Solve Q 3, 4, 5, 6

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\documentclass{article}
% Import for matrices
\usepackage{amsmath}
% Import for therefore symbol
\usepackage{amssymb}
\begin{document}
\title{Mathematics Assignment --- Matrices}
\author{Ahmad Saalim Lone, 2019BCSE017}
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1 & -4 & 11
\end{bmatrix}
\end{equation}
\section{Question 2}
\begin{equation}
A =
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\begin{equation}
AB =
\begin{bmatrix}
1 * 1 + 2 * 2 + 3 * 5 & 1 * 0 + 2 * 1 + 3 * 2 & 1 * 2 + 2 * 2 + 3 * 3 \\
4 * 1 + 5 * 2 + 6 * 5 & 4 * 0 + 5 * 1 + 6 * 2 & 4 * 2 + 5 * 2 + 6 * 3 \\
7 * 1 + 8 * 2 + 9 * 5 & 7 * 0 + 8 * 1 + 9 * 2 & 7 * 2 + 8 * 2 + 9 * 3
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}
\end{equation}
\begin{equation}
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68 & 26 & 57
\end{bmatrix}
\end{equation}
\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 ${(A^{\theta})}^{\theta}$
\[
A =
\begin{bmatrix}
5 & 1 + i \\
-1 + i & 4
\end{bmatrix}
\]
\[
\overline{A^{\theta}} =
\begin{bmatrix}
5 & - 1 - i \\
1 + i & 4
\end{bmatrix}
\]
\[
{(A^{\theta})}^{\theta} =
\begin{bmatrix}
5 & 1 + i \\
-1 + i & 4
\end{bmatrix}
\]
\[
\therefore {(A^{\theta})}^{\theta} = A
\]
\end{document}