IGNOU BMTE-141 Solved Assignment 2024 for Academic Excellence

IGNOU BMTE-141 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTE-141 Assignment Question Paper 2024

IGNOU BMTE-141 Assignment Question Paper 2024

PART – A (30 Marks)

  1. i) Find the angle between the vectors 2 i + 2 j + 2 k 2 i + 2 j + 2 k sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}2i+2j+2k and i + 2 j + 2 k i + 2 j + 2 k i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}i+2j+2k.
ii) Find the vector equation of the plane determined by the points ( 1 , 0 , − 1 ) , ( 0 , 1 , 1 ) ( 1 , 0 , − 1 ) , ( 0 , 1 , 1 ) (1,0,-1),(0,1,1)(1,0,-1),(0,1,1)(1,0,−1),(0,1,1) and ( − 1 , 1 , 0 ) ( − 1 , 1 , 0 ) (-1,1,0)(-1,1,0)(−1,1,0).
iii) Check whether W = { ( x , y , z ) ∈ R 3 ∣ x + y − z = 0 } W = ( x , y , z ) ∈ R 3 ∣ x + y − z = 0 W={(x,y,z)inR^(3)∣x+y-z=0}W=\left\{(x, y, z) \in \mathbb{R}^3 \mid x+y-z=0\right\}W={(x,y,z)∈R3∣x+y−z=0} is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3.
iv) Check whether the set of vectors { 1 + x , x + x 2 , 1 + x 3 } 1 + x , x + x 2 , 1 + x 3 {1+x,x+x^(2),1+x^(3)}\left\{1+x, x+x^2, 1+x^3\right\}{1+x,x+x2,1+x3} is a linearly independent set of vectors in P 3 P 3 P_(3)\mathbf{P}_3P3, the vector space of polynomials of degree ≤ 3 ≤ 3 <= 3\leq 3≤3.
v) Check whether T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2, defined by T ( x , y ) = ( − y , x ) T ( x , y ) = ( − y , x ) T(x,y)=(-y,x)T(x, y)=(-y, x)T(x,y)=(−y,x) is a linear transformation.
vi) If { v 1 , v 2 } v 1 , v 2 {v_(1),v_(2)}\left\{v_1, v_2\right\}{v1,v2} is an ordered basis of R 2 R 2 R^(2)\mathbb{R}^2R2 and { f 1 ( v ) , f 2 ( v ) } f 1 ( v ) , f 2 ( v ) {f_(1)(v),f_(2)(v)}\left\{f_1(v), f_2(v)\right\}{f1(v),f2(v)} is the corresponding dual basis find f 1 ( 2 v 1 + v 2 ) f 1 2 v 1 + v 2 f_(1)(2v_(1)+v_(2))f_1\left(2 v_1+v_2\right)f1(2v1+v2) and f 2 ( v 1 − 2 v 2 ) f 2 v 1 − 2 v 2 f_(2)(v_(1)-2v_(2))f_2\left(v_1-2 v_2\right)f2(v1−2v2).
vii) Find the kernel of the linear transformation T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 defined by T ( x , y ) = ( 2 x + 3 y , 2 x − 3 y ) T ( x , y ) = ( 2 x + 3 y , 2 x − 3 y ) T(x,y)=(2x+3y,2x-3y)T(x, y)=(2 x+3 y, 2 x-3 y)T(x,y)=(2x+3y,2x−3y).
viii) Describe the linear transformation T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 such that
[ T ] B = [ 1 2 2 0 ] [ T ] B = 1      2 2      0 [T]_(B)=[[1,2],[2,0]][T]_B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 0 \end{array}\right][T]B=[1220]
where B B BBB is the standard basis of R 2 R 2 R^(2)\mathbb{R}^2R2.
ix) Find the matrix of the linear transformation T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 defined by T ( x , y ) = ( 2 y , x − y ) T ( x , y ) = ( 2 y , x − y ) T(x,y)=(2y,x-y)T(x, y)=(2 y, x-y)T(x,y)=(2y,x−y) with respect to the ordered basis { ( 0 , − 1 ) , ( − 1 , 0 ) } { ( 0 , − 1 ) , ( − 1 , 0 ) } {(0,-1),(-1,0)}\{(0,-1),(-1,0)\}{(0,−1),(−1,0)}.
x) Let A A AAA be a 2 × 3 2 × 3 2xx32 \times 32×3 matrix, B B BBB be a 3 × 4 3 × 4 3xx43 \times 43×4 matrix and C C CCC be a 3 × 2 3 × 2 3xx23 \times 23×2 matrix and D D DDD be a 3 × 4 3 × 4 3xx43 \times 43×4 matrix. Is A B + C t D A B + C t D AB+C^(t)DA B+C^t DAB+CtD defined? Justify your answer.
xi) Verify Cayley-Hamilton theorem for the matrix A = [ 1 − 1 0 2 ] A = 1 − 1 0 2 A=[[1,-1],[0,2]]A=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]A=[1−102].
xii) Check whether [ 1 1 1 1 ] 1 1 1 1 [[1],[1],[1],[1]]\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right][1111] is an eigenvector for the matrix [ 1 0 0 1 0 1 1 0 0 0 1 1 0 1 0 1 ] 1      0      0      1 0      1      1      0 0      0      1      1 0      1      0      1 [[1,0,0,1],[0,1,1,0],[0,0,1,1],[0,1,0,1]]\left[\begin{array}{llll}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1\end{array}\right][1001011000110101]. What is the corresponding eigenvalue?
xiii) Let C [ 0 , 1 ] C [ 0 , 1 ] C[0,1]C[0,1]C[0,1] be the inner product space of continous real valued functions on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] with the inner product
⟨ f , g ⟩ = ∫ 0 1 f ( t ) g ( t ) d t ⟨ f , g ⟩ = ∫ 0 1   f ( t ) g ( t ) d t (:f,g:)=int_(0)^(1)f(t)g(t)dt\langle f, g\rangle=\int_0^1 f(t) g(t) d t⟨f,g⟩=∫01f(t)g(t)dt
Find the inner product of the functions f ( t ) = 2 t , g ( t ) = 1 t 2 + 5 f ( t ) = 2 t , g ( t ) = 1 t 2 + 5 f(t)=2t,g(t)=(1)/(t^(2)+5)f(t)=2 t, g(t)=\frac{1}{t^2+5}f(t)=2t,g(t)=1t2+5.
