Add assignment three and four

elements_of_mech/assignment-work-and-heat/assign2.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- First Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+First law of thermodynamics suggests that $\sum Q = \sum W$.
+\begin{align*}
+	Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\
+	75 - 40 + Q_3 &= -15 + 44 \\
+	Q_3 &= -6 kJ
+\end{align*}
+i.e.\ from the system to the surroundings.
+\section*{Question 2}
+We know that,
+\begin{align*}
+	Q &= \Delta E + W \\
+	Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\
+	Q &= -8600 kJ \\
+	Q &= -8.6 MJ
+\end{align*}
+\section*{Question 3}
+There is no heat transfer, therefore
+\begin{align*}
+	Q &=  \Delta E + W \\
+	W &= - \Delta E - \Delta V \\
+	W &= \int_{1}^{0} C_v \cdot dT \\
+	W &= -0.718 (T_2 - T_1) \\
+	W &= -50.26 \frac{kJ}{kg} \\
+	\text{Total Work} &= 2 \times (-50.26) = -100 kJ
+\end{align*}
+\section*{Question 4}
+\begin{align*}
+	1000 &= a + b \times 0.2 \\
+	200 &= a + b \times 1.2 \\
+	b &= -800 \\
+	a &= 1000 + 2 \times 800 = 1160 \\
+	\therefore P &= 1160 - 800 V \\
+	W &= \int_{V_1}^{V_2} P \cdot dV \\
+	  &= \int_{0.2}^{1.2} (1160 - 800V)dV \\
+	  &= 6000 kJ \\
+	V_1 &= 1.5 \times  1000 \times \frac{0.2}{1.5} - 85 \\
+		&= 215 \frac{kJ}{kg} \\
+	V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\
+		&= 155 \frac{kJ}{kg} \\
+	\Delta V &= V_2 - V_1 \\
+			 &= 40 \frac{kJ}{kg} \\
+	\Delta U &= m \Delta V  = 60 kJ \\
+	Q &= \Delta U + W \\
+	  &= 60 + 600 \\
+	  &= 660 kJ \\
+	U &= 1.5 PV - 85 \frac{kJ}{kg} \\
+	U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\
+	  &= 800 V^2 - 85 \frac{kJ}{kg} \\
+	\frac{\delta U}{\delta V} &= 1160 - 1600 V \\
+	\text{For maximum V,} \\
+	V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\
+	V &= 0.725 \\
+	u_{\max} = 335.5 \frac{kJ}{kg} \\
+\end{align*}
+\begin{align*}
+	U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\
+\end{align*}
+\section*{Question 5}
+\subsection*{Part a}
+\begin{align*}
+	Q &= \int_{273}^{373} C_p \cdot dT \\
+	t &= T - 273 K \\
+	\therefore t + 100 &= T - 173 \\
+	Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\
+	Q &= 238.32 J
+\end{align*}
+\subsection*{Part b}
+\begin{align*}
+	Q &= \Delta E + \int p\cdot dV \\
+	\Delta E &= Q - p(V_2 - V_1) \\
+	\Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\
+	\Delta E &= 197.79J
+\end{align*}
+\end{document}

elements_of_mech/assignment-work-and-heat/assign3.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- Second Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+We know that the net power output is the difference between the heat input and the heat rejected (cyclic device).
+\begin{align*}
+	W_{net,out} &= Q_H + Q_L \\
+	W_{out} &= 90 - 55 = 35 MW
+\end{align*}
+The thermal efficiency is the ratio of net work output and the heat output.
+\[
+	\eta_{thermal} = W_{out}\frac{W_{out}}{Q_H} =  \frac{35}{90} = 0.3889
+\]
+\section*{Question 2}
+\begin{align*}
+	\text{efficiency} = 1 - \frac{T_1}{T_2}  &= \frac{\text{work input}}{\text{work output}} \;\;\text{(must)}\\
+	1 - \frac{300}{1000} &\implies \frac{6}{1} \;\;\text{(claimed)} \\
+	0.7 &> 0.6 \\
+	\text{output} &> \text{claimed $\implies$ good}
+\end{align*}
+$\therefore$ we can agree to this claim.
+\section*{Question 3}
+\[
+	\eta = \frac{\text{output}}{\text{input}} = \frac{0.65}{0.65 + 0.4} = 0.619 = 61.9%
+\]
+now,
+\begin{align*}
+	\eta &= 1 - \frac{T_2}{T_1} \\
+	T_1 &= \frac{T_2}{1 - \eta} \\
+	T_1 &= 787.5 K
+\end{align*}
+$\therefore$, the temperature at which energy is absorbed is 787.5 K
+\section*{Question 4}
+\begin{align*}
+	\eta &= 1 - \frac{T_1}{T_2} \\
+\end{align*}
+where,
+\begin{align*}
+	\eta &= 0.6 \\
+	T_2 &= 800 K \\
+	T_1 &= T_{sink} \\
+	T_{sink} &= (1 - \eta)T_2 \\
+			  &= 324K
+\end{align*}
+Also,
+\begin{align*}
+	\eta &=  \frac{Q_1 - Q_{\text{rejected}}}{Q_1} \\
+	Q_{\text{rejected}} &= Q_1 (1 - \eta) \\
+							 &= 400(1 - 0.6) \\
+							 &= 160 kJ
+\end{align*}
+\section*{Question 5}
+Max COP will be achieved only in a Carnot refrigerator.
+\[
+	{(COP_{\max})}_{R} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2} = \frac{-20 + 273}{57} = 4.438
+\]
+Minimum Power $\to$ Max COP\@. We know that $P = \frac{Q}{\eta}$.
