Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.
1. Prove Theorem 2.
Composing two general equiareal transformations, we get a transformation of the form below. $$\begin{pmatrix} a_{12}b_{21} + a_{11}b_{11} & a_{12}b_{22} + a_{11}b_{12} & a_{12}b_{23} + a_{11}b_{13} + a_{13} \\ a_{22}b_{21} + a_{21}b_{11} & a_{22}b_{22} + a_{21}b_{12} & a_{22}b_{23} + a_{21}b_{13} + a_{23} \\ 0 & 0 & a_{33}b_{33} \end{pmatrix}$$
Let’s consider the fraction relevant to equiareal transformations $$\frac{(a_{12}b_{21} + a_{11}b_{11})(a_{22}b_{22} + a_{21}b_{12}) - (a_{12}b_{22} + a_{11}b_{12})(a_{22}b_{21} + a_{21}b_{11})}{a_{33}^2b_{33}^2}$$
Expanding this we get $$\frac{(a_{12}a_{21} - a_{11}a_{22})(b_{11}b_{22} - b_{12}b_{21})}{a^2_{33}b^2_{33}}$$ $$\implies \frac{(a_{12}a_{21} - a_{11}a_{22})}{a^2_{33}}\frac{(b_{11}b_{22} - b_{12}b_{21})}{b^2_{33}}$$
As the absolute value of each of these terms is 1, the absolute value of this fraction will also be 1. Thus, two equiareal transformations are closed under the operation of composition.
The inverse of a general equiareal transformation has the form $$\frac{1}{a_{33}(a_{11}a_{22} - a_{12}a_{21})}\begin{pmatrix} a_{22}a_{33} & -a_{12}a_{33} & a_{12}a_{33} - a_{13}a_{22} \\ -a_{21}a_{33} & a_{11}a_{33} & a_{13}a_{21} - a_{11}a_{23} \\ 0 & 0 & a_{11}a_{22} - a_{12}a_{21} \end{pmatrix}$$
Let’s consider the fraction relevant to equiareal transformations $$\frac{a_{33}^2(a_{22}a_{11} - a_{12}a_{21})}{(a_{22}a_{11} - a_{12}a_{21})^2}$$ $$= \frac{a_{33}^2}{(a_{22}a_{11} - a_{12}a_{21})}$$
The absolute value of this is clearly 1. Hence the inverse is also an equiareal transformation.
Finally, it is obvious, from the rules of matrix multiplication, that the composition of equiareal transformations is associative and that the identity is included among the equiareal transformations.
Hence the set of all equiareal transformations is a group under the operation of composition.
2. Are the equiareal transformations of all six types? Give an example of each possible type.
Type I
$a_{33} \ne \pm1$
$\begin{pmatrix} a_{33}^2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & a_{33} \end{pmatrix}$
Type II
$a_{23} \ne 0$
$\begin{pmatrix} a_{11} & 0 & a_{13} \\ 0 & -a_{11} & a_{23} \\ 0 & 0 & -a_{11} \end{pmatrix}$
Type III
$\begin{pmatrix} a_{11} & 0 & 0 \\ 0 & a_{11} & 0 \\ 0 & 0 & -a_{11} \end{pmatrix}$
Type IV
$\begin{pmatrix} a_{11} & 0 & 0 \\ a_{21} & a_{11} & 0 \\ a_{31} & a_{32} & a_{11} \end{pmatrix}$
Type V
$a_{13} \ne 0, a_{23} \ne 0$
$\begin{pmatrix} a_{11} & 0 & a_{13} \\ 0 & a_{11} & a_{23} \\ 0 & 0 & a_{11} \end{pmatrix}$
Type VI
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
3. Determine which, if any, of the following mappings can be accomplished by an equiareal transformation, and when it exists, find the equations of the transformation:
(a) $(0,0,1) \rightarrow (0,0,1)$, $(1,0,1) \rightarrow (1,0,1)$, $(2,1,1) \rightarrow (-1,1,1)$
(b) $(0,0,1) \rightarrow (0,0,1)$, $(2,0,1) \rightarrow (0,3,1)$, $(2,1,1) \rightarrow (1,0,3)$
(c) $(0,0,1) \rightarrow (0,0,1)$, $(2,0,1) \rightarrow (3,1,1)$, $(1,1,1) \rightarrow (1,2,2)$
(a) The two triangles share the same base and their heights are the same. So an equiareal transformation between the two is possible. Using the invariance of $(0, 0, 1)$, we get $a_{13} = a_{23} = 0$ and $a_{33} = 1$. Further, using the invariance of $(1, 0, 1)$, we get $a_{11} = 1$ and $a_{21} = a_{31} = 0$. Finally, using the mapping $(2, 1, 1) \rightarrow (-1, 1, 1)$ we get $a_{12} = -3$ and $a_{22} = 1$. Hence, the matrix of transformation is $$\begin{pmatrix} 1 & -3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
(b) Areas of the two triangles are $\frac{1}{2} \times 2 \times 1 = 1$ and $\frac{1}{2} \times 3 \times \frac{1}{3} = \frac{1}{2}$ respectively. Clearly, the area of the two triangles do not match; hence an equiareal transformation is not possible.
(c) Areas of the two triangles are $\frac{1}{2} \times 2 \times 1 = 1$ and $\frac{1}{2} \times 1 \times \frac{5}{2} = \frac{5}{4}$ respectively. Clearly, the area of the two triangles do not match; hence an equiareal transformation is not possible.
4. If $A$ and $B$ are distinct finite points, and if $A’$ and $B’$ are distinct finite points, show that there is an equiareal transformation which maps $A$ into $A’$ and $B$ into $B’$. Is this transformation unique?
Given two pairs of distint points, it is possible to choose another pair of points C and C’ such that the areas of $\bigtriangleup ABC$ and $\bigtriangleup A’B’C’$ are the same. Hence, it is possible to define an equiareal transformation that achieves the given mapping.
As the choice of $C$ and $C’$ are arbitrary, the equiareal mapping is not unique.
5. Show that for a given equiareal transformation there are in general two directions with the property that the lengths of segments in these directions are unaltered by the transformation.
By corollary 1, every equiareal transformation is an affine transformation. Hence the result in Exercise #8, Sec 5.10 holds for equiareal transformations too.
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