xiv) Find adjoint of the linear operator T : C 2 → C 2 T : C 2 → C 2 T:C^(2)rarrC^(2)T: \mathbb{C}^2 \rightarrow \mathbb{C}^2T:C2→C2 defined by T ( z 1 , z 2 ) = ( z 2 , z 1 + i z 2 ) T z 1 , z 2 = z 2 , z 1 + i z 2 T(z_(1),z_(2))=(z_(2),z_(1)+iz_(2))T\left(z_1, z_2\right)=\left(z_2, z_1+i z_2\right)T(z1,z2)=(z2,z1+iz2) with respect to the standard inner product on C 2 C 2 C^(2)C^2C2.
xv) Find the signature of the quadratic form x 1 2 − 2 x 2 2 + 3 x 3 2 x 1 2 − 2 x 2 2 + 3 x 3 2 x_(1)^(2)-2x_(2)^(2)+3x_(3)^(2)x_1^2-2 x_2^2+3 x_3^2x12−2x22+3x32
Part-B (40 Marks)
  1. a) Let S S SSS be any non-empty set and let V ( S ) V ( S ) V(S)V(S)V(S) be the set of all real valued functions on R R R\mathbb{R}R. Define addition on V ( s ) V ( s ) V(s)V(s)V(s) by ( f + g ) ( x ) = f ( x ) + g ( x ) ( f + g ) ( x ) = f ( x ) + g ( x ) (f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x) and scalar multiplication by ( α ⋅ f ) ( x ) = α f ( x ) ( α ⋅ f ) ( x ) = α f ( x ) (alpha*f)(x)=alpha f(x)(\alpha \cdot f)(x)=\alpha f(x)(α⋅f)(x)=αf(x). Check that ( V ( S ) , + , ⋅ ) ( V ( S ) , + , ⋅ ) (V(S),+,*)(V(S),+, \cdot)(V(S),+,⋅) is a vector space.
b) Check that B = { 1 , 2 x + 1 , ( x − 1 ) 2 } B = 1 , 2 x + 1 , ( x − 1 ) 2 B={1,2x+1,(x-1)^(2)}\boldsymbol{B}=\left\{1,2 x+1,(x-1)^2\right\}B={1,2x+1,(x−1)2} is a basis for P 2 P 2 P_(2)\mathbf{P}_2P2, the vector space of polynomials with real coefficients of degree ≤ 2 ≤ 2 <= 2\leq 2≤2.
  1. a) Let T : R 3 → R 3 T : R 3 → R 3 T:R^(3)rarrR^(3)T: \mathbb{R}^3 \rightarrow \mathbb{R}^3T:R3→R3 be a linear operator and suppose the matrix of the operator with respect to the ordered basis
B = { [ 1 0 0 ] , [ 1 0 1 ] , [ 0 1 0 ] } B = 1 0 0 , 1 0 1 , 0 1 0 B={[[1],[0],[0]],[[1],[0],[1]],[[0],[1],[0]]}\boldsymbol{B}=\left\{\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right\}B={[100],[101],[010]}
is [ 1 0 − 1 0 1 1 1 0 1 ] 1 0 − 1 0 1 1 1 0 1 [[1,0,-1],[0,1,1],[1,0,1]]\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right][10−1011101]. Find the matrix of the linear transformation with respect to the basis
B ′ = { [ 1 0 1 ] , [ 1 0 − 1 ] , [ 0 1 0 ] } B ′ = 1 0 1 , 1 0 − 1 , 0 1 0 B^(‘)={[[1],[0],[1]],[[1],[0],[-1]],[[0],[1],[0]]}\boldsymbol{B}^{\prime}=\left\{\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right\}B′={[101],[10−1],[010]}
b) Show that W = { ( x , 4 x , 3 x ) ∈ R 2 ∣ x ∈ R } W = ( x , 4 x , 3 x ) ∈ R 2 ∣ x ∈ R W={(x,4x,3x)inR^(2)∣x inR}W=\left\{(x, 4 x, 3 x) \in \mathbb{R}^2 \mid x \in \mathbb{R}\right\}W={(x,4x,3x)∈R2∣x∈R} is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3. Also find a basis for subspace U U UUU of R 3 R 3 R^(3)\mathbb{R}^3R3 which satisfies W ⊕ U = R 3 W ⊕ U = R 3 W o+U=R^(3)W \oplus U=\mathbb{R}^3W⊕U=R3.
  1. a) Find the eigenvalues and eigenvectors of the matrix B = [ 1 1 0 − 1 3 0 1 − 1 1 ] B = 1      1      0 − 1      3      0 1      − 1      1 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]B=[110−1301−11]. Is the matrix diagonalisable? Justify your answer.
b) Find Adj ⁡ ( A ) Adj ⁡ ( A ) Adj(A)\operatorname{Adj}(A)Adj⁡(A) where A = [ 3 2 2 − 1 1 0 3 0 1 ] A = 3 2 2 − 1 1 0 3 0 1 A=[[3,2,2],[-1,1,0],[3,0,1]]A=\left[\begin{array}{ccc}3 & 2 & 2 \\ -1 & 1 & 0 \\ 3 & 0 & 1\end{array}\right]A=[322−110301]. Hence find A − 1 A − 1 A^(-1)A^{-1}A−1.
