Explanation:
To show that the matrix A = [cos(2x) sin(2x); sin(2x) -cos(2x)] satisfies the relation A^2 = I, we need to compute the square of the matrix A and demonstrate that it equals the identity matrix I.
Let's compute A^2:
A^2 = A * A
A = [cos(2x) sin(2x); sin(2x) -cos(2x)]
A * A = [cos(2x) sin(2x); sin(2x) -cos(2x)] * [cos(2x) sin(2x); sin(2x) -cos(2x)]
Using matrix multiplication, we have:
A * A = [cos(2x)*cos(2x) + sin(2x)*sin(2x) cos(2x)*sin(2x) + sin(2x)*(-cos(2x));
sin(2x)*cos(2x) + (-cos(2x))*sin(2x) sin(2x)*sin(2x) + (-cos(2x))*(-cos(2x))]
Simplifying further:
A * A = [cos^2(2x) + sin^2(2x) 0;
0 cos^2(2x) + sin^2(2x)]
Using the trigonometric identity cos^2(θ) + sin^2(θ) = 1, we can simplify further:
A * A = [1 0;
0 1] = I
Therefore, we have shown that A^2 = I for any value of x.