second-sem/maths/assignment-matrices/assign4.tex

\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 $A + A^T = 0$, $A$ 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}