Day: September 25, 2020

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202009251201 Homework 1 (Q1)

Solve the system \mathbf{A}\mathbf{x} = \mathbf{0} for each matrix \mathbf{A} below.

(a)

\mathbf{A}= \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \\ \end{bmatrix}

(b)

\mathbf{A}= \begin{bmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & -2 & 0 \\ \end{bmatrix}

(c)

\mathbf{A}= \begin{bmatrix} 4 & 6 & 0 & 1 & -9 \\ 1 & 2 & -4 & 5 & 7 \\ 2 & 3 & 6 & 4 & 2 \\ 1 & 0 & 3 & 2 & -5 \\ \end{bmatrix}

(d)

\mathbf{A}= \begin{bmatrix} 1 & 2 & 3 & 1 & 1 \\ 1 & 4 & 0 & 1 & 2 \\ 0 & 2 & -3 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ \end{bmatrix}

Recall

Observe that \mathbf{A}‘s in (a) and (b) are 3-by-4 matrices, and in (c) and (d) are 4-by-5 matrices. Multiplication of two matrices \mathbf{A} and \mathbf{B} results in a matrix product \mathbf{A}\mathbf{B}. By convention, should an m-by-n matrix \mathbf{A} be written on the left, \mathbf{A} is meant the multiplicand, and should an n-by-p matrix \mathbf{B} be written on the right, \mathbf{B} is meant the multiplier. The matrix product \mathbf{AB} will become an m-by-p matrix.

Solution.

(a)

Let a 4-by-1 column vector \mathbf{x} be

\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}.

Then,

\begin{aligned} \mathbf{A}\mathbf{x} & = \mathbf{0} \\ \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} & = \mathbf{0} \\ \begin{bmatrix} (1)(x_1)+(3)(x_2)+(5)(x_3)+(7)(x_4) \\ (3)(x_1)+(5)(x_2)+(7)(x_3)+(9)(x_4) \\ (5)(x_1)+(7)(x_2)+(9)(x_3)+(1)(x_4) \\ \end{bmatrix} & = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \end{aligned}

Here we have four unknowns but three equations.

\begin{Bmatrix} 15x_1+ 45x_2+75x_3+105x_4 & = 0 \\ 15x_1+25x_2+35x_3+45x_4 & = 0 \\ 15x_1+21x_2+27x_3+3x_4 & = 0 \\ \end{Bmatrix}

It is feasible to carry through all the calculations, but do let us not work in so awkward a manner.

Performing row operations of matrices and using shorthand notations (e.g., R_1 stands for Row 1; 5R_2 stands for 5 times each entry in Row 2; R_1-R_3 means every entry in Row 3 is to be subtracted from the corresponding entry in Row 1.)

(1) R_3-5R_1

\begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 0 & -8 & -16 & -34 \\ \end{bmatrix}

(2) R_2 - 3R_1

\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & -8 & -16 & -34 \\ \end{bmatrix}

(3) R_3 - 2R_2

\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & 0 & 0 & -58 \\ \end{bmatrix}

(4) R_3 \div (-58)

\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

(5) R_2-12R_3

\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

(6) R_1-7R_3

\begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & -4 & -8 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

(7) R_2 \div (-4)

\begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

(8) R_1-3R_2

\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

The matrix above is called the reduced row-echelon form of \mathbf{A}.

Now that the simplification has come handy:

\begin{aligned} x_1 -x_3 &=0 \\ x_2 +2x_3 & = 0 \\ x_4 & = 0 \\ \end{aligned}

and the answer is \mathbf{x}=\begin{bmatrix} x \\ -2x \\ x \\ 0 \end{bmatrix}.

Questions (b), (c), and (d) are left the readers.

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202009250041 Exercise 5.8.1

Find out if any the asymptote(s) of the curve given by:

y=\displaystyle{\frac{x^2-6x+3}{x-3}}.

Attempts.

\begin{aligned} & & x & -3 & & \\\cline{3-5} x & -3 \Big) & x^2 & -6x & +3 & \\ & & x^2 & -3x & & \\ \cline{3-5} & & & -3x & +3 \\ & & & -3x & +9 \\\cline{4-5} & & & & -6 \end{aligned}

The curve is given by

\begin{aligned} y & =\displaystyle{\frac{x^2-6x+3}{x-3}} \\ & = x-3 - \displaystyle{\frac{6}{x-3}} \end{aligned}.

Ans. Of the given curve, x=3 is a vertical asymptote and y=x-3 an oblique asymptote.

