Solve Q 3, 4, 5, 6

maths/assignment-matrices/assign.tex

@@ -1,6 +1,8 @@
 \documentclass{article}
 % Import for matrices
 \usepackage{amsmath}
+% Import for therefore symbol
+\usepackage{amssymb}
 \begin{document}
 \title{Mathematics Assignment --- Matrices}
 \author{Ahmad Saalim Lone, 2019BCSE017}
@@ -50,6 +52,8 @@
 		1 & -4 & 11
 	\end{bmatrix}
 \end{equation}
+
+
 \section{Question 2}
 \begin{equation}
 	A =
@@ -81,9 +85,9 @@
 \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}
@@ -94,4 +98,104 @@
 		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}