+\[
+	P = \frac{3 \times 57}{253} = 0.657 W
+\]
+\section*{Question 6}
+Max COP will be achieved only in a Carnot refrigerator.
+\begin{align*}
+	{(COP)}_{\text{Carnot refrigerator}} &= \frac{T_{low}}{T_{high} - T_{low}} = 1.37 \times 10^{-2} \\
+	{(COP)}_{\text{refrigerator}} &= \frac{Q_L}{W} = 1.37 \times 10^{-2} \\
+	Q_L &= 83.3 \\
+\end{align*}
+\section*{Question 7}
+For process $1-2$
+\begin{align*}
+	Q_{1-2} &= U_2 - U_1 + W_{1-2} \\
+	-732 &= U_2 - U_1 - 2.8 \times 3600 \\
+	U_2 - U_1 &= 9348kJ
+\end{align*}
+For process $2-1$
+\begin{align*}
+	Q_{2-1} &= U_1 - U_2 + W_{2-1} \\
+	-732 &= -9348 - 2.8 \times 3600 \\
+	Q_{2-1} &= -708 kJ
+\end{align*}
+Now,
+\[
+	Q_{2-1} = U_1 - U_2 + W_{2-1}
+\]
+For maximum work, $Q_{2-1} = 0$.
+\[
+	\therefore {(W_{2-1})}_{\max} = U_2 - U_1 = 9348kJ
+\]
+\section*{Question 8}
+\begin{align*}
+	{(COP)}_R &= \frac{268}{298 - 268} = 8.933 \\
+	{(COP)}_R &= \frac{5}{W} \\
+	W &= 0.56KW
+\end{align*}
+\section*{Question 9}
+\begin{align*}
+	\eta &= 1 - \frac{273+60}{273+671} = 0.64725 \\
+	\eta &= 0.3236 \\
+	\implies 1 - 0.3236 &= 0.6764 \\
+	\text{Ideal COP} &= \frac{305.2}{305.2 - 266.4} = 7.866 \\
+	\text{Actual COP} &= 3.923 = \frac{Q_3}{W} \\
+	if \;Q_3 &= 1kJ \\
+	\therefore W &= \frac{Q_3}{3.923} = 0.2549kJ \\
+	W &= Q_{\text{output}} - Q_{\text{input}} = \frac{1}{3.923} Q_1 - 0.6764Q_1 = 0.2549 \\
+	Q_{\text{out}} &= \frac{0.2549}{1 - 0.6764} = 0.7877 kJ \\
+\end{align*}
+\section*{Question 10}
+\begin{align*}
+	\frac{10}{W} &= \frac{293}{293-273} \\
+	or\;W &= 683 W
+\end{align*}
+\end{document}

engg_mech/assignment/t_four.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+Calculate Reactions:
+\begin{align*}
+	\sum M_J &= 0 \\
+	(12 kN)(4.8) + (12 kN)(2.4) - B(9.6) &= 0 \\
+	B &= 9 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	9 kN - 12 kN - 12 kN + J &= 0 \\
+	J &= 15kN
+\end{align*}
+Member CD:\@
+\begin{align*}
+	\sum F_y &= 0 \\
+	9 kN + F_{CD} &= 0 \\
+	F_{CD} &= 9 kN \to \text{compression}
+\end{align*}
+Member DF:\@
+\begin{align*}
+	\sum M_c &= 0 \\
+	F_{DF}1.8 m - 9kN \times 2.4m &= 0 \\
+	F_{DF} &= 12kN  \to \text{Tension}
+\end{align*}
+\section*{Question 2}
+Reactions:\@
+\[
+	A = N = 0
+\]
+DF member:\@
+\begin{align*}
+	\sum M_E &= 0 \\
+	(16 kN)(6m) - \frac{3}{5} F_{DF} (4m) &= 0 \\
+	F_{DF} &= 40 kN \to \text{Tension}
+\end{align*}
+
+EF member:\@
+\begin{align*}
+	\sum F&= 0 \\
+	16 kN \sin{\beta} - F_{EF} \cos{\beta} &= 0 \\
+	F_{EF} &= 16 \tan{\beta} \\
+		   &= 12 kN \to \text{Tension}
+\end{align*}
+
+EG member:\@
+\begin{align*}
+	\sum M_F &= 0 \\
+	16kN \times 9m + \frac{4}{5}F_{EG} \times 3m &= 0 \\
+	F_{EG} &= -60 kN \to \text{Compression}
+\end{align*}
+\section*{Question 3}
+Reactions
+\begin{align*}
+	\sum M_k &= 0 \\
+	36\times 2.4 - B \times 13.5 + 20 \times 9 + 20 \times 4.5 &= 0 \\
+	B &= 26.4kN \\
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	K_x &= 36 \\
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	26.4 - 20 -20 + K_y &= 0 \\
+	K_y &=  13.6 kN \uparrow \\
+\end{align*}
+\begin{align*}
+	\sum M_C &= 0 \\
+	36 \times 1.2 - 26.4 \times 2.25 - F_{AD} \times 1.2 &= 0 \\
+	F_{AD} &= 13.5 kN \to \text{compression} \\
+\end{align*}
+\begin{align*}
+	\sum M_A &= 0 \\
+	\left( \frac{8}{17}F_{CD}\right)(4.5) &= 0 \\
+	F_{CD} &= 0 \\
+\end{align*}
+\begin{align*}
+	\sum M_D &= 9 \\
+	\frac{15}{17} \times F_{CE} \times 2.4 - 26.4 \times 4.5 &= 0 \\
+	F_{CE} &= 56.1 kN \to \text{Tension}
+\end{align*}
+\section*{Question 4}
+
+Support reactions
+\begin{align*}
+	\sum M_I &= 0 \\
+	2\times 12 + 5\times 8 3\times 6 + 2\times 4 - A_y \times 16 &= 0 \\
+	A_y &= 5.625 kN
+\end{align*}
+\begin{align*}
+	\sum A_x &= 0 \\
+	A_x &= 0
+\end{align*}
+Method of joints: By inspection, members BN, NC, DO, OC, HJ, LE \& JG are zero force members \\
+Method of sections:
+\begin{align*}
+	\sum M_M &= 0 \\
+	4F_{CD} - 5.625 \times 4 &= 0 \\
+	F_{CD} &= 5.625 kN \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum M_A &= 0 \\
+	4F_{CM} - 2\times 4 &= 0 \\
+	F_{CM} &= 2 kN \to \text{Tension}
+\end{align*}
+\end{document}

engg_mech/assignment/t_three.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+Applying the equations of the equilibrium to the FBD of the entire truss, we have
+\begin{align*}
+	\sum M_A &= 0 \\
+	N_C (2 + 2) - 4(2) - 3(1.5) &= 0 \\
+	N_C &= 3.125 kN
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	3 - A_x &= 0 \\
+	A_x &= 3 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	A_y + 3.125 - 4 &= 0 \\
+	A_y &= 0.875 kN
+\end{align*}
+Method of joints \\
+Joint C:\@ Just assume it to be in equilibrium
+\begin{align*}
+	\sum F_y &= 0 \\
+	3.125 - F_{CD} \frac{3}{5} &= 0 \\
+	F_{CD} &= 5.21 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	5.208 \times \frac{4}{5} - F_{CB} &= 0 \\
+	F_{CB} &= 4.17kN  \to \text{Tension}
