Let f be the linear transformation represented by by the matrix

M = 0 2
1 -1

(a) State what effect f has on areas, and whether f changes orientaton.

(b) Find the matrix that represents the inverse of f.

(c) (i) Use the matrix that you found in part b to find the image f (ξ ) of the unit circle ξ under f, in the form

ax^2 +bxy +cy^2 = d

where a, b, c and d are integers.

What is the area enclosed by f ( ξ )

2. Originally Posted by fair_lady0072002
Let f be the linear transformation represented by by the matrix

M = 0 2
1 -1

(a) State what effect f has on areas, and whether f changes orientaton.
Assuming the matrix representation acts on column vectors, the images of
the unit vectors $\displaystyle {1 \brack 0}$ and $\displaystyle {0 \brack 1}$ are $\displaystyle {0 \brack 1}$ and $\displaystyle {2 \brack -1}$ respectively.

So the unit square is transformed into a parallelogram with vertices:

$\displaystyle {0 \brack 1}$,$\displaystyle {0 \brack 0}$,$\displaystyle {2 \brack -1}$, and $\displaystyle {2 \brack 0}$, which has area $\displaystyle 2$.

So the image of a figure under $\displaystyle f$ has twice the area of the figure (as $\displaystyle f$ is linear).

RonL

3. Originally Posted by fair_lady0072002
Let f be the linear transformation represented by by the matrix

M = 0 2
1 -1

(b) Find the matrix that represents the inverse of f.
By Cramer's rule if:

$\displaystyle A=\left[ \begin{array}{cc} a&b\\c&d \end{array} \right]$

then:

$\displaystyle A^{-1}=\left[ \begin{array}{cc} a&b\\c&d \end{array} \right]^{-1}= \left \left[ \begin{array}{cc} d&-b\\-c&a \end{array} \right] \right/ \det(A)=$$\displaystyle \left \left[ \begin{array}{cc} -1&-2\\-1&0 \end{array} \right] \right/ (-2)$

RonL

4. Originally Posted by fair_lady0072002
Let f be the linear transformation represented by by the matrix

M = 0 2
1 -1

(c) (i) Use the matrix that you found in part b to find the image f (ξ ) of the unit circle ξ under f, in the form

ax^2 +bxy +cy^2 = d

where a, b, c and d are integers.

What is the area enclosed by f ( ξ )
Let $\displaystyle {x \brack y}$ be a point on $\displaystyle f(\xi)$, then:

$\displaystyle \left \left[ \begin{array}{cc}-1&-2\\-1&0\end{array} \right]{x \brack y}\right/ (-2)={x/2+y \brack x/2}$,

is a point on the unit circle, so:

$\displaystyle \left(\frac{x}{2}+y\right)^2+\left(\frac{x}{2} \right)^2=1$,

or on rearrangement:

$\displaystyle x^2+2xy+2y=2$.

The area enclosed by the unit circle is $\displaystyle \pi$ so the area
enclosed by $\displaystyle f(\xi)$ is $\displaystyle 2 \pi$.

RonL

5. In general a linear transformation (of a real Euclidean space) multiplies areas by the determinant. If the determinant is negative, that means that orientation is reversed.

6. Originally Posted by rgep
In general a linear transformation (of a real Euclidean space) multiplies areas by the determinant. If the determinant is negative, that means that orientation is reversed.
I missed where it asked about orientation

Informally if we take the vertices of the unit square in clockwise order the
corresponding vertices of the image of the unit square are in counter-
clockwise order, so f changes the orientations.

RonL