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Sagot :
To find the determinant of the matrix [tex]\( A \)[/tex], we first define the matrix [tex]\( A \)[/tex] as follows:
[tex]\[ A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \][/tex]
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is calculated using the formula:
[tex]\[ \text{det}(A) = ad - bc \][/tex]
For our specific matrix [tex]\( A \)[/tex]:
[tex]\[ a = \cos \theta, \quad b = \sin \theta, \quad c = -\sin \theta, \quad d = \cos \theta \][/tex]
Substitute these values into the determinant formula:
[tex]\[ \text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \][/tex]
Simplify the expression:
[tex]\[ \text{det}(A) = \cos^2 \theta + \sin^2 \theta \][/tex]
We know from trigonometric identities that:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
Therefore:
[tex]\[ \text{det}(A) = 1 \][/tex]
Hence, the determinant of the matrix [tex]\( A \)[/tex] is:
[tex]\[ |A| = 1 \][/tex]
[tex]\[ A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \][/tex]
The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is calculated using the formula:
[tex]\[ \text{det}(A) = ad - bc \][/tex]
For our specific matrix [tex]\( A \)[/tex]:
[tex]\[ a = \cos \theta, \quad b = \sin \theta, \quad c = -\sin \theta, \quad d = \cos \theta \][/tex]
Substitute these values into the determinant formula:
[tex]\[ \text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \][/tex]
Simplify the expression:
[tex]\[ \text{det}(A) = \cos^2 \theta + \sin^2 \theta \][/tex]
We know from trigonometric identities that:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
Therefore:
[tex]\[ \text{det}(A) = 1 \][/tex]
Hence, the determinant of the matrix [tex]\( A \)[/tex] is:
[tex]\[ |A| = 1 \][/tex]
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