diff --git a/maths/assignment-matrices/assign2.tex b/maths/assignment-matrices/assign2.tex new file mode 100644 index 0000000..d8b030a --- /dev/null +++ b/maths/assignment-matrices/assign2.tex @@ -0,0 +1,323 @@ +\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}