude of
A \mathbf{A} ,|A| |\mathbf{A}| :|A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))=sqrt10 |\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2} = \sqrt{10} -
Magnitude of
B \mathbf{B} ,|B| |\mathbf{B}| :|B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))=sqrt5 |\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{5}
- Non-emptiness:
W W must contain the zero vector ofR^(3) \mathbb{R}^3 . - Closed under addition: For any two vectors
vec(u) \vec{u} andvec(v) \vec{v} inW W , the sumvec(u)+ vec(v) \vec{u} + \vec{v} must also be inW W . - Closed under scalar multiplication: For any vector
vec(u) \vec{u} inW W and any scalarc c , the productc vec(u) c\vec{u} must also be inW W .
-
Non-emptiness:
The zero vector inR^(3) \mathbb{R}^3 is(0,0,0) (0, 0, 0) . For this vector,x=0 x = 0 ,y=0 y = 0 , andz=0 z = 0 , sox+y-z=0+0-0=0 x + y – z = 0 + 0 – 0 = 0 . Therefore, the zero vector is inW W , satisfying the non-emptiness condition. -
Closed under addition:
Letvec(u)=(x_(1),y_(1),z_(1)) \vec{u} = (x_1, y_1, z_1) andvec(v)=(x_(2),y_(2),z_(2)) \vec{v} = (x_2, y_2, z_2) be any two vectors inW W . This meansx_(1)+y_(1)-z_(1)=0 x_1 + y_1 – z_1 = 0 andx_(2)+y_(2)-z_(2)=0 x_2 + y_2 – z_2 = 0 . We need to check if their sumvec(u)+ vec(v)=(x_(1)+x_(2),y_(1)+y_(2),z_(1)+z_(2)) \vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2) also satisfies the condition ofW W .Forvec(u)+ vec(v) \vec{u} + \vec{v} , we have:(x_(1)+x_(2))+(y_(1)+y_(2))-(z_(1)+z_(2))=(x_(1)+y_(1)-z_(1))+(x_(2)+y_(2)-z_(2))=0+0=0 (x_1 + x_2) + (y_1 + y_2) – (z_1 + z_2) = (x_1 + y_1 – z_1) + (x_2 + y_2 – z_2) = 0 + 0 = 0 Therefore,vec(u)+ vec(v) \vec{u} + \vec{v} is inW W , satisfying the closure under addition. -
Closed under scalar multiplication:
Letvec(u)=(x,y,z) \vec{u} = (x, y, z) be any vector inW W andc c be any scalar. Sincevec(u) \vec{u} is inW W ,x+y-z=0 x + y – z = 0 . We need to check ifc vec(u)=(cx,cy,cz) c\vec{u} = (cx, cy, cz) also satisfies the condition ofW W .Forc vec(u) c\vec{u} , we have:cx+cy-cz=c(x+y-z)=c*0=0 cx + cy – cz = c(x + y – z) = c \cdot 0 = 0 Therefore,c vec(u) c\vec{u} is inW W , satisfying the closure under scalar multiplication.
c_(1)+c_(3)=0 c_1 + c_3 = 0 c_(1)+c_(2)=0 c_1 + c_2 = 0 c_(2)=0 c_2 = 0 c_(3)=0 c_3 = 0
- Additivity:
T( vec(u)+ vec(v))=T( vec(u))+T( vec(v)) T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) - Homogeneity:
T(c vec(u))=cT( vec(u)) T(c\vec{u}) = cT(\vec{u})
-
Additivity:
Letvec(u)=(x_(1),y_(1)) \vec{u} = (x_1, y_1) andvec(v)=(x_(2),y_(2)) \vec{v} = (x_2, y_2) be any vectors inR^(2) \mathbb{R}^2 . We need to check ifT( vec(u)+ vec(v))=T( vec(u))+T( vec(v)) T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) .First, calculatevec(u)+ vec(v) \vec{u} + \vec{v} :vec(u)+ vec(v)=(x_(1)+x_(2),y_(1)+y_(2)) \vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2) Then, applyT T tovec(u)+ vec(v) \vec{u} + \vec{v} :T( vec(u)+ vec(v))=T(x_(1)+x_(2),y_(1)+y_(2))=(-(y_(1)+y_(2)),x_(1)+x_(2)) T(\vec{u} + \vec{v}) = T(x_1 + x_2, y_1 + y_2) = (-(y_1 + y_2), x_1 + x_2) Now, calculateT( vec(u)) T(\vec{u}) andT( vec(v)) T(\vec{v}) :T( vec(u))=T(x_(1),y_(1))=(-y_(1),x_(1)) T(\vec{u}) = T(x_1, y_1) = (-y_1, x_1) T( vec(v))=T(x_(2),y_(2))=(-y_(2),x_(2)) T(\vec{v}) = T(x_2, y_2) = (-y_2, x_2) Finally, addT( vec(u)) T(\vec{u}) andT( vec(v)) T(\vec{v}) :T( vec(u))+T( vec(v))=(-y_(1),x_(1))+(-y_(2),x_(2))=(-(y_(1)+y_(2)),x_(1)+x_(2)) T(\vec{u}) + T(\vec{v}) = (-y_1, x_1) + (-y_2, x_2) = (-(y_1 + y_2), x_1 + x_2) SinceT( vec(u)+ vec(v))=T( vec(u))+T( vec(v)) T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) , the additivity property is satisfied. -
Homogeneity:
Letvec(u)=(x,y) \vec{u} = (x, y) be any vector inR^(2) \mathbb{R}^2 andc c be any scalar. We need to check ifT(c vec(u))=cT( vec(u)) T(c\vec{u}) = cT(\vec{u}) .First, calculatec vec(u) c\vec{u} :c vec(u)=c(x,y)=(cx,cy) c\vec{u} = c(x, y) = (cx, cy) Then, applyT T toc vec(u) c\vec{u} :T(c vec(u))=T(cx,cy)=(-cy,cx) T(c\vec{u}) = T(cx, cy) = (-cy, cx) Now, calculateT( vec(u)) T(\vec{u}) and then multiply byc c :T( vec(u))=T(x,y)=(-y,x) T(\vec{u}) = T(x, y) = (-y, x) cT( vec(u))=c(-y,x)=(-cy,cx) cT(\vec{u}) = c(-y, x) = (-cy, cx) SinceT(c vec(u))=cT( vec(u)) T(c\vec{u}) = cT(\vec{u}) , the homogeneity property is satisfied.
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