Add maths assignment 4
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maths/assignment-matrices/assign4.tex
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maths/assignment-matrices/assign4.tex
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\begin{document}
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\title{Mathematics Assignment 4 --- Matrices}
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\author{Ahmad Saalim Lone, 2019BCSE017}
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\date{16 May, 2020}
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\maketitle
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\section*{Question 1}
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\begin{align*}
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A &=
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\begin{bmatrix}
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1 & -1 & -1 & 2 \\
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4 & 2 & 2 & -1 \\
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2 & 2 & 0 & -2
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\end{bmatrix}
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\\
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PAQ &= \text{Normal form} \\
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I_3AI_4 &= A_{3 \times 4} \\
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{bmatrix}
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A
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\begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 0 & 0 & 1
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\end{bmatrix}
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&=
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\begin{bmatrix}
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1 & -1 & -1 & 2 \\
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4 & 2 & 2 & -1 \\
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2 & 2 & 0 & -2
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\end{bmatrix}
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\end{align*}
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Applying the following row tranformations on $I_3$ and column tranformations on $I_4$.
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\begin{align*}
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C_4 &\to C_4 + C_2 \\
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R_3 &\to \frac{R_3}{2} \\
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R_2 &\to \frac{R_2 + 2R_1}{3} \\
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R_1 &\to R_1 + R_3 \\
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C_1 &\to C_1 - 2 C_4 \\
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C_4 &\to C_4 + C_3 \\
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C_2 &\to C_2 - C_1 \\
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C_3 &\rightleftharpoons C_1 \\
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C_2 &\rightleftharpoons C_4 \\
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R_1 &\rightleftharpoons -R_1
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\end{align*}
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we get
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\begin{align*}
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P &=
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\begin{bmatrix}
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-1 & 0 & -\frac{1}{2} \\[6pt]
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\frac{2}{3} & \frac{1}{3} & 0 \\[6pt]
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0 & 0 & \frac{1}{2}
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\end{bmatrix}
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\\
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Q &=
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\begin{bmatrix}
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0 & 0 & 1 & -1 \\
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0 & 0 & -2 & 3 \\
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1 & 1 & 0 & 0 \\
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0 & 1 & -2 & 2
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\end{bmatrix}
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\\
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\text{Normal Form of A} &=
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\begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 0
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\end{bmatrix}
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\end{align*}
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\section*{Question 2}
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\[
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x^2yz = xy^2z^3 = x^3y^2z = e
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\]
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Taking $\ln$ on both sides
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\begin{align*}
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2\ln{x} + \ln{y} \ln{z} &= 1 \\
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\ln{x} + 2 \ln{y} + 3 \ln{z} &= 1 \\
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3\ln{x} + 2\ln{y} + \ln{z} &= 1
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\end{align*}
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\[
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\text{Augemented Matrix} = [A:B] =
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\begin{bmatrix}
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2 & 1 & 1 & : & 1 \\
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1 & 2 & 3 & : & 1 \\
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3 & 2 & 1 & : & 1
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\end{bmatrix}
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\]
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After ppling the following tranformations
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\begin{align*}
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R_3 &\to R_3 - R_1 \\
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R_2 &\to R_2 - R_1 \\
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R_2 &\to R_2 - R_1 \\
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R_1 &\to R_1 -R_3
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\end{align*}
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We get
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\[
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\begin{bmatrix}
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1 & 0 & 1 & : & 1 \\
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-3 & 0 & 1 & : & -1 \\
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1 & 1 & 0 & : & 0
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\end{bmatrix}
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\]
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$Rank(A) = Rank(A:B) = 3 = n$\\
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$\therefore$ unique solution.