  1. a) Solve the folowing set of simultaneous equations using Cramer’s rule:
x + 2 y + z = 3 2 x − y + 2 z = 1 3 x + y + z = 0 x + 2 y + z = 3 2 x − y + 2 z = 1 3 x + y + z = 0 {:[x+2y+z=3],[2x-y+2z=1],[3x+y+z=0]:}\begin{aligned} x+2 y+z & =3 \\ 2 x-y+2 z & =1 \\ 3 x+y+z & =0 \end{aligned}x+2y+z=32x−y+2z=13x+y+z=0
b) Find the minimal polynomial of the matrix
[ 2 1 0 1 − 1 0 0 1 − 2 − 2 − 1 3 0 0 0 1 ] 2      1      0      1 − 1      0      0      1 − 2      − 2      − 1      3 0      0      0      1 [[2,1,0,1],[-1,0,0,1],[-2,-2,-1,3],[0,0,0,1]]\left[\begin{array}{rrrr} 2 & 1 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ -2 & -2 & -1 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right][2101−1001−2−2−130001]
Part C (30 marks)
  1. a) Let V V VVV be the vector space of all real valued functions that are twice differentiable in R R R\mathbb{R}R and
S = { cos ⁡ x , sin ⁡ x , x cos ⁡ x , x sin ⁡ x } . S = { cos ⁡ x , sin ⁡ x , x cos ⁡ x , x sin ⁡ x } . S={cos x,sin x,x cos x,x sin x}.S=\{\cos x, \sin x, x \cos x, x \sin x\} .S={cos⁡x,sin⁡x,xcos⁡x,xsin⁡x}.
Check that S S SSS is a linearly independent set over R R R\mathbb{R}R. (Hint: Consider the equation
a 0 cos ⁡ x + a 1 sin ⁡ x + a 2 x cos ⁡ x + a 3 x sin ⁡ x . a 0 cos ⁡ x + a 1 sin ⁡ x + a 2 x cos ⁡ x + a 3 x sin ⁡ x . a_(0)cos x+a_(1)sin x+a_(2)x cos x+a_(3)x sin x.a_0 \cos x+a_1 \sin x+a_2 x \cos x+a_3 x \sin x .a0cos⁡x+a1sin⁡x+a2xcos⁡x+a3xsin⁡x.
(Put x = 0 , π , π 2 , π 4 x = 0 , π , π 2 , π 4 x=0,pi,(pi)/(2),(pi)/(4)x=0, \pi, \frac{\pi}{2}, \frac{\pi}{4}x=0,π,π2,π4, etc. and find a i a i a_(i)a_iai.)
b) Consider the linear operator T : C 3 → C 3 T : C 3 → C 3 T:C^(3)rarrC^(3)T: \mathbb{C}^3 \rightarrow \mathbb{C}^3T:C3→C3, defined by
T ( z 1 , z 2 , z 3 ) = ( z 1 − i z 2 , i z 1 − 2 z 2 + i z 3 , − i z 2 + z 3 ) . T z 1 , z 2 , z 3 = z 1 − i z 2 , i z 1 − 2 z 2 + i z 3 , − i z 2 + z 3 . T(z_(1),z_(2),z_(3))=(z_(1)-iz_(2),iz_(1)-2z_(2)+iz_(3),-iz_(2)+z_(3)).T\left(z_1, z_2, z_3\right)=\left(z_1-i z_2, i z_1-2 z_2+i z_3,-i z_2+z_3\right) .T(z1,z2,z3)=(z1−iz2,iz1−2z2+iz3,−iz2+z3).
i) Compute T ∗ T ∗ T^(**)T^*T∗ and check whether T T TTT is self-adjoint.
ii) Check whether T T TTT is unitary.
  1. a) Let ( x 1 , x 2 , x 3 ) x 1 , x 2 , x 3 (x_(1),x_(2),x_(3))\left(x_1, x_2, x_3\right)(x1,x2,x3) and ( y 1 , y 2 , y 3 ) y 1 , y 2 , y 3 (y_(1),y_(2),y_(3))\left(y_1, y_2, y_3\right)(y1,y2,y3) represent the coordinates with respect to the bases B 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 ) } B_(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)\}B1={(1,0,0),(1,1,0),(0,0,1)},B2={(1,0,0),(0,1,1),(0,0,1)}. If
Q ( X ) = x 1 2 − 4 x 1 x 2 + 2 x 2 x 3 + x 2 2 + x 3 2 Q ( X ) = x 1 2 − 4 x 1 x 2 + 2 x 2 x 3 + x 2 2 + x 3 2 Q(X)=x_(1)^(2)-4x_(1)x_(2)+2x_(2)x_(3)+x_(2)^(2)+x_(3)^(2)Q(X)=x_1^2-4 x_1 x_2+2 x_2 x_3+x_2^2+x_3^2Q(X)=x12−4x1x2+2x2x3+x22+x32
find the representation of Q Q QQQ in terms of ( y 1 , y 2 , y 3 ) y 1 , y 2 , y 3 (y_(1),y_(2),y_(3))\left(y_1, y_2, y_3\right)(y1,y2,y3).
b) Find the orthogonal canonical reduction of the quadratic form − x 2 + y 2 + z 2 + 4 x y + 4 x z − x 2 + y 2 + z 2 + 4 x y + 4 x z -x^(2)+y^(2)+z^(2)+4xy+4xz-x^2+y^2+z^2+4 x y+4 x z−x2+y2+z2+4xy+4xz. Also, find its principal axes.
  1. Which of the following statements are true and which are false? Justify your answer with a short proof or a counterexample.
    i) If W 1 W 1 W_(1)W_1W1 and W 2 W 2 W_(2)W_2W2 are proper subspaces of a non-zero, finite dimensional, vector space V V VVV and dim ⁡ ( W 1 ) > dim ⁡ ( V ) 2 , dim ⁡ ( W 2 ) > dim ⁡ ( V ) 2 dim ⁡ W 1 > dim ⁡ ( V ) 2 , dim ⁡ W 2 > dim ⁡ ( V ) 2 dim(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}dim⁡(W1)>dim⁡(V)2,dim⁡(W2)>dim⁡(V)2, the W 1 ∩ W 2 ≠ { 0 } W 1 ∩ W 2 ≠ { 0 } W_(1)nnW_(2)!={0}W_1 \cap W_2 \neq\{0\}W1∩W2≠{0}.
ii) If V V VVV is a vector space and S = { v 1 , v 2 , … , v n } ⊂ V , n ≥ 3 S = v 1 , v 2 , … , v n ⊂ V , n ≥ 3 S={v_(1),v_(2),dots,v_(n)}sub V,n >= 3S=\left\{v_1, v_2, \ldots, v_n\right\} \subset V, n \geq 3S={v1,v2,…,vn}⊂V,n≥3, is such that v i ≠ v j v i ≠ v j v_(i)!=v_(j)v_i \neq v_jvi≠vj if i ≠ j i ≠ j i!=ji \neq ji≠j, then S S SSS is a linearly independent set.
