From f2741293b7a9f93864bfb53d2f4f4cdd80cf2670 Mon Sep 17 00:00:00 2001 From: Ceda EI Date: Sun, 17 May 2020 18:50:08 +0530 Subject: [PATCH] Add maths assignment 4 --- maths/assignment-matrices/assign4.tex | 601 ++++++++++++++++++++++++++ 1 file changed, 601 insertions(+) create mode 100644 maths/assignment-matrices/assign4.tex diff --git a/maths/assignment-matrices/assign4.tex b/maths/assignment-matrices/assign4.tex new file mode 100644 index 0000000..fc554df --- /dev/null +++ b/maths/assignment-matrices/assign4.tex @@ -0,0 +1,601 @@ +\documentclass{article} +\usepackage{amsmath} +\usepackage{amssymb} +\begin{document} +\title{Mathematics Assignment 4 --- Matrices} +\author{Ahmad Saalim Lone, 2019BCSE017} +\date{16 May, 2020} +\maketitle + +\section*{Question 1} +\begin{align*} + A &= + \begin{bmatrix} + 1 & -1 & -1 & 2 \\ + 4 & 2 & 2 & -1 \\ + 2 & 2 & 0 & -2 + \end{bmatrix} + \\ + PAQ &= \text{Normal form} \\ + I_3AI_4 &= A_{3 \times 4} \\ + \begin{bmatrix} + 1 & 0 & 0 \\ + 0 & 1 & 0 \\ + 0 & 0 & 1 + \end{bmatrix} + A + \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{bmatrix} + &= + \begin{bmatrix} + 1 & -1 & -1 & 2 \\ + 4 & 2 & 2 & -1 \\ + 2 & 2 & 0 & -2 + \end{bmatrix} +\end{align*} + +Applying the following row tranformations on $I_3$ and column tranformations on $I_4$. + +\begin{align*} + C_4 &\to C_4 + C_2 \\ + R_3 &\to \frac{R_3}{2} \\ + R_2 &\to \frac{R_2 + 2R_1}{3} \\ + R_1 &\to R_1 + R_3 \\ + C_1 &\to C_1 - 2 C_4 \\ + C_4 &\to C_4 + C_3 \\ + C_2 &\to C_2 - C_1 \\ + C_3 &\rightleftharpoons C_1 \\ + C_2 &\rightleftharpoons C_4 \\ + R_1 &\rightleftharpoons -R_1 +\end{align*} + +we get +\begin{align*} + P &= + \begin{bmatrix} + -1 & 0 & -\frac{1}{2} \\[6pt] + \frac{2}{3} & \frac{1}{3} & 0 \\[6pt] + 0 & 0 & \frac{1}{2} + \end{bmatrix} + \\ + Q &= + \begin{bmatrix} + 0 & 0 & 1 & -1 \\ + 0 & 0 & -2 & 3 \\ + 1 & 1 & 0 & 0 \\ + 0 & 1 & -2 & 2 + \end{bmatrix} + \\ + \text{Normal Form of A} &= + \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 + \end{bmatrix} +\end{align*} +\section*{Question 2} +\[ + x^2yz = xy^2z^3 = x^3y^2z = e +\] + +Taking $\ln$ on both sides +\begin{align*} + 2\ln{x} + \ln{y} \ln{z} &= 1 \\ + \ln{x} + 2 \ln{y} + 3 \ln{z} &= 1 \\ + 3\ln{x} + 2\ln{y} + \ln{z} &= 1 +\end{align*} + +\[ + \text{Augemented Matrix} = [A:B] = +\begin{bmatrix} +2 & 1 & 1 & : & 1 \\ +1 & 2 & 3 & : & 1 \\ +3 & 2 & 1 & : & 1 +\end{bmatrix} +\] +After ppling the following tranformations +\begin{align*} + R_3 &\to R_3 - R_1 \\ + R_2 &\to R_2 - R_1 \\ + R_2 &\to R_2 - R_1 \\ + R_1 &\to R_1 -R_3 +\end{align*} + +We get +\[ +\begin{bmatrix} +1 & 0 & 1 & : & 1 \\ +-3 & 0 & 1 & : & -1 \\ +1 & 1 & 0 & : & 0 +\end{bmatrix} +\] +$Rank(A) = Rank(A:B) = 3 = n$\\ +$\therefore$ unique solution. +\section*{Question 3} +\begin{align*} + x - cy - bz &= 0 \\ + cx - y + az &= 0 \\ + bx + ay - z &= 0 \\ + \text{Augemented matrix} &= +\begin{bmatrix} +1 & -c & -b & : & 0 \\ +c & -1 & a & : & 0 \\ +b & a & -1 & : & 0 +\end{bmatrix} +\end{align*} + +Applying the following tranformations +\begin{align*} + R_2 &\to R_2 - cR_1 \\ + R_3 &\to R_3 - bR_1 \\ + R_2 &\to \frac{R_2}{c^2 - 1} \\ + R_3 &\to R_3 - (a + bc)R_2 +\end{align*} +we get +\[ + C = +\begin{bmatrix} + 1 & -c & -b & : & 0 \\[6pt] + 0 & 1 & \frac{bc+c}{c^2 - 1} & : & 0 \\[6pt] +0 & 0 & b^2 - 1 - \frac{{(a+bc)}^2}{c^2 - 1} & : & 0 +\end{bmatrix} +\] + +For non trivial solutions $Rank(A) = Rank(c) \ne number\;of\;unknows$ +\begin{align*} + b^2 - 1 \frac{{(a+bc)}^2}{c^2 -1} &= 0 \\ + \implies a^2 + b^2 + c^2 + 2abc &= 1 \\ +\end{align*} +Now +\begin{align*} + x - cy -bz &= 0 \\ + y + \left(\frac{bc + a}{c^2 - 1}z\right) &= 0 \\ +\end{align*} +We get +\begin{align*} + x &= \frac{ac + b}{1-c^2}z \\ + y &= \frac{bc + a}{1 - c^2}z \\ + z &= z \\ + x : y : z &= \sqrt{|1 - a^2|} : \sqrt{|1 - b^2|} : \sqrt{|1 -c^2|} +\end{align*} + +\section{Question 4} +\begin{align*} + A &= +\begin{bmatrix} +3 & -4 & 4 \\ +1 & -2 & 44 \\ +1 & -1 & 3 +\end{bmatrix} \\ + |A - \lambda I| &= 0\\ +\begin{vmatrix} +3 - \lambda & -4 & 4 \\ +1 & -2 - \lambda & 44 \\ +1 & -1 & 3 - \lambda +\end{vmatrix} &= 0 +\end{align*} + +\( \lambda = -1, 2, 3 \) are Eigen Values + +For $\lambda = -1$ +\begin{align*} + \begin{bmatrix} + 4 & -4 & 4 \\ + 1 & -1 & 4 \\ + 1 & -1 & 4 + \end{bmatrix} + \begin{bmatrix} + x \\ + y \\ + z + \end{bmatrix} +&= +\begin{bmatrix} + 0 \\ + 0 \\ + 0 +\end{bmatrix} \\ + \text{Eigen Vector} &= + \begin{bmatrix} + 1 \\ + 1 \\ + 0 + \end{bmatrix} +\end{align*} +For $\lambda = 2$ +\begin{align*} +\begin{bmatrix} +1 & -4 & 4 \\ +1 & -4 & 4 \\ +1 & -1 & 1 +\end{bmatrix} + \begin{bmatrix} + x \\ + y \\ + z + \end{bmatrix} +&= +\begin{bmatrix} + 0 \\ + 0 \\ + 0 +\end{bmatrix} \\ + \text{Eigen Vector} &= + \begin{bmatrix} + 0 \\ + 1 \\ + 1 + \end{bmatrix} +\end{align*} +For $\lambda = 3$ +\begin{align*} +\begin{bmatrix} +0 & -4 & 4 \\ +1 & -5 & 4 \\ +1 & -1 & 0 +\end{bmatrix} + \begin{bmatrix} + x \\ + y \\ + z + \end{bmatrix} +&= +\begin{bmatrix} + 0 \\ + 0 \\ + 0 +\end{bmatrix} \\ + \text{Eigen Vector} &= + \begin{bmatrix} + 