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\section*{Question 3}
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\begin{align*}
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x - cy - bz &= 0 \\
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cx - y + az &= 0 \\
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bx + ay - z &= 0 \\
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\text{Augemented matrix} &=
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\begin{bmatrix}
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1 & -c & -b & : & 0 \\
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c & -1 & a & : & 0 \\
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b & a & -1 & : & 0
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\end{bmatrix}
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\end{align*}
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Applying the following tranformations
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\begin{align*}
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R_2 &\to R_2 - cR_1 \\
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R_3 &\to R_3 - bR_1 \\
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R_2 &\to \frac{R_2}{c^2 - 1} \\
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R_3 &\to R_3 - (a + bc)R_2
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\end{align*}
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we get
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\[
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C =
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\begin{bmatrix}
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1 & -c & -b & : & 0 \\[6pt]
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0 & 1 & \frac{bc+c}{c^2 - 1} & : & 0 \\[6pt]
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0 & 0 & b^2 - 1 - \frac{{(a+bc)}^2}{c^2 - 1} & : & 0
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\end{bmatrix}
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\]
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For non trivial solutions $Rank(A) = Rank(c) \ne number\;of\;unknows$
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\begin{align*}
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b^2 - 1 \frac{{(a+bc)}^2}{c^2 -1} &= 0 \\
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\implies a^2 + b^2 + c^2 + 2abc &= 1 \\
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\end{align*}
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Now
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\begin{align*}
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x - cy -bz &= 0 \\
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y + \left(\frac{bc + a}{c^2 - 1}z\right) &= 0 \\
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\end{align*}
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We get
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\begin{align*}
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x &= \frac{ac + b}{1-c^2}z \\
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y &= \frac{bc + a}{1 - c^2}z \\
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z &= z \\
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x : y : z &= \sqrt{|1 - a^2|} : \sqrt{|1 - b^2|} : \sqrt{|1 -c^2|}
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\end{align*}
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\section{Question 4}
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\begin{align*}
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A &=
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\begin{bmatrix}
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3 & -4 & 4 \\
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1 & -2 & 44 \\
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1 & -1 & 3
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\end{bmatrix} \\
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|A - \lambda I| &= 0\\
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\begin{vmatrix}
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3 - \lambda & -4 & 4 \\
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1 & -2 - \lambda & 44 \\
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1 & -1 & 3 - \lambda
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\end{vmatrix} &= 0
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\end{align*}
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\( \lambda = -1, 2, 3 \) are Eigen Values
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For $\lambda = -1$
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\begin{align*}
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\begin{bmatrix}
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4 & -4 & 4 \\
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1 & -1 & 4 \\
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1 & -1 & 4
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\end{bmatrix}
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\begin{bmatrix}
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x \\
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y \\
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z
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\end{bmatrix}
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&=
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\begin{bmatrix}
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0 \\
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0 \\
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0
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\end{bmatrix} \\
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\text{Eigen Vector} &=
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\begin{bmatrix}
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1 \\
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1 \\
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0
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\end{bmatrix}
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\end{align*}
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For $\lambda = 2$
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\begin{align*}
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\begin{bmatrix}
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1 & -4 & 4 \\
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1 & -4 & 4 \\
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1 & -1 & 1
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\end{bmatrix}
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\begin{bmatrix}
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x \\
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y \\
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z
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\end{bmatrix}
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&=
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\begin{bmatrix}
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0 \\
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0 \\
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0
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\end{bmatrix} \\
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\text{Eigen Vector} &=
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\begin{bmatrix}
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0 \\
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1 \\
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1
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\end{bmatrix}
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\end{align*}
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For $\lambda = 3$
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\begin{align*}
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\begin{bmatrix}
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0 & -4 & 4 \\
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1 & -5 & 4 \\
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1 & -1 & 0
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\end{bmatrix}
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\begin{bmatrix}
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x \\
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y \\
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z
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\end{bmatrix}
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&=
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\begin{bmatrix}
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0 \\
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0 \\
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0
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\end{bmatrix} \\
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\text{Eigen Vector} &=
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\begin{bmatrix}
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1 \\
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1 \\
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1
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\end{bmatrix}
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\end{align*}
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\section*{Question 5}
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\begin{align*}
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A &=
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\begin{bmatrix}
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2 & 2 & 0 \\
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2 & 1 & 1 \\
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-7 & 2 & -3
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\end{bmatrix}
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\\
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|A - \lambda I| &= 0 \\
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\begin{vmatrix}
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2 - \lambda & 2 & 0 \\
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2 & 1 - \lambda & 1 \\
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-7 & 2 & -3 - \lambda
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\end{vmatrix} &= 0
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\end{align*}
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Eigen values $\lambda = 1, 3, -4$ \\
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1\textsuperscript{st} eigen value of $A^2 -2 A + I = 1^2 - 2(1) + 1 = 0$
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2\textsuperscript{nd} eigen value of $A^2 -2 A + I = 3^2 - 2(3) + 1 = 4$
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3\textsuperscript{rd} eigen value of $A^2 -2 A + I = {(-4)}^2 - 2(-4) + 1 = 25$
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\section*{Question 6}
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\begin{align*}
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A &=
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\begin{bmatrix}
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7 & 3 \\
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2 & 6
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\end{bmatrix}
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\\
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| A - \lambda I| &= 0 \\
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\begin{vmatrix}
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7 - \lambda & 3 \\
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2 & 6 - \lambda
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\end{vmatrix} &= 0 \\
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(7 - \lambda)(6 - \lambda) - 5 &= 0 \\
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(\lambda - 4)(\lambda - 9) &= 0 \\
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A^2 - 13 A + 36 &= 0 \\
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A^2 &= 13 A - 36 \\
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A^2\cdot A &= (13 A - 36)A \\
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A^3 &= 13 A^2 - 36A \\
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A^3 &=
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\begin{bmatrix}
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715 & 507 \\
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338 & 546
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\end{bmatrix}
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-
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\begin{bmatrix}
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252 & 108 \\
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72 & 216
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\end{bmatrix} \\
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A^3 &=
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\begin{bmatrix}
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463 & 399 \\
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266 & 330
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\end{bmatrix}
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\end{align*}
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\section*{Question 7}
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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 $A^T = A$,
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\begin{gather*}
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\overline{V}^T AV = \overline{V}^t(AV) = \overline{V}(\lambda V) = \lambda( \overline{V}V ) \\
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\overline{V}^T AV = {(A \overline{V})}^T V = {( \overline{\lambda}\;\overline{V} )}^T V = \overline{\lambda} ( \overline{V} V )
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\end{gather*}
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Since $V \ne 0$, we have $ \overline{V}V \ne 0$. Thus $\lambda = \overline{\lambda}$, which means $\lambda \in R$.