iii) If T 1 , T 2 : V → V T 1 , T 2 : V → V T_(1),T_(2):V rarr VT_1, T_2: V \rightarrow VT1,T2:V→V are linear operators on a finite dimensional vector space V V VVV and T 1 ∘ T 2 T 1 ∘ T 2 T_(1)@T_(2)T_1 \circ T_2T1∘T2 is invertible, T 2 ∘ T 1 T 2 ∘ T 1 T_(2)@T_(1)T_2 \circ T_1T2∘T1 is also invertible.
iv) If an n × n n × n n xx nn \times nn×n square matrix, n ≥ 2 n ≥ 2 n >= 2n \geq 2n≥2 is diagonalisable then it has the same minimal polynomial and characteristic polynomial.
v) If T 1 , T 2 : V → V T 1 , T 2 : V → V T_(1),T_(2):V rarr VT_1, T_2: V \rightarrow VT1,T2:V→V are self adjoint operators on a finite dimensional inner product space V V VVV, then T 1 + T 2 T 1 + T 2 T_(1)+T_(2)T_1+T_2T1+T2 is also a self adjoint operator.
\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)

BMTE-141 Sample Solution 2024

bmte-141-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

bmte-141-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

  1. i) Find the angle between the vectors 2 i + 2 j + 2 k 2 i + 2 j + 2 k sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}2i+2j+2k and i + 2 j + 2 k i + 2 j + 2 k i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}i+2j+2k.
Answer:
To find the angle between two vectors, we use the dot product formula. The dot product of two vectors A A A\mathbf{A}A and B B B\mathbf{B}B is given by A ⋅ B = | A | | B | cos ⁡ ( θ ) A ⋅ B = | A | | B | cos ⁡ ( θ ) A*B=|A||B|cos(theta)\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta)A⋅B=|A||B|cos⁡(θ), where θ θ theta\thetaθ is the angle between the vectors, and | A | | A | |A||\mathbf{A}||A| and | B | | B | |B||\mathbf{B}||B| are the magnitudes of the vectors.
The vectors given are A = 2 i + 2 j + 2 k A = 2 i + 2 j + 2 k A=sqrt2i+2j+2k\mathbf{A} = \sqrt{2} \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k}A=2i+2j+2k and B = i + 2 j + 2 k B = i + 2 j + 2 k B=i+sqrt2j+sqrt2k\mathbf{B} = \mathbf{i} + \sqrt{2} \mathbf{j} + \sqrt{2} \mathbf{k}B=i+2j+2k.
  1. Calculate the Dot Product A ⋅ B A ⋅ B A*B\mathbf{A} \cdot \mathbf{B}A⋅B:
    A ⋅ B = ( 2 i + 2 j + 2 k ) ⋅ ( i + 2 j + 2 k ) A ⋅ B = ( 2 i + 2 j + 2 k ) ⋅ ( i + 2 j + 2 k ) 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})A⋅B=(2i+2j+2k)⋅(i+2j+2k)
    = 2 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 2 = 2 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 2 =sqrt2*1+2*sqrt2+2*sqrt2= \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2}=2⋅1+2⋅2+2⋅2
  2. Calculate the Magnitudes of A A A\mathbf{A}A and B B B\mathbf{B}B:
    • Magnitude of A A A\mathbf{A}A, | A | | A | |A||\mathbf{A}||A|: | A | = ( 2 ) 2 + 2 2 + 2 2 | A | = ( 2 ) 2 + 2 2 + 2 2 |A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))|\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2}|A|=(2)2+22+22
    • Magnitude of B B B\mathbf{B}B, | B | | B | |B||\mathbf{B}||B|: | B | = 1 2 + ( 2 ) 2 + ( 2 ) 2 | B | = 1 2 + ( 2 ) 2 + ( 2 ) 2 |B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))|\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2}|B|=12+(2)2+(2)2
  3. Find the Angle θ θ theta\thetaθ:
    cos ⁡ ( θ ) = A ⋅ B | A | | B | cos ⁡ ( θ ) = A ⋅ B | A | | B | cos(theta)=(A*B)/(|A||B|)\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}cos⁡(θ)=A⋅B|A||B|
    θ = cos − 1 ⁡ ( A ⋅ B | A | | B | ) θ = cos − 1 ⁡ A ⋅ B | A | | B | theta=cos^(-1)((A*B)/(|A||B|))\theta = \cos^{-1}\left(\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}\right)θ=cos−1⁡(A⋅B|A||B|)
    We will calculate θ θ theta\thetaθ after obtaining the dot product and magnitudes.
Let’s start by calculating the dot product and the magnitudes.
After calculating, we have the following results:
  1. Dot Product A ⋅ B A ⋅ B A*B\mathbf{A} \cdot \mathbf{B}A⋅B:
    A ⋅ B = 2 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 2 = 5 2 A ⋅ B = 2 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 2 = 5 2 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}A⋅B=2⋅1+2⋅2+2⋅2=52
  2. Magnitude of A A A\mathbf{A}A, | A | | A | |A||\mathbf{A}||A|:
    | A | = ( 2 ) 2 + 2 2 + 2 2 = 10 | A | = ( 2 ) 2 + 2 2 + 2 2 = 10 |A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))=sqrt10|\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2} = \sqrt{10}|A|=(2)2+22+22=10
  3. Magnitude of B B B\mathbf{B}B, | B | | B | |B||\mathbf{B}||B|:
    | B | = 1 2 + ( 2 ) 2 + ( 2 ) 2 = 5 | B | = 1 2 + ( 2 ) 2 + ( 2 ) 2 = 5 |B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))=sqrt5|\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{5}|B|=12+(2)2+(2)2=5
Now, we can find the angle θ θ theta\thetaθ between the vectors:
cos ⁡ ( θ ) = A ⋅ B | A | | B | = 5 2 10 ⋅ 5 = 1 cos ⁡ ( θ ) = A ⋅ B | A | | B | = 5 2 10 ⋅ 5 = 1 cos(theta)=(A*B)/(|A||B|)=(5sqrt2)/(sqrt10*sqrt5)=1\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{5\sqrt{2}}{\sqrt{10} \cdot \sqrt{5}}=1cos⁡(θ)=A⋅B|A||B|=5210⋅5=1
After calculating, we find that the angle θ θ theta\thetaθ between the vectors is 0 0 000 radians, which is equivalent to 0 ∘ 0 ∘ 0^(@)0^\circ0∘. This means the vectors are parallel to each other.