Retrospectively, for f(x)=\displaystyle{\frac{x^2-6x+3}{x-3}}, maybe one could also have checked the conditions as follow:

i. \displaystyle{\lim_{x\to 0}}f(x);

ii. \displaystyle{\lim_{x\to 3^{+}}}f(x);

iii. \displaystyle{\lim_{x\to 3^{-}}}f(x);

iv. \displaystyle{\lim_{x\to +\infty}}f(x); and

iv. \displaystyle{\lim_{x\to -\infty}}f(x).

i.

\begin{aligned} \lim_{x\to 0}f(x) & = \lim_{x\to 0} \frac{x^2-6x+3}{x-3} \\ & = \frac{(0)^2-6(0)+3}{(0)-3} \\ & = -1 \end{aligned}

iii.

\begin{aligned} \lim_{x\to 3^{-}}f(x) & = \lim_{x\to 3^{-}} \frac{x^2-6x+3}{x-3} \\ & = \lim_{x\to 3^{-}} \bigg( x-3-\frac{6}{x-3} \bigg) \\ & = 3^{-} -3-\frac{6}{3^{-}-3} \\ & = (3^{-}-3) + \bigg( \frac{6}{3-3^{-}} \bigg) \\ & \\ \hline & \\ & \because 3^{-} < 3 \\ & \therefore 3^{-}-3<0\qquad \textrm{and}\qquad 3-3^{-}>0 \\ & \dots \textrm{excuse me for writing with no more ado,}\dots \\ & \\ \hline & \\ & = 0^{-} + \frac{6}{0^{+}} \\ & = +\infty \\ \end{aligned}

ii., iv., and iv. are not checked.

Just work with it like whom I was taught:

First, find the slope of tangent to the curve,

\begin{aligned} f(x) & = \frac{x^2-6x+3}{x-3} \\ f'(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x) \big) \\ & = \frac{(x-3)(2x-6)-(x^2-6x+3)(1)}{(x-3)^2} \\ & = \frac{(2x^2-12x+18)-(x^2-6x+3)}{(x-3)^2} \\ & = \frac{x^2-6x+15}{x^2-6x+9} \\ & = 1 + \frac{6}{x^2-6x+9} \end{aligned}

The purpose is to find out any crests or troughs, because at the tip or bottom of a plotted curve, the slope of tangent will be zero, i.e., f'(x)=0.

Secondly, find the change of slope of tangent to the curve,

\begin{aligned} f''(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big[ f'(x) \big] \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( 1+\frac{6}{x^2-6x+9} \bigg) \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \big( 1 \big) + \frac{\mathrm{d}}{\mathrm{d}x} \bigg( \frac{6}{x^2-6x+9} \bigg) \\ & = 0 + \frac{(x^2-6x+9)(0)+(6)(2x-6)}{(x^2-6x+9)^2} \\ & = \frac{12x-36}{(x^2-6x+9)^2} \\ \end{aligned}

The purpose is to find out if there were any inflexion points (i.e., f''(x)=0) upon which curvature changes sign.

It begins, assuming x\neq 3, I will set f(x)=0:

\begin{aligned} f(x) & = 0 \\ \Leftrightarrow \qquad\qquad x^2-6x+3 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(3)}}{2(1)} \\ & = 3 \pm \sqrt{6} \\ \end{aligned}

The x-intercepts are thus two, 3+\sqrt{6} and 3-\sqrt{6} .

Then I needs set f'(x)=0 with the same assumption x\neq 3.

\begin{aligned} f'(x) & = 0 \\ \Leftrightarrow \qquad \qquad 1+\frac{6}{x^2-6x+9} & = 0 \\ & \\ \textrm{\dots \quad by assumption }x\neq 3 & \enspace \textrm{jump legitimately to the next step\quad \dots} \\ & \\ x^2-6x+9+6 & = 0 \\ x^2 -6x +15 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(15)}}{2(1)} \\ & = 3 \pm \sqrt{6}\, \textrm{i} \end{aligned}

As curvature here permits of no nonreal points, there are neither convex point(s) nor concave point(s).

Besides, there are none any one point of inflexion because if on one hand f''(x)=0 and on the other x\neq 3 be assumed,

\begin{aligned} \frac{12x-36}{(x^2-6x+9)^2} & = 0 \\ \Rightarrow 12x-36 & = 0 \\ \Rightarrow x & =3\qquad\qquad\qquad \bot \\ \end{aligned}

it would have resulted in contradiction.

In conclusion, this exercise was overdone.