• May 22nd 2006, 12:56 PM
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 )
• May 27th 2006, 03:20 AM
CaptainBlack
Quote:

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 ${1 \brack 0}$ and ${0 \brack 1}$ are ${0 \brack 1}$ and ${2 \brack -1}$ respectively.

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

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

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

RonL
• May 27th 2006, 05:56 AM
CaptainBlack
Quote:

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:

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

then:

$
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)=$
$
\left \left[ \begin{array}{cc}
-1&-2\\-1&0
\end{array} \right] \right/ (-2)

$

RonL
• May 27th 2006, 06:28 AM
CaptainBlack
Quote:

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 ${x \brack y}$ be a point on $f(\xi)$, then:

$
\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:

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

or on rearrangement:

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

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

RonL
• May 27th 2006, 08:07 AM
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.
• May 27th 2006, 08:09 AM
CaptainBlack
Quote:

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.