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

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

101.00

Details For BMTE-141 Solved Assignment

## 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 $\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$$\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}$\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$ and $\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$$\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}$\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$.
ii) Find the vector equation of the plane determined by the points $\left(1,0,-1\right),\left(0,1,1\right)$$\left(1,0,-1\right),\left(0,1,1\right)$(1,0,-1),(0,1,1)(1,0,-1),(0,1,1)$\left(1,0,-1\right),\left(0,1,1\right)$ and $\left(-1,1,0\right)$$\left(-1,1,0\right)$(-1,1,0)(-1,1,0)$\left(-1,1,0\right)$.
iii) Check whether $W=\left\{\left(x,y,z\right)\in {\mathbb{R}}^{3}\mid x+y-z=0\right\}$$W=\left\{\left(x,y,z\right)\in {\mathbb{R}}^{3}\mid x+y-z=0\right\}$W={(x,y,z)inR^(3)∣x+y-z=0}W=\left\{(x, y, z) \in \mathbb{R}^3 \mid x+y-z=0\right\}$W=\left\{\left(x,y,z\right)\in {\mathbb{R}}^{3}\mid x+y-z=0\right\}$ is a subspace of ${\mathbb{R}}^{3}$${\mathbb{R}}^{3}$R^(3)\mathbb{R}^3${\mathbb{R}}^{3}$.
iv) Check whether the set of vectors $\left\{1+x,x+{x}^{2},1+{x}^{3}\right\}$$\left\{1+x,x+{x}^{2},1+{x}^{3}\right\}${1+x,x+x^(2),1+x^(3)}\left\{1+x, x+x^2, 1+x^3\right\}$\left\{1+x,x+{x}^{2},1+{x}^{3}\right\}$ is a linearly independent set of vectors in ${\mathbf{P}}_{3}$${\mathbf{P}}_{3}$P_(3)\mathbf{P}_3${\mathbf{P}}_{3}$, the vector space of polynomials of degree $\le 3$$\le 3$<= 3\leq 3$\le 3$.
v) Check whether $T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$, defined by $T\left(x,y\right)=\left(-y,x\right)$$T\left(x,y\right)=\left(-y,x\right)$T(x,y)=(-y,x)T(x, y)=(-y, x)$T\left(x,y\right)=\left(-y,x\right)$ is a linear transformation.
vi) If $\left\{{v}_{1},{v}_{2}\right\}$$\left\{{v}_{1},{v}_{2}\right\}${v_(1),v_(2)}\left\{v_1, v_2\right\}$\left\{{v}_{1},{v}_{2}\right\}$ is an ordered basis of ${\mathbb{R}}^{2}$${\mathbb{R}}^{2}$R^(2)\mathbb{R}^2${\mathbb{R}}^{2}$ and $\left\{{f}_{1}\left(v\right),{f}_{2}\left(v\right)\right\}$$\left\{{f}_{1}\left(v\right),{f}_{2}\left(v\right)\right\}${f_(1)(v),f_(2)(v)}\left\{f_1(v), f_2(v)\right\}$\left\{{f}_{1}\left(v\right),{f}_{2}\left(v\right)\right\}$ is the corresponding dual basis find ${f}_{1}\left(2{v}_{1}+{v}_{2}\right)$${f}_{1}\left(2{v}_{1}+{v}_{2}\right)$f_(1)(2v_(1)+v_(2))f_1\left(2 v_1+v_2\right)${f}_{1}\left(2{v}_{1}+{v}_{2}\right)$ and ${f}_{2}\left({v}_{1}-2{v}_{2}\right)$${f}_{2}\left({v}_{1}-2{v}_{2}\right)$f_(2)(v_(1)-2v_(2))f_2\left(v_1-2 v_2\right)${f}_{2}\left({v}_{1}-2{v}_{2}\right)$.
