Sure! Let's find the determinants for the matrices representing rotations and reflections. We will calculate each determinant step by step.
### Determinant of Matrix [tex]\( Q_1 \)[/tex]:
Given the matrix [tex]\( Q_1 \)[/tex]:
[tex]\[
Q_1 = \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{bmatrix}
\][/tex]
The determinant of a 2x2 matrix [tex]\(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)[/tex] is given by:
[tex]\[
\text{det}(Q_1) = a \cdot d - b \cdot c
\][/tex]
In this case, we have:
[tex]\[
a = \cos \theta, \quad b = -\sin \theta, \quad c = \sin \theta, \quad d = \cos \theta
\][/tex]
Now, substitute these values into the determinant formula:
[tex]\[
\text{det}(Q_1) = (\cos \theta) (\cos \theta) - (-\sin \theta) (\sin \theta)
\][/tex]
Simplify the expression:
[tex]\[
\text{det}(Q_1) = \cos^2 \theta + \sin^2 \theta
\][/tex]
We know from the trigonometric identity that:
[tex]\[
\cos^2 \theta + \sin^2 \theta = 1
\][/tex]
Thus,
[tex]\[
\text{det}(Q_1) = 1
\][/tex]
### Determinant of Matrix [tex]\( Q_2 \)[/tex]:
Given the matrix [tex]\( Q_2 \)[/tex]:
[tex]\[
Q_2 = \begin{bmatrix}
1 - 2 \cos^2 \theta & -2 \cos \theta \sin \theta \\
-2 \cos \theta \sin \theta & 1 - 2 \sin^2 \theta
\end{bmatrix}
\][/tex]
The determinant of a 2x2 matrix [tex]\(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)[/tex] is given by:
[tex]\[
\text{det}(Q_2) = a \cdot d - b \cdot c
\][/tex]
In this case, we have:
[tex]\[
a = 1 - 2 \cos^2 \theta, \quad b = -2 \cos \theta \sin \theta, \quad c = -2 \cos \theta \sin \theta, \quad d = 1 - 2 \sin^2 \theta
\][/tex]
Now, substitute these values into the determinant formula:
[tex]\[
\text{det}(Q_2) = (1 - 2 \cos^2 \theta) (1 - 2 \sin^2 \theta) - (-2 \cos \theta \sin \theta) (-2 \cos \theta \sin \theta)
\][/tex]
Simplify the expression:
[tex]\[
\text{det}(Q_2) = (1 - 2 \cos^2 \theta) (1 - 2 \sin^2 \theta) - (4 \cos^2 \theta \sin^2 \theta)
\][/tex]
Expand the product:
[tex]\[
\text{det}(Q_2) = 1 - 2 \cos^2 \theta - 2 \sin^2 \theta + 4 \cos^2 \theta \sin^2 \theta - 4 \cos^2 \theta \sin^2 \theta
\][/tex]
Combine like terms and simplify further:
[tex]\[
\text{det}(Q_2) = 1 - 2 (\cos^2 \theta + \sin^2 \theta)
\][/tex]
We know from the trigonometric identity that:
[tex]\[
\cos^2 \theta + \sin^2 \theta = 1
\][/tex]
Thus,
[tex]\[
\text{det}(Q_2) = 1 - 2 \cdot 1
\][/tex]
[tex]\[
\text{det}(Q_2) = 1 - 2
\][/tex]
[tex]\[
\text{det}(Q_2) = -1
\][/tex]
Therefore, the determinants are:
[tex]\[
\text{det}(Q_1) = 1 \quad \text{and} \quad \text{det}(Q_2) = -1
\][/tex]