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\section*{Question 8}
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Quadratic Form $ax_1^2 + cx_2^2 - 2bx_1x_2$
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\[
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\begin{bmatrix}
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a & -b \\
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-b & c
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\end{bmatrix}
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\]
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Convert it to diagonal matrix by applying
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\begin{align*}
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R_2 &\to R_2 + \frac{b}{a}R_1 \\
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C_2 &\to C_2 + \frac{b}{a}C_1
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\end{align*}
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Now, we get
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\[
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\begin{bmatrix}
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a & 0 \\
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0 & c - \frac{b^2}{a}
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\end{bmatrix}
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\]
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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.
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\section*{Question 9}
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\(
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A =
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\begin{bmatrix}
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\lambda & 1 & 1 \\
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1 & \lambda & -1 \\
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1 & -1 & \lambda
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\end{bmatrix}
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\) is a symmetric matrix obtained when compared to Quadratic form $\lambda(x^2+ y^2 + z^2) + 2xy + 2zx -2yz$.
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Now, convert $A$ into diagonal matrix by:
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\begin{align*}
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C_1 &\to C_1 + C_3 \\
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C_1 &\to \frac{C_1}{\lambda + 1} \\
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R_3 &\to R_3 - R_1 \\
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C_3 &\to C_3 - C_1 \\
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C_2 &\to C_2 -C_1 \\
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C_3 &\to C_3 + \frac{C_2}{\lambda} \\
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R_3 &\to R_3 + \frac{2R_2}{\lambda}
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\end{align*}
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we get,
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\[
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & \lambda & 0 \\
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0 & 0 & \lambda - \frac{2}{\lambda} - 1
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\end{bmatrix}
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\]
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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 $\lambda \in (2, \infty)$.
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\section*{Question 10}
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Multiplication of all the eigen values = determinant of the matrix. For singular matrix, determinant value = 0.
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\begin{align*}
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\text{Eigen Values} &= 2, 3, a \\
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6a &= 0 \\
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a &= 0
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\end{align*}
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\section*{Question 11}
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Quadratic Form $x_2^2 + 2x_2^2 - 5x_3^2$
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\[
|
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\begin{bmatrix}
|
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1 & 0 & 0 \\
|
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0 & 2 & 0 \\
|
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0 & 0 & -5
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\end{bmatrix}
|
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\]
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\begin{align*}
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index(s) &= 2\;\text{(No of positive terms)} \\
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rank(r) &= 3 \\
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signature &= 2s -r = 4 - 3 = 1
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\end{align*}
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\section*{Question 12}
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Quadratic form $ax^2 + 2bcy + cy^2$.
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\[
|
||||
\begin{bmatrix}
|
||||
a & b \\
|
||||
b & c
|
||||
\end{bmatrix}
|
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\]
|
||||
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
|
||||
\[
|
||||
\begin{bmatrix}
|
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a & 0 \\
|
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0 & c - \frac{b^2}{a}
|
||||
\end{bmatrix}
|
||||
\]
|
||||
For positive definite, $a>0$ and $ac -b^2 > 0$. \\
|
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For negative definite, $a<0$ and $ac -b^2 > 0$. \\
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Roots of the quadratic equation ($ax^2 + 2bx + c = 0$) are imaginary when $D < 0$. \\
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||||
${(2b)}^2 - 4ac = 4( b^2 - ac )$ is always negative
|
||||
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\section*{Question 13}
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\begin{align*}
|
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X_1 &=
|
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\begin{bmatrix}
|
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1 \\
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2 \\
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-3 \\
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4
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||||
\end{bmatrix}
|
||||
\\
|
||||
X_2 &=
|
||||
\begin{bmatrix}
|
||||
1 \\
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-5 \\
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8 \\
|
||||
-7
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||||
\end{bmatrix}
|
||||
\\
|
||||
X_3 &=
|
||||
\begin{bmatrix}
|
||||
1 \\
|
||||
-5 \\
|
||||
8 \\
|
||||
-7
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||||
\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}
|
||||
Reference in New Issue
Block a user