+\end{align*}
+Joint A:\@
+\begin{align*}
+	\sum F_y &= 0 \\
+	0.875 - F_{AD} \times \frac{3}{5} &= 0 \\
+	F_{AD} &= 1.46 kN  \to \text{Compression}
+\end{align*}
+
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AB} - 3 - 1.458 \times \frac{4}{5} &= 0 \\
+	F_{AB} &= 4.167 kN  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BD} &= 4 kN
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	4.167 - 4.167 &= 0
+\end{align*}
+\section*{Question 2}
+Analyze equilibrium of joint D, C \& E.
+Joint D
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{DE} \times \frac{3}{5} - 600 &= 0 \\
+	F_{DE} &= 1 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	1000 \times \frac{4}{5} - F_{DC} &= 0 \\
+	F_{DC} &= 800 N  \to \text{Tension}
+\end{align*}
+Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 900 N  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CB} &= 800 N  \to \text{Tension}
+\end{align*}
+Joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{EB} &= 750 N  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BA} &= 1.75 kN  \to \text{Compression}
+\end{align*}
+\section*{Question 3}
+Joint A
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{AL} &= 28.28 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AB} &= 20 kN  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BC} &= 20 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BL} &= 0
+\end{align*}
+Joint L
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{LC} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{LK} &= 28.28 kN \to \text{Compression}
+\end{align*}
+Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CD} &= 20 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CK} &= 10 kN  \to \text{Tension}
+\end{align*}
+Joint K
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{KD} &= 7.454 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{KJ} &= 23.57 kN  \to \text{Compression}
+\end{align*}
+Joint J
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{IJ} &= 23.57 kN
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{JD} &= 33.3 kN  \to \text{Tension}
+\end{align*}
+Now we know that there exists symmetry,
+\begin{gather*}
+	F_{AL} = F_{GH} = F{LK} = F{HI} = 28.3 kN \\
+	F_{AB} = F_{GF} = F_{BC} = F_{FE} = F_{CD} = F_{ED} = 20 kN \\
+	F_{BL} = F_{FH} = F_{LC} = F_{HE} = 0 \\
+	F_{CK} = F_{EI} = 10 kN \\
+	F_{KJ} = F_{IJ} = 23.6 kN \\
+	F_{KD} = F_{ID} = 7.45 kN
+\end{gather*}
+\section*{Question 4}
+To evaluate support reactions
+\begin{align*}
+	\sum M_E &= 0 \\
+	A_y &= \frac{4}{3} P
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	E_y &= \frac{4}{3} P
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	E_x &= P
+\end{align*}
+Methods of joints: By inspecting joint C, members CB \& CD are zero force members. Hence
+\[
+	F_{CB} = F_{CD} = 0
+\]
+Joint A
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{AB} &= 2.40 P  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{AF} &= 2P  \to \text{Tension}
+\end{align*}
+Joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	2.404P \times \frac{1.5}{\sqrt{3.25}} - P - F_{BF} \times \frac{0.5}{\sqrt{1.25}} - F_{BD} \times \frac{0.5}{\sqrt{1.25}} &= 0 \\
+	P - 0.447 F_{BF} - 0.447 F_{BD} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	2.404P \times \frac{1}{\sqrt{3.25}} - F_{BF} \times \frac{1}{\sqrt{1.25}} + F_{BD} \times \frac{1}{\sqrt{1.25}} &= 0 \\
+	1.33P - 0.8944 F_{BF} + 0.8944 F_{BD} &= 0
+\end{align*}
+Solving the above equations, we get
+\begin{align*}
+	F_{BD} &= 0.3727 P  \to \text{Compression}\\
+	F_{BF} &= 1.863 P  \to \text{Tension}
+\end{align*}
+Joint D
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{DE} &= 0.3727 P  \to \text{Compression}
+\end{align*}
+\section*{Question 5}
+FBD of Joint A
+\begin{gather*}
+	\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{AB} = 2.29 kN  \to \text{Tension} \\
+	F_{AC} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of Joint F
+\begin{gather*}
+	\frac{F_{DF}}{2.29} = \frac{F_{EF}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{DF} = 2.29 kN  \to \text{Tension} \\
+	F_{EF} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of Joint D
+\begin{gather*}
+	\frac{F_{BD}}{2.21} = \frac{F_{DE}}{0.6} = \frac{2.29}{2.29} kN \\
+	F_{DE} = 0.6 kN  \to \text{Compression}\\
+	F_{EF} = 2.21 kN \to \text{Tension}
+\end{gather*}
+FBD of Joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 2.21 kN  \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CH} &= 1.2 kN  \to \text{Compression}
+\end{align*}
+FBD of Joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BH} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{EJ} &= 1.2 kN  \to \text{Compression}
+\end{align*}
+\section*{Question 6}
+Zero Force Members
+Analyzing joint F:\@ Note that $DF$ and $EF$ are zero force members.
