\documentclass{article}
% Import for matrices
\usepackage{amsmath}
% Import for therefore symbol
\usepackage{amssymb}
\begin{document}
\title{Mathematics Assignment --- Matrices}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{05 May, 2020}
\maketitle
\section{Question 1}
If
\(
A =
\begin{bmatrix}
	2 & -5 & -1 \\
	-2 & -1 & 4
\end{bmatrix}
B =
\begin{bmatrix}
	3 & 4 & 0 \\
	5 & -2 & 3
\end{bmatrix}
\) find:

\begin{enumerate}
	\item $A + B$
	\item $2A + B$
\end{enumerate}

\subsection{Part 1}
\[
	A + B =
	\begin{bmatrix}
		2 + 1 & -5 + 4 & -1 + 0 \\
		-2 + 5 & -1 -2 & 4 + 3
	\end{bmatrix}
\]
\[
	A + B =
	\begin{bmatrix}
		3 & -1 & -1 \\
		3 & -3 & 7
	\end{bmatrix}
\]
\subsection{Part 2}
\[
	2A + B =
	\begin{bmatrix}
		4 + 1 & -10 + 4 & -2 + 0 \\
		-4 + 5 & -2 -2 & 8 + 3
	\end{bmatrix}
\]
\[
	2A + B =
	\begin{bmatrix}
		5 & -6 & -2 \\
		1 & -4 & 11
	\end{bmatrix}
\]


\section{Question 2}
If \(
A =
\begin{bmatrix}
	1 & 2 & 3 \\
	4 & 5 & 6 \\
	7 & 8 & 9
\end{bmatrix}
B =
\begin{bmatrix}
	1 & 0 & 2 \\
	2 & 1 & 2 \\
	5 & 2 & 3
\end{bmatrix}
\), find $AB$.
\[
	AB =
	\begin{bmatrix}
		1 & 2 & 3 \\
		4 & 5 & 6 \\
		7 & 8 & 9
	\end{bmatrix}
	\begin{bmatrix}
		1 & 0 & 2 \\
		2 & 1 & 2 \\
		5 & 2 & 3
	\end{bmatrix}
\]
\[
	AB =
	\begin{bmatrix}
		1 \times 1 + 2 \times 2 + 3 \times 5 & 1 \times 0 + 2 \times 1 + 3 \times 2 & 1 \times 2 + 2 \times 2 + 3 \times 3 \\
		4 \times 1 + 5 \times 2 + 6 \times 5 & 4 \times 0 + 5 \times 1 + 6 \times 2 & 4 \times 2 + 5 \times 2 + 6 \times 3 \\
		7 \times 1 + 8 \times 2 + 9 \times 5 & 7 \times 0 + 8 \times 1 + 9 \times 2 & 7 \times 2 + 8 \times 2 + 9 \times 3
	\end{bmatrix}
\]
\[
	AB =
	\begin{bmatrix}
		20 & 8 & 16 \\
		44 & 17 & 36 \\
		68 & 26 & 57
	\end{bmatrix}
\]


\section{Question 3}
If \(
A =
\begin{bmatrix}
	1 & -2 & -3 \\
	-4 & 2 & 5
\end{bmatrix}
B =
\begin{bmatrix}
	2 & 3 \\
	4 & 5 \\
	2 & 1
\end{bmatrix}
\), show that \(AB \ne BA\).

Order of $A$ = $2\times3$

Order of $B$ = $3\times2$

Order of $AB$ = $rows \; of \; A \times columns \; of \; B$ = $2\times2$

Order of $BA$ = $rows \; of \; B \times columns \; of \; A$ = $3\times3$

Matrices of different order can't be equal.

$\therefore AB \ne BA$

\section{Question 4}

Show that \(
A =
\begin{bmatrix}
	3 & 1 + 2i \\
	1-2i & 2
\end{bmatrix}
\) is a hermitian.

For a matrix to be hermitian, each element $a_{i,j}$ needs to be the complex
conjugate of the element at $a_{j,i}$. In given matrix, we have

\begin{itemize}
	\item \(a_{11} = 3\)
	\item \(a_{12} = 1 + 2i\)
	\item \(a_{21} = 1 - 2i\)
	\item \(a_{22} = 2\)
\end{itemize}

The conjugates are as follows

\begin{itemize}
	\item \(\overline{a_{11}} = 3\)
	\item \(\overline{a_{12}} = 1 - 2i\)
	\item \(\overline{a_{21}} = 1 + 2i\)
	\item \(\overline{a_{22}} = 2\)
\end{itemize}

As we can see, \(\overline{a_{11}} = a_{11}\), \(\overline{a_{12}} = a_{21}\), \(\overline{a_{21}} = a_{12}\) and \(\overline{a_{22}} = a_{22}\).

$\therefore A$ is hermitian.

