202 lines
3.5 KiB
TeX
202 lines
3.5 KiB
TeX
\documentclass{article}
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% Import for matrices
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\usepackage{amsmath}
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% Import for therefore symbol
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\usepackage{amssymb}
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\begin{document}
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\title{Mathematics Assignment --- Matrices}
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\author{Ahmad Saalim Lone, 2019BCSE017}
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\date{05 May, 2020}
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\maketitle
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\section{Question 1}
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\begin{equation}
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A =
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\begin{bmatrix}
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2 & -5 & -1 \\
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-2 & -1 & 4
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\end{bmatrix}
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B =
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\begin{bmatrix}
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3 & 4 & 0 \\
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5 & -2 & 3
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\end{bmatrix}
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\end{equation}
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\subsection{Part i}
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\begin{equation}
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A + B =
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\begin{bmatrix}
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2 + 1 & -5 + 4 & -1 + 0 \\
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-2 + 5 & -1 -2 & 4 + 3
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\end{bmatrix}
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\end{equation}
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\begin{equation}
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A + B =
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\begin{bmatrix}
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3 & -1 & -1 \\
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3 & -3 & 7
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\end{bmatrix}
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\end{equation}
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\subsection{Part ii}
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\begin{equation}
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2A + B =
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\begin{bmatrix}
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4 + 1 & -10 + 4 & -2 + 0 \\
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-4 + 5 & -2 -2 & 8 + 3
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\end{bmatrix}
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\end{equation}
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\begin{equation}
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2A + B =
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\begin{bmatrix}
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5 & -6 & -2 \\
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1 & -4 & 11
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\end{bmatrix}
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\end{equation}
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\section{Question 2}
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\begin{equation}
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A =
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\begin{bmatrix}
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1 & 2 & 3 \\
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4 & 5 & 6 \\
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7 & 8 & 9
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\end{bmatrix}
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B =
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\begin{bmatrix}
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1 & 0 & 2 \\
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2 & 1 & 2 \\
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5 & 2 & 3
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\end{bmatrix}
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\end{equation}
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\begin{equation}
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AB =
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\begin{bmatrix}
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1 & 2 & 3 \\
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4 & 5 & 6 \\
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7 & 8 & 9
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\end{bmatrix}
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\begin{bmatrix}
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1 & 0 & 2 \\
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2 & 1 & 2 \\
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5 & 2 & 3
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\end{bmatrix}
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\end{equation}
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\begin{equation}
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AB =
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\begin{bmatrix}
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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 \\
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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 \\
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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
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\end{bmatrix}
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\end{equation}
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\begin{equation}
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AB =
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\begin{bmatrix}
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20 & 8 & 16 \\
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44 & 17 & 36 \\
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68 & 26 & 57
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\end{bmatrix}
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\end{equation}
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\section{Question 3}
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If \(
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A =
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\begin{bmatrix}
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1 & -2 & -3 \\
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-4 & 2 & 5
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\end{bmatrix}
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B =
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\begin{bmatrix}
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2 & 3 \\
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4 & 5 \\
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2 & 1
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\end{bmatrix}
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\), show that \(AB \ne BA\).
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Order of $A$ = $2\times3$
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Order of $B$ = $3\times2$
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Order of $AB$ = $rows \; of \; A \times columns \; of \; B$ = $2\times2$
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Order of $BA$ = $rows \; of \; B \times columns \; of \; A$ = $3\times3$
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Matrices of different order can't be equal.
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$\therefore AB \ne BA$
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\section{Question 4}
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Show that \(
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A =
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\begin{bmatrix}
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3 & 1 + 2i \\
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1-2i & 2
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\end{bmatrix}
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\) is a hermitian.
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For a matrix to be hermitian, each element $a_{i,j}$ needs to be the complex
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conjugate of the element at $a_{j,i}$. In given matrix, we have
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\begin{itemize}
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\item \(a_{11} = 3\)
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\item \(a_{12} = 1 + 2i\)
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\item \(a_{21} = 1 - 2i\)
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\item \(a_{22} = 2\)
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\end{itemize}
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The conjugates are as follows
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\begin{itemize}
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\item \(\overline{a_{11}} = 3\)
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\item \(\overline{a_{12}} = 1 - 2i\)
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\item \(\overline{a_{21}} = 1 + 2i\)
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\item \(\overline{a_{22}} = 2\)
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\end{itemize}
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As we can see, \(\overline{a_{11}} = a_{11}\), \(\overline{a_{12}} = a_{21}\), \(\overline{a_{21}} = a_{12}\) and \(\overline{a_{22}} = a_{22}\).
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$\therefore A$ is hermitian.
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\section{Question 5}
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If \(
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A =
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\begin{bmatrix}
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5 & 1 + i \\
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-1 + i & 4
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\end{bmatrix}
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\), show that ${(A^{\theta})}^{\theta}$
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\[
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A =
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\begin{bmatrix}
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5 & 1 + i \\
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-1 + i & 4
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\end{bmatrix}
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\]
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\[
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\overline{A^{\theta}} =
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\begin{bmatrix}
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5 & - 1 - i \\
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1 + i & 4
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\end{bmatrix}
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\]
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\[
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{(A^{\theta})}^{\theta} =
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\begin{bmatrix}
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5 & 1 + i \\
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-1 + i & 4
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\end{bmatrix}
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\]
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\[
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\therefore {(A^{\theta})}^{\theta} = A
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\]
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\end{document}
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