vii) Find the kernel of the linear transformation $T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ defined by $T\left(x,y\right)=\left(2x+3y,2x-3y\right)$$T\left(x,y\right)=\left(2x+3y,2x-3y\right)$T(x,y)=(2x+3y,2x-3y)T(x, y)=(2 x+3 y, 2 x-3 y)$T\left(x,y\right)=\left(2x+3y,2x-3y\right)$.
viii) Describe the linear transformation $T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ such that
$\left[T{\right]}_{B}=\left[\begin{array}{ll}1& 2\\ 2& 0\end{array}\right]$$\left[T{\right]}_{B}=\left[\begin{array}{l}1 2\\ 2 0\end{array}\right]$[T]_(B)=[[1,2],[2,0]][T]_B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 0 \end{array}\right]$\left[T{\right]}_{B}=\left[\begin{array}{ll}1& 2\\ 2& 0\end{array}\right]$
where $B$$B$BB$B$ is the standard basis of ${\mathbb{R}}^{2}$${\mathbb{R}}^{2}$R^(2)\mathbb{R}^2${\mathbb{R}}^{2}$.
ix) Find the matrix of the linear transformation $T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ defined by $T\left(x,y\right)=\left(2y,x-y\right)$$T\left(x,y\right)=\left(2y,x-y\right)$T(x,y)=(2y,x-y)T(x, y)=(2 y, x-y)$T\left(x,y\right)=\left(2y,x-y\right)$ with respect to the ordered basis $\left\{\left(0,-1\right),\left(-1,0\right)\right\}$$\left\{\left(0,-1\right),\left(-1,0\right)\right\}${(0,-1),(-1,0)}\{(0,-1),(-1,0)\}$\left\{\left(0,-1\right),\left(-1,0\right)\right\}$.
x) Let $A$$A$AA$A$ be a $2×3$$2×3$2xx32 \times 3$2×3$ matrix, $B$$B$BB$B$ be a $3×4$$3×4$3xx43 \times 4$3×4$ matrix and $C$$C$CC$C$ be a $3×2$$3×2$3xx23 \times 2$3×2$ matrix and $D$$D$DD$D$ be a $3×4$$3×4$3xx43 \times 4$3×4$ matrix. Is $AB+{C}^{t}D$$AB+{C}^{t}D$AB+C^(t)DA B+C^t D$AB+{C}^{t}D$ defined? Justify your answer.
xi) Verify Cayley-Hamilton theorem for the matrix $A=\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]$$A=\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]$A=[[1,-1],[0,2]]A=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]$A=\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]$.
xii) Check whether $\left[\begin{array}{l}1\\ 1\\ 1\\ 1\end{array}\right]$$\left[\begin{array}{l}1\\ 1\\ 1\\ 1\end{array}\right]$[[1],[1],[1],[1]]\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right]$\left[\begin{array}{l}1\\ 1\\ 1\\ 1\end{array}\right]$ is an eigenvector for the matrix $\left[\begin{array}{llll}1& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\\ 0& 1& 0& 1\end{array}\right]$$\left[\begin{array}{l}1 0 0 1\\ 0 1 1 0\\ 0 0 1 1\\ 0 1 0 1\end{array}\right]$[[1,0,0,1],[0,1,1,0],[0,0,1,1],[0,1,0,1]]\left[\begin{array}{llll}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1\end{array}\right]$\left[\begin{array}{llll}1& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 1\\ 0& 1& 0& 1\end{array}\right]$. What is the corresponding eigenvalue?
