To show that the determinant is independent of 0, we need to show that its value remains constant as 0 varies.
Expanding the determinant along the first row, we get:
|x sin 0 cos 0| = x |-sin 0 1| – sin 0 |cos 0 1| |cos 0 x| |-sin 0 x|
= -xsin0(-sin0) – cos0(x)(cos0) + sin0(cos0)(x) = x(sin^2 0 + cos^2 0) = x
Therefore, we can see that the value of the determinant is x, which is independent of 0.
Now, let’s check whether the given matrix satisfies the determinant equation:
|x sin 0 cos 0| |-sin 0 -x 1| |cos 0 1 x|
Expanding the determinant along the first row, we get:
x |-sin 0 1| – sin 0 |cos 0 1| |1 x| |0 cos 0|
= -xsin0(cos0) – sin0(1)(x) + (cos0)(1)(0) = -xsin0
Therefore, we can see that the determinant of the given matrix is -xsin0, which is not equal to x. Hence, the given matrix does not satisfy the determinant equation.
