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)

Re: Isometry and determinant of matrix

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**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)

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

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

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

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

If $\displaystyle d=a$ then by (3), $\displaystyle b=-c$ so, $\displaystyle det(A)=a^2 + b^2 \overset{(1)}{=}1$ and if $\displaystyle d=-a$ then $\displaystyle b=c$ so, $\displaystyle det(A) =-a^2 - b^2 \overset{(1)}{=}-1$.

Re: Isometry and determinant of matrix