xiii) Let $C\left[0,1\right]$$C\left[0,1\right]$C[0,1]C[0,1]$C\left[0,1\right]$ be the inner product space of continous real valued functions on the interval $\left[0,1\right]$$\left[0,1\right]$[0,1][0,1]$\left[0,1\right]$ with the inner product
$⟨f,g⟩={\int }_{0}^{1}f\left(t\right)g\left(t\right)dt$$⟨f,g⟩={\int }_{0}^{1} f\left(t\right)g\left(t\right)dt$(:f,g:)=int_(0)^(1)f(t)g(t)dt\langle f, g\rangle=\int_0^1 f(t) g(t) d t$⟨f,g⟩={\int }_{0}^{1}f\left(t\right)g\left(t\right)dt$
Find the inner product of the functions $f\left(t\right)=2t,g\left(t\right)=\frac{1}{{t}^{2}+5}$$f\left(t\right)=2t,g\left(t\right)=\frac{1}{{t}^{2}+5}$f(t)=2t,g(t)=(1)/(t^(2)+5)f(t)=2 t, g(t)=\frac{1}{t^2+5}$f\left(t\right)=2t,g\left(t\right)=\frac{1}{{t}^{2}+5}$.
xiv) Find adjoint of the linear operator $T:{\mathbb{C}}^{2}\to {\mathbb{C}}^{2}$$T:{\mathbb{C}}^{2}\to {\mathbb{C}}^{2}$T:C^(2)rarrC^(2)T: \mathbb{C}^2 \rightarrow \mathbb{C}^2$T:{\mathbb{C}}^{2}\to {\mathbb{C}}^{2}$ defined by $T\left({z}_{1},{z}_{2}\right)=\left({z}_{2},{z}_{1}+i{z}_{2}\right)$$T\left({z}_{1},{z}_{2}\right)=\left({z}_{2},{z}_{1}+i{z}_{2}\right)$T(z_(1),z_(2))=(z_(2),z_(1)+iz_(2))T\left(z_1, z_2\right)=\left(z_2, z_1+i z_2\right)$T\left({z}_{1},{z}_{2}\right)=\left({z}_{2},{z}_{1}+i{z}_{2}\right)$ with respect to the standard inner product on ${C}^{2}$${C}^{2}$C^(2)C^2${C}^{2}$.
xv) Find the signature of the quadratic form ${x}_{1}^{2}-2{x}_{2}^{2}+3{x}_{3}^{2}$${x}_{1}^{2}-2{x}_{2}^{2}+3{x}_{3}^{2}$x_(1)^(2)-2x_(2)^(2)+3x_(3)^(2)x_1^2-2 x_2^2+3 x_3^2${x}_{1}^{2}-2{x}_{2}^{2}+3{x}_{3}^{2}$
Part-B (40 Marks)
1. a) Let $S$$S$SS$S$ be any non-empty set and let $V\left(S\right)$$V\left(S\right)$V(S)V(S)$V\left(S\right)$ be the set of all real valued functions on $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$. Define addition on $V\left(s\right)$$V\left(s\right)$V(s)V(s)$V\left(s\right)$ by $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$(f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x)$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ and scalar multiplication by $\left(\alpha \cdot f\right)\left(x\right)=\alpha f\left(x\right)$$\left(\alpha \cdot f\right)\left(x\right)=\alpha f\left(x\right)$(alpha*f)(x)=alpha f(x)(\alpha \cdot f)(x)=\alpha f(x)$\left(\alpha \cdot f\right)\left(x\right)=\alpha f\left(x\right)$. Check that $\left(V\left(S\right),+,\cdot \right)$$\left(V\left(S\right),+,\cdot \right)$(V(S),+,*)(V(S),+, \cdot)$\left(V\left(S\right),+,\cdot \right)$ is a vector space.
b) Check that $\mathbit{B}=\left\{1,2x+1,\left(x-1{\right)}^{2}\right\}$$\mathbit{B}=\left\{1,2x+1,\left(x-1{\right)}^{2}\right\}$B={1,2x+1,(x-1)^(2)}\boldsymbol{B}=\left\{1,2 x+1,(x-1)^2\right\}$\mathbit{B}=\left\{1,2x+1,\left(x-1{\right)}^{2}\right\}$ is a basis for ${\mathbf{P}}_{2}$${\mathbf{P}}_{2}$P_(2)\mathbf{P}_2${\mathbf{P}}_{2}$, the vector space of polynomials with real coefficients of degree $\le 2$$\le 2$<= 2\leq 2$\le 2$.