+\[
+	F_{DF} = F_{EF} = 0
+\]
+Analyzing joint D:\@ Note that $BD$ and $DE$ are zero force members.
+\[
+	F_{BD} = F_{DE} = 0
+\]
+FBD of joint A
+\begin{gather*}
+	\frac{F_{AB}}{2.29} = \frac{F_{AC}}{2.29} = \frac{1.2}{1.2} kN \\
+	F_{AB} = 2.29 kN  \to \text{Tension}\\
+	F_{AC} = 2.29 kN  \to \text{Compression}
+\end{gather*}
+FBD of joint B
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{BE} &= 2.7625 kN  \to \text{Tension}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{BC} &= 2.25 kN  \to \text{Compression}
+\end{align*}
+FBD of joint C
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{CE} &= 2.21 kN \to \text{Compression}
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	F_{CH} &= 2.86 kN  \to \text{Compression}
+\end{align*}
+FBD of joint E
+\begin{align*}
+	\sum F_x &= 0 \\
+	F_{EH} &= 0
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	f_{EJ} &= 1.657 kN  \to \text{Tension}
+\end{align*}
+
+\end{document}

engg_mech/assignment/t_two.tex

@@ -0,0 +1,135 @@
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{siunitx}
+\usepackage{graphicx}
+\usepackage{wrapfig}
+\graphicspath{{./images/}}
+\begin{document}
+\title{Engineering Mechanics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{17 May, 2020}
+\maketitle
+\section*{Question 1}
+\subsection*{Question 1.a}
+We shall balance the torque about $C$. $\angle ABC = \theta$, Tension in cable $=T$.
+\begin{align*}
+	240 (0.4) + 240 (0.8) &= T\sin{\theta} \times 0.18 \\
+	T\sin{\theta} &= 1600 \\
+	T \frac{0.24}{0.3} &= 1600 \\
+	T &= 2000
+\end{align*}
+\subsection*{Question 1.b}
+On making FBD of bracket BCD.\@
+
+\begin{align*}
+	x-component &= N_x = T\sin{\theta} = 1600 \\
+	y-component &= N_y = T\cos{\theta} + 240 +240 = 1680
+\end{align*}
+
+\section*{Question 2}
+\subsection*{Question 2.a}
+We shall balance the torque about C
+\begin{align*}
+	P \times 7.5 &= T \times 5 \\
+	T &= 150 lb
+\end{align*}
+\subsection*{Question 2.b}
+\begin{align*}
+	N_x \text{(reaction at C along x-axis)} &= - (P + T\sin{\ang{37}}) = -190 lb \\
+	N_y \text{(reaction at C along y-axis)} &= - T \cos{\ang{37}} = -120 lb
+\end{align*}
+\section*{Question 3}
+\subsection*{Question 3.a}
+\[
+	\alpha = 0
+\]
+Balance torque about B
+\begin{align*}
+	N_a \times 20 &= 75 \times 10 \\
+	N_a &= 37.5 lb
+\end{align*}
+Balance torque about A
+\begin{align*}
+	N_b \times 20 &= 75 \times 10 \\
+	N_b &= 37.5 lb
+\end{align*}
+\subsection*{Question 3.b}
+\[
+	\alpha = \ang{90}
+\]
+Balance torque about A
+\begin{align*}
+	N_b \times 20 &= 75 \times 10 \\
+	N_b &= 37.5 lb
+\end{align*}
+Balance torque about B
+\begin{align*}
+	N_a \times 12 &= 75 \times 10 \\
+	N_a &= 62.5 lb
+\end{align*}
+\subsection*{Question 3.c}
+\[
+	\alpha = \ang{30}
+\]
+Balance torque about the mid point of horizontal rod
+\begin{align*}
+	N_a \times 10 &= {(N_b)}_y \times 10 \\
+	N_a &= {(N_b)}_y
+\end{align*}
+Balance torque about B
+\begin{align*}
+	N_a \times 20 &= 75 \times 10 \\
+	N_a &= 37.5 lb \\\\
+	N_a &= N_b \cos{\ang{30}} \\
+	N_b &= 43.30
+\end{align*}
+
+\section*{Question 4}
+\subsection*{Question 4.a}
+Balance torque about C
+\begin{align*}
+	120 \times 0.28 &= T \times  \frac{150}{250} \times 0.2 + T \times \frac{150}{390} \times 0.36 \\
+	33.6 &= T \times 0.26 \\
+	T &= 129.2 N \approx 130 N
+\end{align*}
+\subsection*{Question 4.b}
+\begin{align*}
+	{(N_c)}_x &= T \times \frac{200}{250} + T \times \frac{360}{390} \\
+	{(N_c)}_x &= 223 N \\
+	{(N_c)}_y &= 120 - \left(T \times  \frac{150}{250} + T \times  \frac{150}{390}\right) \\
+	{(N_c)}_y &= -7.21 N \\
+	N_c &= \sqrt{223^2 + 7.21^2} N \\
+	N_c &= 224 N
+\end{align*}
+
+\section*{Question 5}
