Add maths assignment 2

maths/assignment-matrices/assign2.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Mathematics Assignment 2 --- Matrices}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{05 May, 2020}
+\maketitle
+
+\section*{Question 1}
+If
+\(
+A =
+\begin{bmatrix}
+	3 & 3 & 4 \\
+	2 & -3 & 4 \\
+	0 & -1 & 1
+\end{bmatrix}
+\)
+find $A^{-1}$ and verify that $A^{-1}A = I = AA^{-1}$.
+\subsection*{Solution}
+
+\begin{align*}
+	A &=
+	\begin{bmatrix}
+		3 & 3 & 4 \\
+		2 & -3 & 4 \\
+		0 & -1 & 1
+	\end{bmatrix} \\
+	A^{-1} &= \frac{adj(A)}{|A|} \\
+	|A| &= 3 \times (-3 + 4 ) - 3 \times 2 + 4 \times -2 \\
+	|A| &= -11 \\
+	\text{Cofactor matrix of A } &=
+	\begin{bmatrix}
+		1 & 2 & -2 \\
+		7 & 3 & -3 \\
+		24 & 4 & -15
+	\end{bmatrix} \\
+	adj(A) &=
+	\begin{bmatrix}
+		1 & 7 & 24 \\
+		2 & 3 & 4 \\
+		-2 & -3 & -15
+	\end{bmatrix} \\
+	A^{-1} &=
+	\begin{bmatrix}
+		\frac{1}{11} & \frac{2}{11} & -\frac{2}{11} \\[6pt]
+		\frac{7}{11} & \frac{3}{11} & -\frac{3}{11} \\[6pt]
+		\frac{24}{11} & \frac{4}{11} & -\frac{15}{11}
+	\end{bmatrix}
+\end{align*}
+
+\begin{align*}
+	A \times A^{-1} &=
+	\begin{bmatrix}
+		\frac{1}{11} & \frac{2}{11} & -\frac{2}{11} \\[6pt]
+		\frac{7}{11} & \frac{3}{11} & -\frac{3}{11} \\[6pt]
+		\frac{24}{11} & \frac{4}{11} & -\frac{15}{11}
+	\end{bmatrix}
+	\begin{bmatrix}
+		3 & 3 & 4 \\
+		2 & -3 & 4 \\
+		0 & -1 & 1
+	\end{bmatrix} \\
+					&=
+					\begin{bmatrix}
+						1 & 0 & 0 \\
+						0 & 1 & 0 \\
+						0 & 0 & 1
+					\end{bmatrix} \\
+					&= I
+\end{align*}
+
+Similarly, $A^{-1}\times A = I$
+
+\section*{Question 2}
+
+Find the inverse of
+\(
+A =\begin{bmatrix}
+	1 & 2 & 1 \\
+	3 & 2 & 3 \\
+	1 & 1 & 2
+\end{bmatrix}
+\) by applying E-transformation.
+\subsection*{Solution}
+\[
+	\begin{bmatrix}
+		1 & 2 & 1 \\
+		3 & 2 & 3 \\
+		1 & 1 & 2
+	\end{bmatrix}
+	=
+	\begin{bmatrix}
+		1 & 0 & 0 \\
+		0 & 1 & 0 \\
+		0 & 0 & 1
+	\end{bmatrix}
+	A
+\]
+
+After applying the following transformations
+\begin{align*}
+	C_3 &\to C_3 - C_1 \\
+	C_2 &\to \frac{C_1}{2} \\
+	R_1 &\to R_1 - R_2 \\
+	R_1 &\to \frac{R_1}{-2} \\
+	C_1 &\to C_1 - C_3 \\
+	R_2 &\to R_2 - 3R1 \\
+	R_2 &\to R_3 - \frac{R_2}{2}
+\end{align*}
+
+we get
+
+\begin{align*}
+	\begin{bmatrix}
+		1 & 0 & 0 \\
+		0 & 1 & 0 \\
+		0 & 0 & 1
+	\end{bmatrix}
+&=
+\begin{bmatrix}
+	-1 & \frac{1}{4} & \frac{1}{2} \\[6pt]
+	3 & -\frac{1}{4} & -\frac{1}{2} \\[6pt]
+	-\frac{5}{2} & \frac{1}{8} & \frac{3}{4}
+\end{bmatrix}
+\times A \\
+\therefore A^{-1} &=
+\begin{bmatrix}
+	-1 & \frac{1}{4} & \frac{1}{2} \\[6pt]
+	3 & -\frac{1}{4} & -\frac{1}{2} \\[6pt]
+	-\frac{5}{2} & \frac{1}{8} & \frac{3}{4}
+\end{bmatrix}
+\end{align*}
+
+\section*{Question 3}
+
+Reduce the matrix
+\(
+A =
+\begin{bmatrix}
+	1 & -1 & 2 & -3 \\
+	4 & 1 & 0 & 2 \\
+	0 & 3 & 0 & 4 \\
+	0 & 1 & 0 & 2
+\end{bmatrix}
+\) to the normal form and hence determine its rank.