1. a) Let $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$$T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$T:R^(3)rarrR^(3)T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$ be a linear operator and suppose the matrix of the operator with respect to the ordered basis
$\mathbit{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\}$$\mathbit{B}=\left\{\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right],\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]\right\}$B={[[1],[0],[0]],[[1],[0],[1]],[[0],[1],[0]]}\boldsymbol{B}=\left\{\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right\}$\mathbit{B}=\left\{\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right],\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]\right\}$
is $\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 1\\ 1& 0& 1\end{array}\right]$$\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 1\\ 1& 0& 1\end{array}\right]$[[1,0,-1],[0,1,1],[1,0,1]]\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right]$\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 1\\ 1& 0& 1\end{array}\right]$. Find the matrix of the linear transformation with respect to the basis
${\mathbit{B}}^{\mathrm{\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\}$${\mathbit{B}}^{\mathrm{\prime }}=\left\{\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right],\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]\right\}$B^(‘)={[[1],[0],[1]],[[1],[0],[-1]],[[0],[1],[0]]}\boldsymbol{B}^{\prime}=\left\{\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right\}${\mathbit{B}}^{\mathrm{\prime }}=\left\{\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right],\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]\right\}$
b) Show that $W=\left\{\left(x,4x,3x\right)\in {\mathbb{R}}^{2}\mid x\in \mathbb{R}\right\}$$W=\left\{\left(x,4x,3x\right)\in {\mathbb{R}}^{2}\mid x\in \mathbb{R}\right\}$W={(x,4x,3x)inR^(2)∣x inR}W=\left\{(x, 4 x, 3 x) \in \mathbb{R}^2 \mid x \in \mathbb{R}\right\}$W=\left\{\left(x,4x,3x\right)\in {\mathbb{R}}^{2}\mid x\in \mathbb{R}\right\}$ is a subspace of ${\mathbb{R}}^{3}$${\mathbb{R}}^{3}$R^(3)\mathbb{R}^3${\mathbb{R}}^{3}$. Also find a basis for subspace $U$$U$UU$U$ of ${\mathbb{R}}^{3}$${\mathbb{R}}^{3}$R^(3)\mathbb{R}^3${\mathbb{R}}^{3}$ which satisfies $W\oplus U={\mathbb{R}}^{3}$$W\oplus U={\mathbb{R}}^{3}$W o+U=R^(3)W \oplus U=\mathbb{R}^3$W\oplus U={\mathbb{R}}^{3}$.
1. a) Find the eigenvalues and eigenvectors of the matrix $B=\left[\begin{array}{rrr}1& 1& 0\\ -1& 3& 0\\ 1& -1& 1\end{array}\right]$$B=\left[\begin{array}{r}1 1 0\\ -1 3 0\\ 1 -1 1\end{array}\right]$B=[[1,1,0],[-1,3,0],[1,-1,1]]B=\left[\begin{array}{rrr}1 & 1 & 0 \\ -1 & 3 & 0 \\ 1 & -1 & 1\end{array}\right]$B=\left[\begin{array}{rrr}1& 1& 0\\ -1& 3& 0\\ 1& -1& 1\end{array}\right]$. Is the matrix diagonalisable? Justify your answer.
b) Find $\mathrm{Adj}\left(A\right)$$\mathrm{Adj}\left(A\right)$Adj(A)\operatorname{Adj}(A)$\mathrm{Adj}\left(A\right)$ where $A=\left[\begin{array}{ccc}3& 2& 2\\ -1& 1& 0\\ 3& 0& 1\end{array}\right]$$A=\left[\begin{array}{ccc}3& 2& 2\\ -1& 1& 0\\ 3& 0& 1\end{array}\right]$A=[[3,2,2],[-1,1,0],[3,0,1]]A=\left[\begin{array}{ccc}3 & 2 & 2 \\ -1 & 1 & 0 \\ 3 & 0 & 1\end{array}\right]$A=\left[\begin{array}{ccc}3& 2& 2\\ -1& 1& 0\\ 3& 0& 1\end{array}\right]$. Hence find ${A}^{-1}$${A}^{-1}$A^(-1)A^{-1}${A}^{-1}$.
