# Isometry and determinant of matrix

• Sep 16th 2011, 08:08 AM
H12504106
Isometry and determinant of matrix
Let e_1 = (1,0) and e_2 = (0,1). If T is an isometry of the plane fixing O, let T(e_1) = (a,b) and
T(e_2) = (c,d) and let A be a 2x2 matrix such that the first row is (a c) and second row is (b d). Prove that det(A) = 1 or -1.

Using the definition of isometry i obtained the following equations:
a^2+b^2=1 ----------(1)
c^2+d^2=1 ----------(2)
ac+bd=0 -------------(3)

However, i was unable to continue on to find det(A). Can this problem be solved just by using definition of isometry alone? Or do i need some results from linear algebra. (This problem was part of a group theory course)
• Sep 16th 2011, 10:58 AM
zoek
Re: Isometry and determinant of matrix
Quote:

Originally Posted by H12504106
Let e_1 = (1,0) and e_2 = (0,1). If T is an isometry of the plane fixing O, let T(e_1) = (a,b) and
T(e_2) = (c,d) and let A be a 2x2 matrix such that the first row is (a c) and second row is (b d). Prove that det(A) = 1 or -1.

Using the definition of isometry i obtained the following equations:
a^2+b^2=1 ----------(1)
c^2+d^2=1 ----------(2)
ac+bd=0 -------------(3)

However, i was unable to continue on to find det(A). Can this problem be solved just by using definition of isometry alone? Or do i need some results from linear algebra. (This problem was part of a group theory course)

$\bullet \,\,a=0.$

$\overset{(1)}{\Longrightarrow}b^2 = 1\Longrightarrow b=\pm 1\overset{(3)}{\Longrightarrow}d=0 \overset{(2)}{\Longrightarrow} c^2 = 1 \Longrightarrow c=\pm 1 \Longrightarrow det(A)=1$ or $det(A) = -1$.

$\bullet \,\,a\neq 0$.

$\overset{(3)}{\Longrightarrow} c=-\frac{bd}{a}\overset{(2)}{\Longrightarrow}\left(b^ 2 + a^2 \right)d^2 = a^2\overset{(1)}{\Longrightarrow} d^2=a^2 \Longrightarrow d= \pm a.$

If $d=a$ then by (3), $b=-c$ so, $det(A)=a^2 + b^2 \overset{(1)}{=}1$ and if $d=-a$ then $b=c$ so, $det(A) =-a^2 - b^2 \overset{(1)}{=}-1$.
• Sep 16th 2011, 10:59 PM
H12504106
Re: Isometry and determinant of matrix
Thanks a lot!.