+Balance torque about B
+\begin{align*}
+	{(N_a)}_y \times 8 &= 4000 \times 2 \\
+	{(N_a)}_y &= 1000 N
+\end{align*}
+FBD of Rod
+\begin{align*}
+	4000 &= {(N_A)}_y + {(N_B)}_y \\
+	4000 &= 1000 + {(N_B)}_y \\
+	{(N_B)}_y &= 3000 \\\\
+	{(N_B)}_y &= N_b \sin{\ang{60}} \\
+	N_B &= 3465 \\
+	{(N_A)}_x &= {(N_B)}_x \\
+	{(N_A)}_x &= N_B \sin{\ang{30}} \\
+	{(N_A)}_x &= 1732
+\end{align*}
+\section*{Question 6}
+First we have to calculate reaction at A
+\begin{align*}
+	\sum F_x &= 0 \\
+	4000 \cos{\ang{30}} &= A_x \\
+	A_x &= 3464 N
+\end{align*}
+\begin{align*}
+	\sum F_y &= 0 \\
+	6000 + 4000 \cos{\ang{30}} &= A_y \\
+	A_y &= 8000 N
+\end{align*}
+\end{document}

engg_physics/lab_assignment_1/.gitignore

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+page_00.jpg

engg_physics/lab_assignment_1/generate.sh

@@ -0,0 +1,19 @@
+#!/usr/bin/env bash
+
+text='Name: Ahmad Saalim Lone
+Enrollment No: 2019BCSE017
+Subject: Engineering Physics Lab
+Branch: Computer Science
+Unit: Semiconductor Physics
+No. of pages: 6
+Submission Date: 15 May, 2020
+Name of concerned teacher:
+Dr Mohd Zubair Ansari'
+convert -size 638x877 xc:white \
+	-font 'Fira-Code-Bold' \
+	-pointsize 30 \
+	-fill black \
+	-gravity center \
+	-draw "text 0,0 '$text'" \
+	page_00.jpg
+convert page_{00..05}.jpg assignment.pdf

maths/assignment-matrices/assign4.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Mathematics Assignment 4 --- Matrices}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{16 May, 2020}
+\maketitle
+
+\section*{Question 1}
+\begin{align*}
+	A &=
+	\begin{bmatrix}
+		1 & -1 & -1 & 2 \\
+		4 & 2 & 2 & -1 \\
+		2 & 2 & 0 & -2
+	\end{bmatrix}
+	\\
+	PAQ &= \text{Normal form} \\
+	I_3AI_4 &= A_{3 \times 4} \\
+	\begin{bmatrix}
+		1 & 0 & 0 \\
+		0 & 1 & 0 \\
+		0 & 0 & 1
+	\end{bmatrix}
+	A
+	\begin{bmatrix}
+		1 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 \\
+		0 & 0 & 1 & 0 \\
+		0 & 0 & 0 & 1
+	\end{bmatrix}
+			&=
+			\begin{bmatrix}
+				1 & -1 & -1 & 2 \\
+				4 & 2 & 2 & -1 \\
+				2 & 2 & 0 & -2
+			\end{bmatrix}
+\end{align*}
+
+Applying the following row tranformations on $I_3$ and column tranformations on $I_4$.
+
+\begin{align*}
+	C_4 &\to C_4 + C_2 \\
+	R_3 &\to \frac{R_3}{2} \\
+	R_2 &\to \frac{R_2 + 2R_1}{3} \\
+	R_1 &\to R_1 + R_3 \\
+	C_1 &\to C_1 - 2 C_4 \\
+	C_4 &\to  C_4 + C_3 \\
+	C_2 &\to C_2 - C_1 \\
+	C_3 &\rightleftharpoons C_1 \\
+	C_2 &\rightleftharpoons C_4 \\
+	R_1 &\rightleftharpoons -R_1
+\end{align*}
+
+we get
+\begin{align*}
+	P &=
+	\begin{bmatrix}
+		-1 & 0 & -\frac{1}{2} \\[6pt]
+		\frac{2}{3} & \frac{1}{3} & 0 \\[6pt]
+		0 & 0 & \frac{1}{2}
+	\end{bmatrix}
+	\\
+	Q &=
+	\begin{bmatrix}
+		0 & 0 & 1 & -1 \\
+		0 & 0 & -2 & 3 \\
+		1 & 1 & 0 & 0 \\
+		0 & 1 & -2 & 2
+	\end{bmatrix}
+	\\
+	\text{Normal Form of A} &=
+	\begin{bmatrix}
+		1 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 \\
+		0 & 0 & 1 & 0
+	\end{bmatrix}
+\end{align*}
+\section*{Question 2}
+\[
+	x^2yz = xy^2z^3 = x^3y^2z = e
+\]
+
+Taking $\ln$ on both sides
+\begin{align*}
+	2\ln{x} + \ln{y} \ln{z} &= 1 \\
+	\ln{x} + 2 \ln{y} + 3 \ln{z} &= 1 \\
+	3\ln{x} + 2\ln{y} + \ln{z} &= 1
+\end{align*}
+
+\[
+	\text{Augemented Matrix} = [A:B] =
+\begin{bmatrix}
+2 & 1 & 1 & : & 1 \\
+1 & 2 & 3 & : & 1 \\
+3 & 2 & 1 & : & 1
+\end{bmatrix}
+\]
+After ppling the following tranformations
+\begin{align*}
+	R_3 &\to R_3 - R_1 \\
+	R_2 &\to R_2 - R_1 \\
+	R_2 &\to R_2 - R_1 \\
+	R_1 &\to R_1 -R_3
+\end{align*}
+
+We get
+\[
+\begin{bmatrix}
+1 & 0 & 1 & : & 1 \\
+-3 & 0 & 1 & : & -1 \\
+1 & 1 & 0 & : & 0
+\end{bmatrix}
+\]
+$Rank(A) = Rank(A:B) = 3 = n$\\
+$\therefore$ unique solution.