1. a) Solve the folowing set of simultaneous equations using Cramer’s rule:
$\begin{array}{rl}x+2y+z& =3\\ 2x-y+2z& =1\\ 3x+y+z& =0\end{array}$$\begin{array}{r}x+2y+z=3\\ 2x-y+2z=1\\ 3x+y+z=0\end{array}${:[x+2y+z=3],[2x-y+2z=1],[3x+y+z=0]:}\begin{aligned} x+2 y+z & =3 \\ 2 x-y+2 z & =1 \\ 3 x+y+z & =0 \end{aligned}$\begin{array}{rl}x+2y+z& =3\\ 2x-y+2z& =1\\ 3x+y+z& =0\end{array}$
b) Find the minimal polynomial of the matrix
$\left[\begin{array}{rrrr}2& 1& 0& 1\\ -1& 0& 0& 1\\ -2& -2& -1& 3\\ 0& 0& 0& 1\end{array}\right]$$\left[\begin{array}{r}2 1 0 1\\ -1 0 0 1\\ -2 -2 -1 3\\ 0 0 0 1\end{array}\right]$[[2,1,0,1],[-1,0,0,1],[-2,-2,-1,3],[0,0,0,1]]\left[\begin{array}{rrrr} 2 & 1 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ -2 & -2 & -1 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right]$\left[\begin{array}{rrrr}2& 1& 0& 1\\ -1& 0& 0& 1\\ -2& -2& -1& 3\\ 0& 0& 0& 1\end{array}\right]$
Part C (30 marks)
1. a) Let $V$$V$VV$V$ be the vector space of all real valued functions that are twice differentiable in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ and
$S=\left\{\mathrm{cos}x,\mathrm{sin}x,x\mathrm{cos}x,x\mathrm{sin}x\right\}.$$S=\left\{\mathrm{cos}x,\mathrm{sin}x,x\mathrm{cos}x,x\mathrm{sin}x\right\}.$S={cos x,sin x,x cos x,x sin x}.S=\{\cos x, \sin x, x \cos x, x \sin x\} .$S=\left\{\mathrm{cos}x,\mathrm{sin}x,x\mathrm{cos}x,x\mathrm{sin}x\right\}.$
Check that $S$$S$SS$S$ is a linearly independent set over $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$. (Hint: Consider the equation
${a}_{0}\mathrm{cos}x+{a}_{1}\mathrm{sin}x+{a}_{2}x\mathrm{cos}x+{a}_{3}x\mathrm{sin}x.$${a}_{0}\mathrm{cos}x+{a}_{1}\mathrm{sin}x+{a}_{2}x\mathrm{cos}x+{a}_{3}x\mathrm{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}\mathrm{cos}x+{a}_{1}\mathrm{sin}x+{a}_{2}x\mathrm{cos}x+{a}_{3}x\mathrm{sin}x.$
(Put $x=0,\pi ,\frac{\pi }{2},\frac{\pi }{4}$$x=0,\pi ,\frac{\pi }{2},\frac{\pi }{4}$x=0,pi,(pi)/(2),(pi)/(4)x=0, \pi, \frac{\pi}{2}, \frac{\pi}{4}$x=0,\pi ,\frac{\pi }{2},\frac{\pi }{4}$, etc. and find ${a}_{i}$${a}_{i}$a_(i)a_i${a}_{i}$.)
