Final answer:
It is not correct to assert that det(U) = -1 for a matrix U satisfying U^TU = I; it can only be said that det(U) is either +1 or -1 since U is an orthogonal matrix. The determinant of U^2=I is 1, and since det(U)^2=1, det(U) equals either +1 or -1.
Step-by-step explanation:
In this problem, we are given a square matrix U that satisfies the equation UTU = I, where I is the identity matrix, and UT denotes the transpose of U. We are asked to show that det(U) = -1. However, it's actually not possible to show that det(U) always equals -1 because such a statement is not necessarily true.
If UTU = I, the matrix U is an orthogonal matrix which implies that all columns (and rows) of U are orthonormal vectors. In general, the determinant of an orthonormal matrix can only be ±1. This is because the determinant of the identity matrix is 1, and since UTU = I, it follows that det(UT)det(U) = det(I). And since the determinant of a matrix and its transpose are the same, we have det(U)² = 1, which means det(U) could be either 1 or -1, not necessarily -1.
To 'undo' a square in an equation, as per the example with the Pythagorean Theorem, we would take the square root. In the context of the matrix U, to find the determinant after squaring, we revert to taking the square root of the squared determinant value.