ii) Find the vector equation of the plane determined by the points ( 1 , 0 , − 1 ) , ( 0 , 1 , 1 ) ( 1 , 0 , − 1 ) , ( 0 , 1 , 1 ) (1,0,-1),(0,1,1)(1,0,-1),(0,1,1)(1,0,−1),(0,1,1) and ( − 1 , 1 , 0 ) ( − 1 , 1 , 0 ) (-1,1,0)(-1,1,0)(−1,1,0).
Answer:
To find the vector equation of a plane determined by three points, we first need to find two direction vectors that lie in the plane. These can be obtained by subtracting the coordinates of the points. Let’s denote the points as A = ( 1 , 0 , − 1 ) A = ( 1 , 0 , − 1 ) A=(1,0,-1)A = (1,0,-1)A=(1,0,−1), B = ( 0 , 1 , 1 ) B = ( 0 , 1 , 1 ) B=(0,1,1)B = (0,1,1)B=(0,1,1), and C = ( − 1 , 1 , 0 ) C = ( − 1 , 1 , 0 ) C=(-1,1,0)C = (-1,1,0)C=(−1,1,0). We can find two direction vectors, say A B → A B → vec(AB)\vec{AB}AB→ and A C → A C → vec(AC)\vec{AC}AC→, by subtracting the coordinates of these points.
The vector A B → A B → vec(AB)\vec{AB}AB→ is given by:
A B → = B − A = ( 0 , 1 , 1 ) − ( 1 , 0 , − 1 ) A B → = B − A = ( 0 , 1 , 1 ) − ( 1 , 0 , − 1 ) vec(AB)=B-A=(0,1,1)-(1,0,-1)\vec{AB} = B – A = (0,1,1) – (1,0,-1)AB→=B−A=(0,1,1)−(1,0,−1)
Similarly, the vector A C → A C → vec(AC)\vec{AC}AC→ is:
A C → = C − A = ( − 1 , 1 , 0 ) − ( 1 , 0 , − 1 ) A C → = C − A = ( − 1 , 1 , 0 ) − ( 1 , 0 , − 1 ) vec(AC)=C-A=(-1,1,0)-(1,0,-1)\vec{AC} = C – A = (-1,1,0) – (1,0,-1)AC→=C−A=(−1,1,0)−(1,0,−1)
Next, we find a normal vector to the plane by taking the cross product of A B → A B → vec(AB)\vec{AB}AB→ and A C → A C → vec(AC)\vec{AC}AC→. The cross product of two vectors is perpendicular to both, and hence, will be normal to the plane.
The cross product n → n → vec(n)\vec{n}n→ of A B → A B → vec(AB)\vec{AB}AB→ and A C → A C → vec(AC)\vec{AC}AC→ is given by:
n → = A B → × A C → n → = A B → × A C → vec(n)= vec(AB)xx vec(AC)\vec{n} = \vec{AB} \times \vec{AC}n→=AB→×AC→
= ( − 1 , 1 , 2 ) × ( − 2 , 1 , 1 ) = ( − 1 , 1 , 2 ) × ( − 2 , 1 , 1 ) =(-1,1,2)xx(-2,1,1)= (-1, 1, 2) \times (-2, 1, 1)=(−1,1,2)×(−2,1,1)
Let’s calculate the cross product:
n → = | i j k − 1 1 2 − 2 1 1 | n → = i j k − 1 1 2 − 2 1 1 vec(n)=|[i,j,k],[-1,1,2],[-2,1,1]|\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 2 \\ -2 & 1 & 1 \\ \end{vmatrix}n→=|ijk−112−211|
Expanding this determinant, we get:
n → = i ( 1 ⋅ 1 − 2 ⋅ 1 ) − j ( − 1 ⋅ 1 − 2 ⋅ − 2 ) + k ( − 1 ⋅ 1 − 1 ⋅ − 2 ) n → = i ( 1 ⋅ 1 − 2 ⋅ 1 ) − j ( − 1 ⋅ 1 − 2 ⋅ − 2 ) + k ( − 1 ⋅ 1 − 1 ⋅ − 2 ) vec(n)=i(1*1-2*1)-j(-1*1-2*-2)+k(-1*1-1*-2)\vec{n} = \mathbf{i}(1 \cdot 1 – 2 \cdot 1) – \mathbf{j}(-1 \cdot 1 – 2 \cdot -2) + \mathbf{k}(-1 \cdot 1 – 1 \cdot -2)n→=i(1⋅1−2⋅1)−j(−1⋅1−2⋅−2)+k(−1⋅1−1⋅−2)
So, the normal vector n → n → vec(n)\vec{n}n→ is:
n → = ( − 1 , − 3 , 1 ) n → = ( − 1 , − 3 , 1 ) vec(n)=(-1,-3,1)\vec{n} = (-1, -3, 1)n→=(−1,−3,1)
The vector equation of the plane can be written as:
n → ⋅ ( r → − A → ) = 0 n → ⋅ ( r → − A → ) = 0 vec(n)*( vec(r)- vec(A))=0\vec{n} \cdot (\vec{r} – \vec{A}) = 0n→⋅(r→−A→)=0
where r → = ( x , y , z ) r → = ( x , y , z ) vec(r)=(x,y,z)\vec{r} = (x, y, z)r→=(x,y,z) is any point on the plane, and A → A → vec(A)\vec{A}A→ is one of the given points, say ( 1 , 0 , − 1 ) ( 1 , 0 , − 1 ) (1,0,-1)(1,0,-1)(1,0,−1).