b) Consider the linear operator $T:{\mathbb{C}}^{3}\to {\mathbb{C}}^{3}$$T:{\mathbb{C}}^{3}\to {\mathbb{C}}^{3}$T:C^(3)rarrC^(3)T: \mathbb{C}^3 \rightarrow \mathbb{C}^3$T:{\mathbb{C}}^{3}\to {\mathbb{C}}^{3}$, defined by
$T\left({z}_{1},{z}_{2},{z}_{3}\right)=\left({z}_{1}-i{z}_{2},i{z}_{1}-2{z}_{2}+i{z}_{3},-i{z}_{2}+{z}_{3}\right).$$T\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).$T(z_(1),z_(2),z_(3))=(z_(1)-iz_(2),iz_(1)-2z_(2)+iz_(3),-iz_(2)+z_(3)).T\left(z_1, z_2, z_3\right)=\left(z_1-i z_2, i z_1-2 z_2+i z_3,-i z_2+z_3\right) .$T\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).$
i) Compute ${T}^{\ast }$${T}^{\ast }$T^(**)T^*${T}^{\ast }$ and check whether $T$$T$TT$T$ is self-adjoint.
ii) Check whether $T$$T$TT$T$ is unitary.
1. a) Let $\left({x}_{1},{x}_{2},{x}_{3}\right)$$\left({x}_{1},{x}_{2},{x}_{3}\right)$(x_(1),x_(2),x_(3))\left(x_1, x_2, x_3\right)$\left({x}_{1},{x}_{2},{x}_{3}\right)$ and $\left({y}_{1},{y}_{2},{y}_{3}\right)$$\left({y}_{1},{y}_{2},{y}_{3}\right)$(y_(1),y_(2),y_(3))\left(y_1, y_2, y_3\right)$\left({y}_{1},{y}_{2},{y}_{3}\right)$ represent the coordinates with respect to the bases ${B}_{1}=\left\{\left(1,0,0\right),\left(1,1,0\right),\left(0,0,1\right)\right\},{B}_{2}=\left\{\left(1,0,0\right),\left(0,1,1\right),\left(0,0,1\right)\right\}$${B}_{1}=\left\{\left(1,0,0\right),\left(1,1,0\right),\left(0,0,1\right)\right\},{B}_{2}=\left\{\left(1,0,0\right),\left(0,1,1\right),\left(0,0,1\right)\right\}$B_(1)={(1,0,0),(1,1,0),(0,0,1)},B_(2)={(1,0,0),(0,1,1),(0,0,1)}B_1=\{(1,0,0),(1,1,0),(0,0,1)\}, B_2=\{(1,0,0),(0,1,1),(0,0,1)\}${B}_{1}=\left\{\left(1,0,0\right),\left(1,1,0\right),\left(0,0,1\right)\right\},{B}_{2}=\left\{\left(1,0,0\right),\left(0,1,1\right),\left(0,0,1\right)\right\}$. If
$Q\left(X\right)={x}_{1}^{2}-4{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+{x}_{2}^{2}+{x}_{3}^{2}$$Q\left(X\right)={x}_{1}^{2}-4{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+{x}_{2}^{2}+{x}_{3}^{2}$Q(X)=x_(1)^(2)-4x_(1)x_(2)+2x_(2)x_(3)+x_(2)^(2)+x_(3)^(2)Q(X)=x_1^2-4 x_1 x_2+2 x_2 x_3+x_2^2+x_3^2$Q\left(X\right)={x}_{1}^{2}-4{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+{x}_{2}^{2}+{x}_{3}^{2}$
find the representation of $Q$$Q$QQ$Q$ in terms of $\left({y}_{1},{y}_{2},{y}_{3}\right)$$\left({y}_{1},{y}_{2},{y}_{3}\right)$(y_(1),y_(2),y_(3))\left(y_1, y_2, y_3\right)$\left({y}_{1},{y}_{2},{y}_{3}\right)$.