+
+\subsection*{Solution}
+
+To reduce the matrix we apply the following operations
+
+\begin{align*}
+	R_1 &\to R_1 + R_3 \\
+	R_4 &\to R_4 - R_2 \\
+	R_2 &\to R_2 + R_4 \\
+	C_2 &\to C_2 - C_3 \\
+	C_4 &\to C_4 - C_2 \\
+	R_3 &\to  R_3 - R_2 \\
+	C_2 &\rightleftharpoons C_3 \\
+	C_2 &\to \frac{C_2}{2} \\
+	R_3 &\to \frac{R_3}{2} \\
+	R_4 &\to \frac{R_4}{2} \\
+	C_4 &\to C_4 - C_2 \\
+	C_3 &\to C_3 - C_4 \\
+	C_4 &\rightleftharpoons C_2 \\
+	R_4 &\to R_4 - R_1 \\
+	R_4 &\to -R_4 \\
+	R_1 &\to R_1 - R_4
+\end{align*}
+
+At the end we arrive at
+\[
+	\begin{bmatrix}
+		1 & 0 & 0 & 0 \\
+		0 & 1 & 0 & 0 \\
+		0 & 0 & 1 & 0 \\
+		0 & 0 & 0 & 1
+	\end{bmatrix}
+\]
+i.e. $[I_4]$
+
+$rank = 4$
+
+\section*{Question 4}
+
+Reduce the matrix \(A =
+\begin{bmatrix}
+	-2 & -1 & -3 & -1 \\
+	1 & 2 & 3 & -1 \\
+	1 & 0 & 1 & 1 \\
+	0 & 1 & 1 & -1
+\end{bmatrix}
+\) to Echelon form and find its rank.
+\subsection*{Solution}
+
+To convert it into Echelon form, we apply the following transformations.
+
+\begin{align*}
+	R_1 &\to R_1 + R_2 + R_3 \\
+	R_2 &\to  R_2 - R_3 \\
+	C_2 &\to C_2 + C_4 \\
+	C_4 &\to C_4 + C_3 \\
+	C_4 &\to  C_4 - 2 C_1 \\
+	C_3 &\to C_3 - C_2 \\
+	C_2 &\to C_2 - C_1 \\
+	R_2 &\to \frac{R_2}{2} \\
+	R_1 &\leftrightharpoons R_4 \\
+	C_4 &\leftrightharpoons C_1 \\
+	R_4 &\to R_4 - R_2
+\end{align*}
+
+We get
+\[
+	\begin{bmatrix}
+		0 & 1 & 0 & 0 \\
+		0 & 0 & 1 & 0 \\
+		0 & 0 & 0 & 1 \\
+		0 & 0 & 0 & 0
+	\end{bmatrix}
+\]
+
+\[
+	rank = 3
+\]
+
+\section*{Question 5}
+Solve
+\begin{align*}
+	x + y + z &= 9 \\
+	2x + 5y + 7z &= 52 \\
+	2x + y - z &= 0
+\end{align*}
+\subsection*{Solution}
+
+\begin{align*}
+	\Delta &=
+	\begin{vmatrix}
+		1 & 1 & 1 \\
+		2 & 5 & 7 \\
+		2 & 1 & -1
+	\end{vmatrix}
+	= -4
+	\\
+	\Delta_1 &=
+	\begin{vmatrix}
+		9 & 1 & 1 \\
+		52 & 5 & 7 \\
+		0 & 1 & -1
+	\end{vmatrix}
+	= -4
+	\\
+	\Delta_2 &=
+	\begin{vmatrix}
+		1 & 9 & 1 \\
+		2 & 52 & 7 \\
+		2 & 0 & -1
+	\end{vmatrix}
+	= -12
+	\\
+	\Delta_3 &=
+	\begin{vmatrix}
+		1 & 1 & 9 \\
+		2 & 5 & 52 \\
+		2 & 1 & 0
+	\end{vmatrix}
+	= -20
+	\\
+	x &= \frac{\Delta_1}{\Delta} = 1 \\
+	y &= \frac{\Delta_2}{\Delta} = 3 \\
+	z &= \frac{\Delta_3}{\Delta} = 5
+\end{align*}
+\section*{Question 6}
+Test the consistency of:
+
+\begin{align*}
+	3x - y + 2z &= 3 \\
+	2x + y + 3z &= 5 \\
+	x - 2y - z &= 1
+\end{align*}
+\subsection*{Solution}
+\[
+	\text{Augemented matrix} =
+\begin{bmatrix}
+3 & -1 & 2 & : & 3 \\
+2 & 1 & 3 & : & 5 \\
+1 & -2 & -1 & : & 1
+\end{bmatrix}
+\]
+\[
+\Delta =
+\begin{vmatrix}
+3 & -1 & 2 \\
+2 & 1 & 3 \\
+1 & -2 & -1
+\end{vmatrix}
+= 10
+\]
+\[
+	\Delta \ne 0 \therefore \text{it is consistent with the unique solution}
+\]
+\section*{Question 7}
+Solve the equations
+\begin{align*}
+	x + 3y - 2z &= 0 \\
+	2x -y + 4z &= 0 \\
+	x - 11y + 14z &= 0
+\end{align*}
+\subsection*{Solution}
+
+\begin{align*}
+	\Delta =\begin{vmatrix}
+	1 & 3 & -2 \\
+	2 & -1 & 4 \\
+	1 & -11 & 14
+	\end{vmatrix}
+	&= 0 \\
+	\Delta_1 = \Delta_2 = \Delta_3 &= 0 \\
+	\therefore \text{unique solution is } x = y = z &= 0
+\end{align*}
+
+
+\end{document}