+\section*{Question 3}
+\begin{align*}
+	x - cy - bz &= 0 \\
+	cx - y + az &= 0 \\
+	bx + ay - z &= 0 \\
+	\text{Augemented matrix} &=
+\begin{bmatrix}
+1 & -c & -b & : & 0 \\
+c & -1 & a & : & 0 \\
+b & a & -1 & : & 0
+\end{bmatrix}
+\end{align*}
+
+Applying the following tranformations
+\begin{align*}
+	R_2 &\to R_2 - cR_1 \\
+	R_3 &\to R_3 - bR_1 \\
+	R_2 &\to \frac{R_2}{c^2 - 1} \\
+	R_3 &\to R_3 - (a + bc)R_2
+\end{align*}
+we get
+\[
+	C =
+\begin{bmatrix}
+	1 & -c & -b & : & 0 \\[6pt]
+	0 & 1 & \frac{bc+c}{c^2 - 1} & : & 0 \\[6pt]
+0 & 0 & b^2 - 1 - \frac{{(a+bc)}^2}{c^2 - 1} & : & 0
+\end{bmatrix}
+\]
+
+For non trivial solutions $Rank(A) = Rank(c) \ne number\;of\;unknows$
+\begin{align*}
+	b^2 - 1 \frac{{(a+bc)}^2}{c^2 -1} &= 0 \\
+	\implies a^2 + b^2 + c^2 + 2abc &= 1 \\
+\end{align*}
+Now
+\begin{align*}
+	x - cy -bz &= 0 \\
+	y + \left(\frac{bc + a}{c^2 - 1}z\right) &= 0 \\
+\end{align*}
+We get
+\begin{align*}
+	x &= \frac{ac + b}{1-c^2}z \\
+	y &= \frac{bc + a}{1 - c^2}z \\
+	z &= z \\
+	x : y : z &= \sqrt{|1 - a^2|} : \sqrt{|1 - b^2|} : \sqrt{|1 -c^2|}
+\end{align*}
+
+\section{Question 4}
+\begin{align*}
+	A &=
+\begin{bmatrix}
+3 & -4 & 4 \\
+1 & -2 & 44 \\
+1 & -1 & 3
+\end{bmatrix} \\
+	|A - \lambda I| &= 0\\
+\begin{vmatrix}
+3 - \lambda & -4 & 4 \\
+1 & -2 - \lambda & 44 \\
+1 & -1 & 3 - \lambda
+\end{vmatrix} &= 0
+\end{align*}
+
+\( \lambda = -1, 2, 3 \) are Eigen Values
+
+For $\lambda = -1$
+\begin{align*}
+	\begin{bmatrix}
+		4 & -4 & 4 \\
+		1 & -1 & 4 \\
+		1 & -1 & 4
+	\end{bmatrix}
+	\begin{bmatrix}
+		x \\
+		y \\
+		z
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	0 \\
+	0 \\
+	0
+\end{bmatrix} \\
+	\text{Eigen Vector} &=
+	\begin{bmatrix}
+		1 \\
+		1 \\
+		0
+	\end{bmatrix}
+\end{align*}
+For $\lambda = 2$
+\begin{align*}
+\begin{bmatrix}
+1 & -4 & 4 \\
+1 & -4 & 4 \\
+1 & -1 & 1
+\end{bmatrix}
+	\begin{bmatrix}
+		x \\
+		y \\
+		z
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	0 \\
+	0 \\
+	0
+\end{bmatrix} \\
+	\text{Eigen Vector} &=
+	\begin{bmatrix}
+		0 \\
+		1 \\
+		1
+	\end{bmatrix}
+\end{align*}
+For $\lambda = 3$
+\begin{align*}
+\begin{bmatrix}
+0 & -4 & 4 \\
+1 & -5 & 4 \\
+1 & -1 & 0
+\end{bmatrix}
+	\begin{bmatrix}
+		x \\
+		y \\
+		z
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	0 \\
+	0 \\
+	0
+\end{bmatrix} \\
+	\text{Eigen Vector} &=
+	\begin{bmatrix}
+		1 \\
+		1 \\
+		1
+	\end{bmatrix}
+\end{align*}
+\section*{Question 5}
+\begin{align*}
+	A &=
+\begin{bmatrix}
+2 & 2 & 0 \\
+2 & 1 & 1 \\
+-7 & 2 & -3
+\end{bmatrix}
+\\
+	|A - \lambda I| &=  0 \\
+\begin{vmatrix}
+2 - \lambda & 2 & 0 \\
+2 & 1 - \lambda & 1 \\
+-7 & 2 & -3 - \lambda
+\end{vmatrix} &= 0
+\end{align*}
+
+Eigen values $\lambda = 1, 3, -4$ \\
+
+1\textsuperscript{st} eigen value of $A^2 -2 A  + I = 1^2 - 2(1) + 1 = 0$
+
+2\textsuperscript{nd} eigen value of $A^2 -2 A  + I = 3^2 - 2(3) + 1 = 4$
+
+3\textsuperscript{rd} eigen value of $A^2 -2 A  + I = {(-4)}^2 - 2(-4) + 1 = 25$
+
+\section*{Question 6}
+\begin{align*}
+	A &=
+\begin{bmatrix}
+7 & 3 \\
+2 & 6
+\end{bmatrix}
+\\
+	| A - \lambda I| &= 0 \\
+\begin{vmatrix}
+	7 - \lambda & 3 \\
+	2 & 6 - \lambda
+\end{vmatrix} &= 0 \\