Substituting n → n → vec(n)\vec{n}n→ and A → A → vec(A)\vec{A}A→ into the equation, we get:
( − 1 , − 3 , 1 ) ⋅ ( ( x , y , z ) − ( 1 , 0 , − 1 ) ) = 0 ( − 1 , − 3 , 1 ) ⋅ ( ( x , y , z ) − ( 1 , 0 , − 1 ) ) = 0 (-1,-3,1)*((x,y,z)-(1,0,-1))=0(-1, -3, 1) \cdot ((x, y, z) – (1, 0, -1)) = 0(−1,−3,1)⋅((x,y,z)−(1,0,−1))=0
( − 1 , − 3 , 1 ) ⋅ ( x − 1 , y , z + 1 ) = 0 ( − 1 , − 3 , 1 ) ⋅ ( x − 1 , y , z + 1 ) = 0 (-1,-3,1)*(x-1,y,z+1)=0(-1, -3, 1) \cdot (x – 1, y, z + 1) = 0(−1,−3,1)⋅(x−1,y,z+1)=0
Expanding this, we have:
− 1 ( x − 1 ) − 3 y + 1 ( z + 1 ) = 0 − 1 ( x − 1 ) − 3 y + 1 ( z + 1 ) = 0 -1(x-1)-3y+1(z+1)=0-1(x – 1) – 3y + 1(z + 1) = 0−1(x−1)−3y+1(z+1)=0
− x + 1 − 3 y + z + 1 = 0 − x + 1 − 3 y + z + 1 = 0 -x+1-3y+z+1=0-x + 1 – 3y + z + 1 = 0−x+1−3y+z+1=0
− x − 3 y + z + 2 = 0 − x − 3 y + z + 2 = 0 -x-3y+z+2=0-x – 3y + z + 2 = 0−x−3y+z+2=0
Therefore, the vector equation of the plane is:
− x − 3 y + z + 2 = 0 − x − 3 y + z + 2 = 0 -x-3y+z+2=0-x – 3y + z + 2 = 0−x−3y+z+2=0
iii) Check whether W = { ( x , y , z ) ∈ R 3 ∣ x + y − z = 0 } W = ( x , y , z ) ∈ R 3 ∣ x + y − z = 0 W={(x,y,z)inR^(3)∣x+y-z=0}W=\left\{(x, y, z) \in \mathbb{R}^3 \mid x+y-z=0\right\}W={(x,y,z)∈R3∣x+y−z=0} is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3.
Answer:
To determine whether a set W W WWW is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3, it must satisfy three conditions:
  1. Non-emptiness: W W WWW must contain the zero vector of R 3 R 3 R^(3)\mathbb{R}^3R3.
  2. Closed under addition: For any two vectors u → u → vec(u)\vec{u}u→ and v → v → vec(v)\vec{v}v→ in W W WWW, the sum u → + v → u → + v → vec(u)+ vec(v)\vec{u} + \vec{v}u→+v→ must also be in W W WWW.
  3. Closed under scalar multiplication: For any vector u → u → vec(u)\vec{u}u→ in W W WWW and any scalar c c ccc, the product c u → c u → c vec(u)c\vec{u}cu→ must also be in W W WWW.
Let’s check these conditions for W = { ( x , y , z ) ∈ R 3 ∣ x + y − z = 0 } W = ( x , y , z ) ∈ R 3 ∣ x + y − z = 0 W={(x,y,z)inR^(3)∣x+y-z=0}W = \left\{ (x, y, z) \in \mathbb{R}^3 \mid x + y – z = 0 \right\}W={(x,y,z)∈R3∣x+y−z=0}.
  1. Non-emptiness:
    The zero vector in R 3 R 3 R^(3)\mathbb{R}^3R3 is ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0, 0, 0)(0,0,0). For this vector, x = 0 x = 0 x=0x = 0x=0, y = 0 y = 0 y=0y = 0y=0, and z = 0 z = 0 z=0z = 0z=0, so x + y − z = 0 + 0 − 0 = 0 x + y − z = 0 + 0 − 0 = 0 x+y-z=0+0-0=0x + y – z = 0 + 0 – 0 = 0x+y−z=0+0−0=0. Therefore, the zero vector is in W W WWW, satisfying the non-emptiness condition.
  2. Closed under addition:
    Let u → = ( x 1 , y 1 , z 1 ) u → = ( x 1 , y 1 , z 1 ) vec(u)=(x_(1),y_(1),z_(1))\vec{u} = (x_1, y_1, z_1)u→=(x1,y1,z1) and v → = ( x 2 , y 2 , z 2 ) v → = ( x 2 , y 2 , z 2 ) vec(v)=(x_(2),y_(2),z_(2))\vec{v} = (x_2, y_2, z_2)v→=(x2,y2,z2) be any two vectors in W W WWW. This means x 1 + y 1 − z 1 = 0 x 1 + y 1 − z 1 = 0 x_(1)+y_(1)-z_(1)=0x_1 + y_1 – z_1 = 0x1+y1−z1=0 and x 2 + y 2 − z 2 = 0 x 2 + y 2 − z 2 = 0 x_(2)+y_(2)-z_(2)=0x_2 + y_2 – z_2 = 0x2+y2−z2=0. We need to check if their sum u → + v → = ( x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) u → + 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))\vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2)u→+v→=(x1+x2,y1+y2,z1+z2) also satisfies the condition of W W WWW.
    For u → + v → u → + v → vec(u)+ vec(v)\vec{u} + \vec{v}u→+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 (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(x1+x2)+(y1+y2)−(z1+z2)=(x1+y1−z1)+(x2+y2−z2)=0+0=0
    Therefore, u → + v → u → + v → vec(u)+ vec(v)\vec{u} + \vec{v}u→+v→ is in W W WWW, satisfying the closure under addition.
  3. Closed under scalar multiplication:
    Let u → = ( x , y , z ) u → = ( x , y , z ) vec(u)=(x,y,z)\vec{u} = (x, y, z)u→=(x,y,z) be any vector in W W WWW and c c ccc be any scalar. Since u → u → vec(u)\vec{u}u→ is in W W WWW, x + y − z = 0 x + y − z = 0 x+y-z=0x + y – z = 0x+y−z=0. We need to check if c u → = ( c x , c y , c z ) c u → = ( c x , c y , c z ) c vec(u)=(cx,cy,cz)c\vec{u} = (cx, cy, cz)cu→=(cx,cy,cz) also satisfies the condition of W W WWW.
