Final answer:
To prove matrix A is orthogonal, we showed the product of matrix A and its transpose equals the identity matrix by using properties of trigonometric functions, thus confirming that matrix A is orthogonal for every angle θ.
Step-by-step explanation:
To prove that matrix A = [ cos(θ) -sin(θ) sin(θ) cos(θ)] is an orthogonal matrix for every angle θ, we need to show that A multiplied by its transpose AT equals the identity matrix I. The transpose AT of matrix A is [ cos(θ) sin(θ) -sin(θ) cos(θ)].
Now, let's compute the product of A and AT:
- Top left: cos(θ) * cos(θ) + (-sin(θ)) * sin(θ) = cos²(θ) + sin²(θ) = 1,
- Top right: cos(θ) * sin(θ) + (-sin(θ)) * cos(θ) = 0,
- Bottom left: sin(θ) * cos(θ) + cos(θ) * (-sin(θ)) = 0,
- Bottom right: sin(θ) * sin(θ) + cos(θ) * cos(θ) = sin²(θ) + cos²(θ) = 1.
Since the product of A and its transpose AT results in the identity matrix with 1's on the diagonal and 0's elsewhere, we can conclude that matrix A is indeed orthogonal for any θ.