+	(7 - \lambda)(6 - \lambda) - 5 &= 0 \\
+	(\lambda - 4)(\lambda - 9) &= 0 \\
+	A^2 - 13 A + 36 &= 0 \\
+	A^2 &= 13 A - 36 \\
+	A^2\cdot A &= (13 A - 36)A \\
+	A^3 &= 13 A^2 - 36A \\
+	A^3 &=
+\begin{bmatrix}
+715 & 507 \\
+338 & 546
+\end{bmatrix}
+-
+\begin{bmatrix}
+252 & 108 \\
+72 & 216
+\end{bmatrix} \\
+	A^3 &=
+\begin{bmatrix}
+463 & 399 \\
+266 & 330
+\end{bmatrix}
+\end{align*}
+
+\section*{Question 7}
+Suppose that $\lambda$ is a (possibly complex) eigen value of the real symmetric matrix $A$. Thus, there is a non-zero vector $V$, also with complex entries such that $AV = \lambda V$. By taking the complex conjugate of both sides and noting that $\overline{A} = A$ since $A$ has real entries, we get $\overline{AV} = \overline{\lambda V} \implies A\overline{V} = \overline{\lambda}\;\overline{V}$. Then using that $A^T = A$,
+\begin{gather*}
+	\overline{V}^T AV = \overline{V}^t(AV) = \overline{V}(\lambda V) = \lambda( \overline{V}V ) \\
+	\overline{V}^T AV = {(A \overline{V})}^T V = {( \overline{\lambda}\;\overline{V} )}^T V = \overline{\lambda} ( \overline{V} V )
+\end{gather*}
+
+Since $V \ne 0$, we have $ \overline{V}V \ne 0$. Thus $\lambda = \overline{\lambda}$, which means $\lambda \in R$.
+
+\section*{Question 8}
+Quadratic Form $ax_1^2 + cx_2^2 - 2bx_1x_2$
+\[
+\begin{bmatrix}
+a & -b \\
+-b & c
+\end{bmatrix}
+\]
+Convert it to diagonal matrix by applying
+\begin{align*}
+	R_2 &\to R_2 + \frac{b}{a}R_1 \\
+	C_2 &\to C_2 + \frac{b}{a}C_1
+\end{align*}
+Now, we get
+\[
+\begin{bmatrix}
+a & 0 \\
+0 & c - \frac{b^2}{a}
+\end{bmatrix}
+\]
+
+Nature $\to$ positive definite $\to$ when $rank(r) = index(s)$ or when all eigen values are positive i.e. $a > 0$ \& $c - \frac{b^2}{a} > 0 \implies ac -b^2 > 0$. Hence proved.
+
+\section*{Question 9}
+\(
+A =
+\begin{bmatrix}
+	\lambda & 1 & 1 \\
+1 & \lambda & -1 \\
+1 & -1 & \lambda
+\end{bmatrix}
+\) is a symmetric matrix obtained when compared to Quadratic form $\lambda(x^2+ y^2 + z^2) + 2xy + 2zx -2yz$.
+
+Now, convert $A$ into diagonal matrix by:
+\begin{align*}
+	C_1 &\to C_1 + C_3 \\
+	C_1 &\to \frac{C_1}{\lambda + 1} \\
+	R_3 &\to R_3 - R_1 \\
+	C_3 &\to C_3 - C_1 \\
+	C_2 &\to C_2 -C_1 \\
+	C_3 &\to C_3 + \frac{C_2}{\lambda} \\
+	R_3 &\to R_3 + \frac{2R_2}{\lambda}
+\end{align*}
+
+we get,
+
+\[
+\begin{bmatrix}
+1 & 0 & 0 \\
+0 & \lambda & 0 \\
+0 & 0 & \lambda - \frac{2}{\lambda} - 1
+\end{bmatrix}
+\]
+For definite positive nature, all Eigen values must be positive i.e. $\lambda > 0$, $\lambda - \frac{2}{\lambda} - 1 > 0$. Taking intersection of these two, we get $\lambda \in (2, \infty)$.
+
+\section*{Question 10}
+
+Multiplication of all the eigen values = determinant of the matrix. For singular matrix, determinant value = 0.
+\begin{align*}
+	\text{Eigen Values} &= 2, 3, a \\
+	6a &= 0 \\
+	a &= 0
+\end{align*}
+
+\section*{Question 11}
+
+Quadratic Form $x_2^2 + 2x_2^2 - 5x_3^2$
+\[
+\begin{bmatrix}
+1 & 0 & 0 \\
+0 & 2 & 0 \\
+0 & 0 & -5
+\end{bmatrix}
+\]
+\begin{align*}
+	index(s) &= 2\;\text{(No of positive terms)} \\
+	rank(r) &= 3 \\
+	signature &= 2s -r = 4 - 3 = 1
+\end{align*}
+
+\section*{Question 12}
+
+Quadratic form $ax^2 + 2bcy + cy^2$.