\section{Question 5}

If \(
A =
\begin{bmatrix}
	5 & 1 + i \\
	-1 + i & 4
\end{bmatrix}
\), show that ${(A^{\theta})}^{\theta}$

\[
	A =
	\begin{bmatrix}
		5 & 1 + i \\
		-1 + i & 4
	\end{bmatrix}
\]

\[
	\overline{A^{\theta}} =
	\begin{bmatrix}
		5 &  - 1 - i \\
		1 + i & 4
	\end{bmatrix}
\]


\[
	{(A^{\theta})}^{\theta} =
	\begin{bmatrix}
		5 & 1 + i \\
		-1 + i & 4
	\end{bmatrix}
\]

\[
	\therefore {(A^{\theta})}^{\theta} = A
\]

\section{Question 6}

$ A =
\begin{bmatrix}
	1 & 0 & -1 \\
	3 & 4 & 5 \\
	0 & -6 & -7
\end{bmatrix}
$, find $ adj. A$

Adjoint of a matrix is the transpose of the cofactor matrix of the original matrix

\[
	Cofactor\;of\;A_{11} =
	\begin{vmatrix}
		4 & 5 \\
		-6 & -7
	\end{vmatrix}
	= 2
\]
\[
	Cofactor\;of\;A_{12} =
	\begin{vmatrix}
		3 & 5 \\
		0 & -7
	\end{vmatrix}
	= -21
\]
\[
	Cofactor\;of\;A_{13} =
	\begin{vmatrix}
		3 & 4 \\
		0 & -6
	\end{vmatrix}
	= -18
\]
\[
	Cofactor\;of\;A_{21} =
	\begin{vmatrix}
		0 & -1 \\
		-6 & -7
	\end{vmatrix}
	= -6
\]
\[
	Cofactor\;of\;A_{22} =
	\begin{vmatrix}
		1 & -1 \\
		0 & -7
	\end{vmatrix}
	= -7
\]
\[
	Cofactor\;of\;A_{23} =
	\begin{vmatrix}
		1 & -1 \\
		0 & -6
	\end{vmatrix}
	= -6
\]
\[
	Cofactor\;of\;A_{31} =
	\begin{vmatrix}
		0 & -1 \\
		4 & 5
	\end{vmatrix}
	= 4
\]
\[
	Cofactor\;of\;A_{32} =
	\begin{vmatrix}
		1 & -1 \\
		3 & 5
	\end{vmatrix}
	= 8
\]
\[
	Cofactor\;of\;A_{33} =
	\begin{vmatrix}
		1 & 0 \\
		3 & 4
	\end{vmatrix}
	= 4
\]

\[
	Cofactor\;Matrix =
	\begin{bmatrix}
		2 & -21 & -18 \\
		-6 & -7 & -6 \\
		4 & 8 & 4
	\end{bmatrix}
\]

Adjoint matrix is the transpose of Cofactor Matrix.
\[
	\therefore adj.A =
	\begin{bmatrix}
		2 & -6 & 4 \\
		-21 & -7 & 8 \\
		-18 & -6 & 4
	\end{bmatrix}
\]

\section{Question 7}

\( A =
\begin{bmatrix}
	0 & 0 & 1 \\
	0 & 1 & 0 \\
	1 & 0 & 0
\end{bmatrix}
\), show that $A^{-1} = A$.

We know that
\[
	A^{-1} = \frac{adj.(A)}{|A|}
\]

\begin{align*}
	Cofactor\;of\;A_{11} &= 0 \\
	Cofactor\;of\;A_{12} &= 0 \\
	Cofactor\;of\;A_{13} &= -1 \\
	Cofactor\;of\;A_{21} &= 0 \\
	Cofactor\;of\;A_{22} &= -1 \\
	Cofactor\;of\;A_{23} &= 0 \\
	Cofactor\;of\;A_{31} &= -1 \\
	Cofactor\;of\;A_{32} &= 0 \\
	Cofactor\;of\;A_{33} &= 0
\end{align*}

\[
	Cofactor\;Matrix =
	\begin{bmatrix}
		0 & 0 & -1 \\
		0 & -1 & 0 \\
		-1 & 0 & 0
	\end{bmatrix}
\]
\[
	adj.A =
	\begin{bmatrix}
		0 & 0 & -1 \\
		0 & -1 & 0 \\
		-1 & 0 & 0
	\end{bmatrix}
\]

\[
	|A| =
	\begin{vmatrix}
		0 & 0 & 1 \\
		0 & 1 & 0 \\
		1 & 0 & 0
	\end{vmatrix}
	= 1 \times
	\begin{vmatrix}
		0 & 1 \\
		1 & 0
	\end{vmatrix}
	= 1 \times -1 = -1
\]

\[
	A^{-1} = \frac{adj.A}{|A|}
\]
\[
	A^{-1} =
	\begin{bmatrix}
		0 & 0 & 1 \\
		0 & 1 & 0 \\
		1 & 0 & 0
	\end{bmatrix}
\]
\[
	\therefore A^{-1} = A
\]
\end{document}