b) Find the orthogonal canonical reduction of the quadratic form $-{x}^{2}+{y}^{2}+{z}^{2}+4xy+4xz$$-{x}^{2}+{y}^{2}+{z}^{2}+4xy+4xz$-x^(2)+y^(2)+z^(2)+4xy+4xz-x^2+y^2+z^2+4 x y+4 x z$-{x}^{2}+{y}^{2}+{z}^{2}+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}$${W}_{1}$W_(1)W_1${W}_{1}$ and ${W}_{2}$${W}_{2}$W_(2)W_2${W}_{2}$ are proper subspaces of a non-zero, finite dimensional, vector space $V$$V$VV$V$ and $\mathrm{dim}\left({W}_{1}\right)>\frac{\mathrm{dim}\left(V\right)}{2},\mathrm{dim}\left({W}_{2}\right)>\frac{\mathrm{dim}\left(V\right)}{2}$$\mathrm{dim}\left({W}_{1}\right)>\frac{\mathrm{dim}\left(V\right)}{2},\mathrm{dim}\left({W}_{2}\right)>\frac{\mathrm{dim}\left(V\right)}{2}$dim(W_(1)) > (dim(V))/(2),dim(W_(2)) > (dim(V))/(2)\operatorname{dim}\left(W_1\right)>\frac{\operatorname{dim}(V)}{2}, \operatorname{dim}\left(W_2\right)>\frac{\operatorname{dim}(V)}{2}$\mathrm{dim}\left({W}_{1}\right)>\frac{\mathrm{dim}\left(V\right)}{2},\mathrm{dim}\left({W}_{2}\right)>\frac{\mathrm{dim}\left(V\right)}{2}$, the ${W}_{1}\cap {W}_{2}\ne \left\{0\right\}$${W}_{1}\cap {W}_{2}\ne \left\{0\right\}$W_(1)nnW_(2)!={0}W_1 \cap W_2 \neq\{0\}${W}_{1}\cap {W}_{2}\ne \left\{0\right\}$.
ii) If $V$$V$VV$V$ is a vector space and $S=\left\{{v}_{1},{v}_{2},\dots ,{v}_{n}\right\}\subset V,n\ge 3$$S=\left\{{v}_{1},{v}_{2},\dots ,{v}_{n}\right\}\subset V,n\ge 3$S={v_(1),v_(2),dots,v_(n)}sub V,n >= 3S=\left\{v_1, v_2, \ldots, v_n\right\} \subset V, n \geq 3$S=\left\{{v}_{1},{v}_{2},\dots ,{v}_{n}\right\}\subset V,n\ge 3$, is such that ${v}_{i}\ne {v}_{j}$${v}_{i}\ne {v}_{j}$v_(i)!=v_(j)v_i \neq v_j${v}_{i}\ne {v}_{j}$ if $i\ne j$$i\ne j$i!=ji \neq j$i\ne j$, then $S$$S$SS$S$ is a linearly independent set.
iii) If ${T}_{1},{T}_{2}:V\to V$${T}_{1},{T}_{2}:V\to V$T_(1),T_(2):V rarr VT_1, T_2: V \rightarrow V${T}_{1},{T}_{2}:V\to V$ are linear operators on a finite dimensional vector space $V$$V$VV$V$ and ${T}_{1}\circ {T}_{2}$${T}_{1}\circ {T}_{2}$T_(1)@T_(2)T_1 \circ T_2${T}_{1}\circ {T}_{2}$ is invertible, ${T}_{2}\circ {T}_{1}$${T}_{2}\circ {T}_{1}$T_(2)@T_(1)T_2 \circ T_1${T}_{2}\circ {T}_{1}$ is also invertible.
iv) If an $n×n$$n×n$n xx nn \times n$n×n$ square matrix, $n\ge 2$$n\ge 2$n >= 2n \geq 2$n\ge 2$ is diagonalisable then it has the same minimal polynomial and characteristic polynomial.
v) If ${T}_{1},{T}_{2}:V\to V$${T}_{1},{T}_{2}:V\to V$T_(1),T_(2):V rarr VT_1, T_2: V \rightarrow V${T}_{1},{T}_{2}:V\to V$ are self adjoint operators on a finite dimensional inner product space $V$$V$VV$V$, then ${T}_{1}+{T}_{2}$${T}_{1}+{T}_{2}$T_(1)+T_(2)T_1+T_2${T}_{1}+{T}_{2}$ is also a self adjoint operator.
$$cos\:2\theta =2\:cos^2\theta -1$$

## 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 $\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$$\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}$\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$ and $\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$$\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}$\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$.
