1. ## matrices problem

Question:
A transformation T is represented by
$\displaystyle \begin{pmatrix}x'\\ y'\end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ where $\displaystyle b \epsilon \mathbb{R}^{+}$
a) Draw a diagram showing the unit square and its image under T
[b) Show that area is invariant under T
c) Show that T maps the curve $\displaystyle y =1/x$ onto itself
d) Show that T maps the region bounded by the curve $\displaystyle y = 1/x$, the lines x=1 and x =a and the x-axis onto the region bounded by the same curve the lines x =b and x =ab and the x-axis.
e) hence show that $\displaystyle \int_{1}^{a} \frac{1}{x} dx = \int_{a}^{ab} \frac{1}{x} dx$
f)Given that $\displaystyle F(t) = \int_{1}^{t} \frac{1}{x} dx$ show that $\displaystyle F(ab)=F(a)+F(b)$

my attempt:a)
the unit square $\displaystyle = \begin{pmatrix} 0 &1 & 1 &0 \\ 0& 0 & 1 &1 \end{pmatrix}$
transformed by T ;

$\displaystyle \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix} b &0 \\ 0 & 1/b \end{pmatrix} \begin{pmatrix} 0 &1 & 1 &0 \\ 0& 0 & 1 &1 \end{pmatrix} = \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix} 0 &b &b & 0\\ 0 &0 & 1/b &1/b \end{pmatrix}$ did i draw the right diagram?

b) the area is invariant because the area of the transformed image is $\displaystyle \frac{1}{b} \times b = 1$

the determinant of the unit square = 1
Since the area of the transformed = 1 and the determinant of the unit square = 1 therefore, the area is invariant.

c)$\displaystyle \begin{pmatrix}x'\\ y'\end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$

since it maps $\displaystyle y = \frac{1}{x}$ onto itself thus

$\displaystyle \begin{pmatrix}x' \\ \frac{1}{x}' \end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x \\ \frac{1}{x}\end{pmatrix}$

$\displaystyle \begin{pmatrix}x' \\ \frac{1}{x}' \end{pmatrix} = \begin{pmatrix}bx\\ \frac{b}{x}\end{pmatrix}$

T maps points on y =1/x to new points on the curve y =1/x

d) I need help this part i dont know what to do
e)
$\displaystyle \int_{1}^{a} \frac{1}{x} dx = \int_{a}^{ab} \frac{1}{x} dx$
we know that $\displaystyle \int \frac{1}{x}dx = ln \left | x \right |+c$
thus, $\displaystyle \int_{1}^{a} \frac{1}{x} dx = ln \left | a \right | - ln \left | 1 \right | =ln \left | a \right | - 0 = ln \left | a \right |$
$\displaystyle \int_{a}^{ab} \frac{1}{x} dx = ln \left | ab \right | - ln \left | a \right | = \frac{ln \left | ab \right |}{ln \left | a \right |}= \frac{ln \left | a \right |ln \left | b \right |}{ln \left | a \right |}= ln \left | b \right |$

a =b
i am not confident if i did this part right. I would like to know if i did this part correctly
f) can i please have some help doing part F?
Please I would like help with this question

2. ## Re: matrices problem

In your second last line in (e) you are saying that ln|ab| - ln |a| = ln|ab|/ln|a|, in other words the subtraction of logs is the same as division of logs?? Rethink that!

The rule is: ln(m/n) = ln m - ln n . Can you see what you have done wrong? It is a common mistake!