    For c u → c u → c vec(u)c\vec{u}cu→, we have:
    c x + c y − c z = c ( x + y − z ) = c ⋅ 0 = 0 c x + c y − c z = c ( x + y − z ) = c ⋅ 0 = 0 cx+cy-cz=c(x+y-z)=c*0=0cx + cy – cz = c(x + y – z) = c \cdot 0 = 0cx+cy−cz=c(x+y−z)=c⋅0=0
    Therefore, c u → c u → c vec(u)c\vec{u}cu→ is in W W WWW, satisfying the closure under scalar multiplication.
Since W W WWW satisfies all three conditions, it is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3.
iv) Check whether the set of vectors { 1 + x , x + x 2 , 1 + x 3 } 1 + x , x + x 2 , 1 + x 3 {1+x,x+x^(2),1+x^(3)}\left\{1+x, x+x^2, 1+x^3\right\}{1+x,x+x2,1+x3} is a linearly independent set of vectors in P 3 P 3 P_(3)\mathbf{P}_3P3, the vector space of polynomials of degree ≤ 3 ≤ 3 <= 3\leq 3≤3.
Answer:
To determine whether the set of vectors { 1 + x , x + x 2 , 1 + x 3 } { 1 + x , x + x 2 , 1 + x 3 } {1+x,x+x^(2),1+x^(3)}\{1+x, x+x^2, 1+x^3\}{1+x,x+x2,1+x3} is linearly independent in the vector space P 3 P 3 P_(3)\mathbf{P}_3P3 (the space of polynomials of degree ≤ 3 ≤ 3 <= 3\leq 3≤3), we need to check if the only solution to the linear combination equation
c 1 ( 1 + x ) + c 2 ( x + x 2 ) + c 3 ( 1 + x 3 ) = 0 c 1 ( 1 + x ) + c 2 ( x + x 2 ) + c 3 ( 1 + x 3 ) = 0 c_(1)(1+x)+c_(2)(x+x^(2))+c_(3)(1+x^(3))=0c_1(1+x) + c_2(x+x^2) + c_3(1+x^3) = 0c1(1+x)+c2(x+x2)+c3(1+x3)=0
is c 1 = c 2 = c 3 = 0 c 1 = c 2 = c 3 = 0 c_(1)=c_(2)=c_(3)=0c_1 = c_2 = c_3 = 0c1=c2=c3=0, where c 1 , c 2 , c 1 , c 2 , c_(1),c_(2),c_1, c_2,c1,c2, and c 3 c 3 c_(3)c_3c3 are constants.
Expanding the equation, we get:
c 1 + c 1 x + c 2 x + c 2 x 2 + c 3 + c 3 x 3 = 0 c 1 + c 1 x + c 2 x + c 2 x 2 + c 3 + c 3 x 3 = 0 c_(1)+c_(1)x+c_(2)x+c_(2)x^(2)+c_(3)+c_(3)x^(3)=0c_1 + c_1x + c_2x + c_2x^2 + c_3 + c_3x^3 = 0c1+c1x+c2x+c2x2+c3+c3x3=0
This can be rearranged as:
( c 1 + c 3 ) + ( c 1 + c 2 ) x + c 2 x 2 + c 3 x 3 = 0 ( c 1 + c 3 ) + ( c 1 + c 2 ) x + c 2 x 2 + c 3 x 3 = 0 (c_(1)+c_(3))+(c_(1)+c_(2))x+c_(2)x^(2)+c_(3)x^(3)=0(c_1 + c_3) + (c_1 + c_2)x + c_2x^2 + c_3x^3 = 0(c1+c3)+(c1+c2)x+c2x2+c3x3=0
For this polynomial to be identically zero, each coefficient must be zero. Therefore, we have the system of equations:
  1. c 1 + c 3 = 0 c 1 + c 3 = 0 c_(1)+c_(3)=0c_1 + c_3 = 0c1+c3=0
  2. c 1 + c 2 = 0 c 1 + c 2 = 0 c_(1)+c_(2)=0c_1 + c_2 = 0c1+c2=0
  3. c 2 = 0 c 2 = 0 c_(2)=0c_2 = 0c2=0
  4. c 3 = 0 c 3 = 0 c_(3)=0c_3 = 0c3=0
From equation 3, we immediately have c 2 = 0 c 2 = 0 c_(2)=0c_2 = 0c2=0. Substituting this into equation 2 gives c 1 = 0 c 1 = 0 c_(1)=0c_1 = 0c1=0. Finally, substituting c 1 = 0 c 1 = 0 c_(1)=0c_1 = 0c1=0 into equation 1 gives c 3 = 0 c 3 = 0 c_(3)=0c_3 = 0c3=0.
Since the only solution to the system is c 1 = c 2 = c 3 = 0 c 1 = c 2 = c 3 = 0 c_(1)=c_(2)=c_(3)=0c_1 = c_2 = c_3 = 0c1=c2=c3=0, the set of vectors { 1 + x , x + x 2 , 1 + x 3 } { 1 + x , x + x 2 , 1 + x 3 } {1+x,x+x^(2),1+x^(3)}\{1+x, x+x^2, 1+x^3\}{1+x,x+x2,1+x3} is linearly independent in P 3 P 3 P_(3)\mathbf{P}_3P3.
v) Check whether T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2, defined by T ( x , y ) = ( − y , x ) T ( x , y ) = ( − y , x ) T(x,y)=(-y,x)T(x, y)=(-y, x)T(x,y)=(−y,x) is a linear transformation.
Answer:
To determine whether a function T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 is a linear transformation, it must satisfy two properties for all vectors u → , v → ∈ R 2 u → , v → ∈ R 2 vec(u), vec(v)inR^(2)\vec{u}, \vec{v} \in \mathbb{R}^2u→,v→∈R2 and any scalar c c ccc:
  1. Additivity: T ( u → + v → ) = T ( u → ) + T ( v → ) T ( u → + v → ) = T ( u → ) + T ( v → ) T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})T(u→+v→)=T(u→)+T(v→)
  2. Homogeneity: T ( c u → ) = c T ( u → ) T ( c u → ) = c T ( u → ) T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u})T(cu→)=cT(u→)
Let’s check these properties for T T TTT defined by T ( x , y ) = ( − y , x ) T ( x , y ) = ( − y , x ) T(x,y)=(-y,x)T(x, y) = (-y, x)T(x,y)=(−y,x).
