Answer :
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]
### 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]