1 \\ + 1 \\ + 1 + \end{bmatrix} +\end{align*} +\section*{Question 5} +\begin{align*} + A &= +\begin{bmatrix} +2 & 2 & 0 \\ +2 & 1 & 1 \\ +-7 & 2 & -3 +\end{bmatrix} +\\ + |A - \lambda I| &= 0 \\ +\begin{vmatrix} +2 - \lambda & 2 & 0 \\ +2 & 1 - \lambda & 1 \\ +-7 & 2 & -3 - \lambda +\end{vmatrix} &= 0 +\end{align*} + +Eigen values $\lambda = 1, 3, -4$ \\ + +1\textsuperscript{st} eigen value of $A^2 -2 A + I = 1^2 - 2(1) + 1 = 0$ + +2\textsuperscript{nd} eigen value of $A^2 -2 A + I = 3^2 - 2(3) + 1 = 4$ + +3\textsuperscript{rd} eigen value of $A^2 -2 A + I = {(-4)}^2 - 2(-4) + 1 = 25$ + +\section*{Question 6} +\begin{align*} + A &= +\begin{bmatrix} +7 & 3 \\ +2 & 6 +\end{bmatrix} +\\ + | A - \lambda I| &= 0 \\ +\begin{vmatrix} + 7 - \lambda & 3 \\ + 2 & 6 - \lambda +\end{vmatrix} &= 0 \\ + (7 - \lambda)(6 - \lambda) - 5 &= 0 \\ + (\lambda - 4)(\lambda - 9) &= 0 \\ + A^2 - 13 A + 36 &= 0 \\ + A^2 &= 13 A - 36 \\ + A^2\cdot A &= (13 A - 36)A \\ + A^3 &= 13 A^2 - 36A \\ + A^3 &= +\begin{bmatrix} +715 & 507 \\ +338 & 546 +\end{bmatrix} +- +\begin{bmatrix} +252 & 108 \\ +72 & 216 +\end{bmatrix} \\ + A^3 &= +\begin{bmatrix} +463 & 399 \\ +266 & 330 +\end{bmatrix} +\end{align*} + +\section*{Question 7} +Suppose that $\lambda$ is a (possibly complex) eigen value of the real symmetric matrix $A$. Thus, there is a non-zero vector $V$, also with complex entries such that $AV = \lambda V$. By taking the complex conjugate of both sides and noting that $\overline{A} = A$ since $A$ has real entries, we get $\overline{AV} = \overline{\lambda V} \implies A\overline{V} = \overline{\lambda}\;\overline{V}$. Then using that $A^T = A$, +\begin{gather*} + \overline{V}^T AV = \overline{V}^t(AV) = \overline{V}(\lambda V) = \lambda( \overline{V}V ) \\ + \overline{V}^T AV = {(A \overline{V})}^T V = {( \overline{\lambda}\;\overline{V} )}^T V = \overline{\lambda} ( \overline{V} V ) +\end{gather*} + +Since $V \ne 0$, we have $ \overline{V}V \ne 0$. Thus $\lambda = \overline{\lambda}$, which means $\lambda \in R$. + +\section*{Question 8} +Quadratic Form $ax_1^2 + cx_2^2 - 2bx_1x_2$ +\[ +\begin{bmatrix} +a & -b \\ +-b & c +\end{bmatrix} +\] +Convert it to diagonal matrix by applying +\begin{align*} + R_2 &\to R_2 + \frac{b}{a}R_1 \\ + C_2 &\to C_2 + \frac{b}{a}C_1 +\end{align*} +Now, we get +\[ +\begin{bmatrix} +a & 0 \\ +0 & c - \frac{b^2}{a} +\end{bmatrix} +\] + +Nature $\to$ positive definite $\to$ when $rank(r) = index(s)$ or when all eigen values are positive i.e. $a > 0$ \& $c - \frac{b^2}{a} > 0 \implies ac -b^2 > 0$. Hence proved. + +\section*{Question 9} +\( +A = +\begin{bmatrix} + \lambda & 1 & 1 \\ +1 & \lambda & -1 \\ +1 & -1 & \lambda +\end{bmatrix} +\) is a symmetric matrix obtained when compared to Quadratic form $\lambda(x^2+ y^2 + z^2) + 2xy + 2zx -2yz$. + +Now, convert $A$ into diagonal matrix by: +\begin{align*} + C_1 &\to C_1 + C_3 \\ + C_1 &\to \frac{C_1}{\lambda + 1} \\ + R_3 &\to R_3 - R_1 \\ + C_3 &\to C_3 - C_1 \\ + C_2 &\to C_2 -C_1 \\ + C_3 &\to C_3 + \frac{C_2}{\lambda} \\ + R_3 &\to R_3 + \frac{2R_2}{\lambda} +\end{align*} + +we get, + +\[ +\begin{bmatrix} +1 & 0 & 0 \\ +0 & \lambda & 0 \\ +0 & 0 & \lambda - \frac{2}{\lambda} - 1 +\end{bmatrix} +\] +For definite positive nature, all Eigen values must be positive i.e. $\lambda > 0$, $\lambda - \frac{2}{\lambda} - 1 > 0$. Taking intersection of these two, we get $\lambda \in (2, \infty)$. + +\section*{Question 10} + +Multiplication of all the eigen values = determinant of the matrix. For singular matrix, determinant value = 0. +\begin{align*} + \text{Eigen Values} &= 2, 3, a \\ + 6a &= 0 \\ + a &= 0 +\end{align*} + +\section*{Question 11} + +Quadratic Form $x_2^2 + 2x_2^2 - 5x_3^2$ +\[ +\begin{bmatrix} +1 & 0 & 0 \\ +0 & 2 & 0 \\ +0 & 0 & -5 +\end{bmatrix} +\] +\begin{align*} + index(s) &= 2\;\text{(No of positive terms)} \\ + rank(r) &= 3 \\ + signature &= 2s -r = 4 - 3 = 1 +\end{align*} + +\section*{Question 12} + +Quadratic form $ax^2 + 2bcy + cy^2$. +\[ +\begin{bmatrix} +a & b \\ +b & c +\end{bmatrix} +\] +Convert it into diagonal matrix by doing: +\begin{align*} + R_2 &\to R_2 - \left(\frac{b}{a}\right)R_1 \\ + C_2 &\to C_2 - \left(\frac{b}{a}\right)C_1 +\end{align*} + +Finally, we get +\[ +\begin{bmatrix} +a & 0 \\ +0 & c - \frac{b^2}{a} +\end{bmatrix} +\] +For positive definite, $a>0$ and $ac -b^2 > 0$. \\ +For negative definite, $a<0$ and $ac -b^2 > 0$. \\ +Roots of the quadratic equation ($ax^2 + 2bx + c = 0$) are imaginary when $D < 0$. \\ +${(2b)}^2 - 4ac = 4( b^2 - ac )$ is always negative + +\section*{Question 13} +\begin{align*} + X_1 &= +\begin{bmatrix} +1 \\ +2 \\ +-3 \\ +4 +\end{bmatrix} +\\ + X_2 &= +\begin{bmatrix} +1 \\ +-5 \\ +8 \\ +-7 +\end{bmatrix} +\\ + X_3 &= +\begin{bmatrix} +1 \\ +-5 \\ +8 \\ +-7 +\end{bmatrix} +\\ + \lambda_1 X_1 + \lambda_2 X_2 + \lambda_3 X_3 &= 0 \\ +\begin{bmatrix} +1 & 1 & 1 \\ +2 & -5 & -5 \\ +-3 & 8 & 8 \\ +4 & -7 & -7 +\end{bmatrix} +\begin{bmatrix} +\lambda_1 \\ +\lambda_2 \\ +\lambda_3 +\end{bmatrix} &= +\begin{bmatrix} +0 \\ +0 \\ +0 \\ +0 +\end{bmatrix} +\end{align*} +Applying the following tranformations +\begin{align*} + R_2 &\to R_2 - 2R_1 \\ + R_2 &\to -\frac{R_2}{7} \\ + R_3 &\to R_3 + R_4 \\ + R_3 &\to R_3 - R_1 \\ + R_1 &\to R_1 - R_2 \\ + R_3 &\to R_3 - 4 R_1 \\ + R_4 &\to -\frac{R_4}{7} \\ + R_4 &\to R_4 - R_2 +\end{align*} +we get +\begin{align*} +\begin{bmatrix} +1 & 2 & 0 \\ +0 & 1 & 1 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{bmatrix} +\begin{bmatrix} +\lambda_1 \\ +\lambda_2 \\ +\lambda_3 +\end{bmatrix} +&= +\begin{bmatrix} +0 \\ +0 \\ +0 \\ +0 +\end{bmatrix} \\ + \lambda_1 + 2\lambda_2 &= 0 \\ + \lambda_2 + \lambda_3 &= 0 + \therefore \\ + \lambda_1 &= t \\ + \lambda_2 &= -\frac{t}{2} \\ + \lambda_3 &= \frac{t}{2} +\end{align*} +Putting the values, we get + +\[ + 2 X_1 - X_2 + X_3 = 0 +\] + +\section*{Question 14} + +\begin{align*} + (2 - \lambda)x_1 + (-2)x_2 + x_3 &= 0 \\ + 2x_1 - (\lambda + 3) x_2 + 2x_3 &= 0 \\ + -x_1 + 2x_2 - \lambda x_3 &= 0 +\end{align*} + +\( + Rank(A) = Rank(\text{augemented matrix}) < 3 +\) for non trivial solutions. + +Check the determinant first, $\Delta = 0$ + +$\Delta = 0$ gets us $\lambda = 1, 3$ + +Now, for augemented matrix $[A:B]$, put $\lambda = 1, -3$ in \( +\begin{bmatrix} +2-\lambda & -2 & 1 & : & 0 \\ +2 & -(\lambda + 3) & 2 & : & 0 \\ +-1 & 2 & -\lambda & : & 0 +\end{bmatrix} +\) + +For $\lambda = 1$ +\[ +\begin{bmatrix} +1 & -2 & 1 & : & 0 \\ +2 & -4 & 2 & : & 0 \\ +-1 & 2 & -1 & : & 0 +\end{bmatrix} +\] + +Doing transformations to form row echelon form, we get +\[ +\begin{bmatrix} +1 & -2 & 1 & : & 0 \\ +0 & 0 & 0 & : & 0 \\ +0 & 0 & 0 & : & 0 +\end{bmatrix} +\] +\begin{align*} + x_1 -2x_2 + x_3 &= 0 \\ + if\;x_2 = k,\;x_3 = t, then\;x_1 &= 2k - t +\end{align*} +For $\lambda = -3$ +\[ +\begin{bmatrix} +5 & -2 & 1 & : & 0 \\ +2 & 0 & 2 & : & 0 \\ +-1 & 2 & 3 & : & 0 +\end{bmatrix} +\] + +Doing transformations to form row echelon form, we get +\[ +\begin{bmatrix} +5 & -2 & 1 & : & 0 \\ +0 & \frac{4}{5} & \frac{8}{5} & : & 0 \\ +0 & 0 & 0 & 0 & 0 +\end{bmatrix} +\] +\begin{align*} + 5x_1 - 2x_2 + x_3 &= 0 \\ + \frac{4x_2}{5} + \frac{8x_3}{5} &= 0 \\ + if\;x_3 = t, then \\ + x_1 &= - t \\ + x_2 &= -2t \\ + x_3 &= t +\end{align*} + +\section*{Question 15} +\subsection*{Part i} +Since $A + A^{-1} = 0$, $A$ must either be skew symmetric. If A is skew symmetric, we know that the rank of an odd order skew symmetric matrix must be even. $\therefore Rank \leq 2020$ +\subsection*{Part ii} +Inverse does not exist as $A$ is singular matrix. +\end{document}