  1. Additivity:
    Let u → = ( x 1 , y 1 ) u → = ( x 1 , y 1 ) vec(u)=(x_(1),y_(1))\vec{u} = (x_1, y_1)u→=(x1,y1) and v → = ( x 2 , y 2 ) v → = ( x 2 , y 2 ) vec(v)=(x_(2),y_(2))\vec{v} = (x_2, y_2)v→=(x2,y2) be any vectors in R 2 R 2 R^(2)\mathbb{R}^2R2. We need to check if T ( u → + v → ) = T ( u → ) + T ( v → ) T ( u → + v → ) = T ( u → ) + T ( v → ) T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})T(u→+v→)=T(u→)+T(v→).
    First, calculate u → + v → u → + v → vec(u)+ vec(v)\vec{u} + \vec{v}u→+v→:
    u → + v → = ( x 1 + x 2 , y 1 + y 2 ) u → + v → = ( x 1 + x 2 , y 1 + y 2 ) vec(u)+ vec(v)=(x_(1)+x_(2),y_(1)+y_(2))\vec{u} + \vec{v} = (x_1 + x_2, y_1 + y_2)u→+v→=(x1+x2,y1+y2)
    Then, apply T T TTT to u → + v → u → + v → vec(u)+ vec(v)\vec{u} + \vec{v}u→+v→:
    T ( u → + v → ) = T ( x 1 + x 2 , y 1 + y 2 ) = ( − ( y 1 + y 2 ) , x 1 + x 2 ) T ( u → + 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))T(\vec{u} + \vec{v}) = T(x_1 + x_2, y_1 + y_2) = (-(y_1 + y_2), x_1 + x_2)T(u→+v→)=T(x1+x2,y1+y2)=(−(y1+y2),x1+x2)
    Now, calculate T ( u → ) T ( u → ) T( vec(u))T(\vec{u})T(u→) and T ( v → ) T ( v → ) T( vec(v))T(\vec{v})T(v→):
    T ( u → ) = T ( x 1 , y 1 ) = ( − y 1 , x 1 ) T ( u → ) = T ( x 1 , y 1 ) = ( − y 1 , x 1 ) 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(u→)=T(x1,y1)=(−y1,x1)
    T ( v → ) = T ( x 2 , y 2 ) = ( − y 2 , x 2 ) T ( v → ) = T ( x 2 , y 2 ) = ( − y 2 , x 2 ) T( vec(v))=T(x_(2),y_(2))=(-y_(2),x_(2))T(\vec{v}) = T(x_2, y_2) = (-y_2, x_2)T(v→)=T(x2,y2)=(−y2,x2)
    Finally, add T ( u → ) T ( u → ) T( vec(u))T(\vec{u})T(u→) and T ( v → ) T ( v → ) T( vec(v))T(\vec{v})T(v→):
    T ( u → ) + T ( v → ) = ( − y 1 , x 1 ) + ( − y 2 , x 2 ) = ( − ( y 1 + y 2 ) , x 1 + x 2 ) T ( u → ) + T ( 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))T(\vec{u}) + T(\vec{v}) = (-y_1, x_1) + (-y_2, x_2) = (-(y_1 + y_2), x_1 + x_2)T(u→)+T(v→)=(−y1,x1)+(−y2,x2)=(−(y1+y2),x1+x2)
    Since T ( u → + v → ) = T ( u → ) + T ( v → ) T ( u → + v → ) = T ( u → ) + T ( v → ) T( vec(u)+ vec(v))=T( vec(u))+T( vec(v))T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})T(u→+v→)=T(u→)+T(v→), the additivity property is satisfied.
  2. Homogeneity:
    Let u → = ( x , y ) u → = ( x , y ) vec(u)=(x,y)\vec{u} = (x, y)u→=(x,y) be any vector in R 2 R 2 R^(2)\mathbb{R}^2R2 and c c ccc be any scalar. We need to check if T ( c u → ) = c T ( u → ) T ( c u → ) = c T ( u → ) T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u})T(cu→)=cT(u→).
    First, calculate c u → c u → c vec(u)c\vec{u}cu→:
    c u → = c ( x , y ) = ( c x , c y ) c u → = c ( x , y ) = ( c x , c y ) c vec(u)=c(x,y)=(cx,cy)c\vec{u} = c(x, y) = (cx, cy)cu→=c(x,y)=(cx,cy)
    Then, apply T T TTT to c u → c u → c vec(u)c\vec{u}cu→:
    T ( c u → ) = T ( c x , c y ) = ( − c y , c x ) T ( c u → ) = T ( c x , c y ) = ( − c y , c x ) T(c vec(u))=T(cx,cy)=(-cy,cx)T(c\vec{u}) = T(cx, cy) = (-cy, cx)T(cu→)=T(cx,cy)=(−cy,cx)
    Now, calculate T ( u → ) T ( u → ) T( vec(u))T(\vec{u})T(u→) and then multiply by c c ccc:
    T ( u → ) = T ( x , y ) = ( − y , x ) T ( u → ) = T ( x , y ) = ( − y , x ) T( vec(u))=T(x,y)=(-y,x)T(\vec{u}) = T(x, y) = (-y, x)T(u→)=T(x,y)=(−y,x)
    c T ( u → ) = c ( − y , x ) = ( − c y , c x ) c T ( u → ) = c ( − y , x ) = ( − c y , c x ) cT( vec(u))=c(-y,x)=(-cy,cx)cT(\vec{u}) = c(-y, x) = (-cy, cx)cT(u→)=c(−y,x)=(−cy,cx)
    Since T ( c u → ) = c T ( u → ) T ( c u → ) = c T ( u → ) T(c vec(u))=cT( vec(u))T(c\vec{u}) = cT(\vec{u})T(cu→)=cT(u→), the homogeneity property is satisfied.
Since T T TTT satisfies both the additivity and homogeneity properties, it is a linear transformation.
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