+\[
+\begin{bmatrix}
+a & b \\
+b & c
+\end{bmatrix}
+\]
+Convert it into diagonal matrix by doing:
+\begin{align*}
+	R_2 &\to R_2 - \left(\frac{b}{a}\right)R_1 \\
+	C_2 &\to C_2 - \left(\frac{b}{a}\right)C_1
+\end{align*}
+
+Finally, we get
+\[
+\begin{bmatrix}
+a & 0 \\
+0 & c - \frac{b^2}{a}
+\end{bmatrix}
+\]
+For positive definite, $a>0$ and $ac -b^2 > 0$. \\
+For negative definite, $a<0$ and $ac -b^2 > 0$. \\
+Roots of the quadratic equation ($ax^2 + 2bx + c = 0$) are imaginary when $D < 0$. \\
+${(2b)}^2 - 4ac = 4( b^2 - ac )$ is always negative
+
+\section*{Question 13}
+\begin{align*}
+	X_1 &=
+\begin{bmatrix}
+1 \\
+2 \\
+-3 \\
+4
+\end{bmatrix}
+\\
+	X_2 &=
+\begin{bmatrix}
+1 \\
+-5 \\
+8 \\
+-7
+\end{bmatrix}
+\\
+	X_3 &=
+\begin{bmatrix}
+1 \\
+-5 \\
+8 \\
+-7
+\end{bmatrix}
+\\
+	\lambda_1 X_1 + \lambda_2 X_2 + \lambda_3 X_3 &= 0 \\
+\begin{bmatrix}
+1 & 1 & 1 \\
+2 & -5 & -5 \\
+-3 & 8 & 8 \\
+4 & -7 & -7
+\end{bmatrix}
+\begin{bmatrix}
+\lambda_1 \\
+\lambda_2 \\
+\lambda_3
+\end{bmatrix} &=
+\begin{bmatrix}
+0 \\
+0 \\
+0 \\
+0
+\end{bmatrix}
+\end{align*}
+Applying the following tranformations
+\begin{align*}
+	R_2 &\to R_2 - 2R_1 \\
+	R_2 &\to -\frac{R_2}{7} \\
+	R_3 &\to R_3 + R_4 \\
+	R_3 &\to R_3 - R_1 \\
+	R_1 &\to R_1 - R_2 \\
+	R_3 &\to R_3 - 4 R_1 \\
+	R_4 &\to -\frac{R_4}{7} \\
+	R_4 &\to R_4 - R_2
+\end{align*}
+we get
+\begin{align*}
+\begin{bmatrix}
+1 & 2 & 0 \\
+0 & 1 & 1 \\
+0 & 0 & 0 \\
+0 & 0 & 0
+\end{bmatrix}
+\begin{bmatrix}
+\lambda_1 \\
+\lambda_2 \\
+\lambda_3
+\end{bmatrix}
+&=
+\begin{bmatrix}
+0 \\
+0 \\
+0 \\
+0
+\end{bmatrix} \\
+	\lambda_1 + 2\lambda_2 &= 0 \\
+	\lambda_2 + \lambda_3 &= 0
+	\therefore \\
+	\lambda_1 &= t \\
+	\lambda_2 &= -\frac{t}{2} \\
+	\lambda_3 &= \frac{t}{2}
+\end{align*}
+Putting the values, we get
+
+\[
+	2 X_1 - X_2 + X_3 = 0
+\]
+
+\section*{Question 14}
+
+\begin{align*}
+	(2 - \lambda)x_1 + (-2)x_2 + x_3 &= 0 \\
+	2x_1 - (\lambda + 3) x_2 + 2x_3 &= 0 \\
+	-x_1 + 2x_2 - \lambda x_3 &= 0
+\end{align*}
+
+\(
+	Rank(A) = Rank(\text{augemented matrix}) < 3
+\) for non trivial solutions.
+
+Check the determinant first, $\Delta = 0$
+
+$\Delta = 0$ gets us $\lambda = 1, 3$
+
+Now, for augemented matrix $[A:B]$, put $\lambda = 1, -3$ in \(
+\begin{bmatrix}
+2-\lambda & -2 & 1 & : & 0 \\
+2 & -(\lambda + 3) & 2 & : & 0 \\
+-1 & 2 & -\lambda & : & 0
+\end{bmatrix}
+\)
+
+For $\lambda = 1$
+\[
+\begin{bmatrix}
+1 & -2 & 1 & : & 0 \\
+2 & -4 & 2 & : & 0 \\
+-1 & 2 & -1 & : & 0
+\end{bmatrix}
+\]
+
+Doing transformations to form row echelon form, we get
+\[
+\begin{bmatrix}
+1 & -2 & 1 & : & 0 \\
+0 & 0 & 0 & : & 0 \\
+0 & 0 & 0 & : & 0
+\end{bmatrix}
+\]
+\begin{align*}
+	x_1 -2x_2 + x_3 &= 0 \\
+	if\;x_2 = k,\;x_3 = t, then\;x_1 &= 2k - t
+\end{align*}
+For $\lambda = -3$
+\[
+\begin{bmatrix}
+5 & -2 & 1 & : & 0 \\
+2 & 0 & 2 & : & 0 \\
+-1 & 2 & 3 & : & 0
+\end{bmatrix}
+\]
+
+Doing transformations to form row echelon form, we get
+\[
+\begin{bmatrix}
+5 & -2 & 1 & : & 0 \\
+0 & \frac{4}{5} & \frac{8}{5} & : & 0 \\
+0 & 0 & 0 & 0 & 0
+\end{bmatrix}
+\]
+\begin{align*}
+	5x_1 - 2x_2 + x_3 &= 0 \\
+	\frac{4x_2}{5} + \frac{8x_3}{5} &= 0 \\
+	if\;x_3 = t, then \\
+	x_1 &= - t \\
+	x_2 &= -2t \\
+	x_3 &= t
+\end{align*}
+
+\section*{Question 15}
+\subsection*{Part i}
+Since $A + A^T = 0$, $A$ must either be skew symmetric. If A is skew symmetric, we know that the rank of an odd order skew symmetric matrix must be even. $\therefore Rank \leq 2020$
+\subsection*{Part ii}
+Inverse does not exist as $A$ is singular matrix.
+\end{document}