To find the angle between two vectors, we use the dot product formula. The dot product of two vectors $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$ and $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$ is given by $\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|\mathrm{cos}\left(\theta \right)$$\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|\mathrm{cos}\left(\theta \right)$A*B=|A||B|cos(theta)\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta)$\mathbf{A}\cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}|\mathrm{cos}\left(\theta \right)$, where $\theta$$\theta$theta\theta$\theta$ is the angle between the vectors, and $|\mathbf{A}|$$|\mathbf{A}|$|A||\mathbf{A}|$|\mathbf{A}|$ and $|\mathbf{B}|$$|\mathbf{B}|$|B||\mathbf{B}|$|\mathbf{B}|$ are the magnitudes of the vectors.
The vectors given are $\mathbf{A}=\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$$\mathbf{A}=\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$A=sqrt2i+2j+2k\mathbf{A} = \sqrt{2} \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k}$\mathbf{A}=\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}$ and $\mathbf{B}=\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$$\mathbf{B}=\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$B=i+sqrt2j+sqrt2k\mathbf{B} = \mathbf{i} + \sqrt{2} \mathbf{j} + \sqrt{2} \mathbf{k}$\mathbf{B}=\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}$.
1. Calculate the Dot Product $\mathbf{A}\cdot \mathbf{B}$$\mathbf{A}\cdot \mathbf{B}$A*B\mathbf{A} \cdot \mathbf{B}$\mathbf{A}\cdot \mathbf{B}$:
$\mathbf{A}\cdot \mathbf{B}=\left(\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right)\cdot \left(\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}\right)$$\mathbf{A}\cdot \mathbf{B}=\left(\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right)\cdot \left(\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}\right)$A*B=(sqrt2i+2j+2k)*(i+sqrt2j+sqrt2k)\mathbf{A} \cdot \mathbf{B} = (\sqrt{2}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{i} + \sqrt{2}\mathbf{j} + \sqrt{2}\mathbf{k})$\mathbf{A}\cdot \mathbf{B}=\left(\sqrt{2}\mathbf{i}+2\mathbf{j}+2\mathbf{k}\right)\cdot \left(\mathbf{i}+\sqrt{2}\mathbf{j}+\sqrt{2}\mathbf{k}\right)$
$=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}$$=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}$=sqrt2*1+2*sqrt2+2*sqrt2= \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2}$=\sqrt{2}\cdot 1+2\cdot \sqrt{2}+2\cdot \sqrt{2}$
2. Calculate the Magnitudes of $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$ and $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$:
• Magnitude of $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$, $|\mathbf{A}|$$|\mathbf{A}|$|A||\mathbf{A}|$|\mathbf{A}|$:$|\mathbf{A}|=\sqrt{\left(\sqrt{2}{\right)}^{2}+{2}^{2}+{2}^{2}}$$|\mathbf{A}|=\sqrt{\left(\sqrt{2}{\right)}^{2}+{2}^{2}+{2}^{2}}$|A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))|\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2}$|\mathbf{A}|=\sqrt{\left(\sqrt{2}{\right)}^{2}+{2}^{2}+{2}^{2}}$
• Magnitude of $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$, $|\mathbf{B}|$$|\mathbf{B}|$|B||\mathbf{B}|$|\mathbf{B}|$:$|\mathbf{B}|=\sqrt{{1}^{2}+\left(\sqrt{2}{\right)}^{2}+\left(\sqrt{2}{\right)}^{2}}$$|\mathbf{B}|=\sqrt{{1}^{2}+\left(\sqrt{2}{\right)}^{2}+\left(\sqrt{2}{\right)}^{2}}$|B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))|\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2}