show that if A= A inverse, then det(A) = +/- 1

if A(transpose) = A(inverse), what is det(A)?

To show that if A = A inverse, then det(A) = +/-1, we can use the properties of determinants and inverse matrices.

First, let's assume that A is a square matrix of size n x n, and A is equal to its own inverse, which means A * A = I, where I is the identity matrix.

Next, let's recall the properties of inverse matrices:

1. A * A^(-1) = A^(-1) * A = I (Definition of inverse matrix)
2. det(A * B) = det(A) * det(B) (Property of determinants)
3. det(I) = 1 (Determinant of identity matrix is 1)

Now, to find the determinant of A, we can multiply both sides of A * A = I by the determinant of A:

det(A * A) = det(I)

Using property 2 above, we get:

det(A) * det(A) = 1

Taking the square root of both sides, we have:

det(A) = +/- 1

Therefore, if A is equal to its own inverse, then the determinant of A is either +1 or -1.

Now, let's consider the case where A(transpose) = A(inverse), and we want to find det(A).

If A(transpose) = A(inverse), it means that A * A = I, and A = A(transpose).

To find det(A), we can use the same approach as before:

det(A * A) = det(I)

Since A = A(transpose), we can rewrite it as:

det(A * A(transpose)) = det(I)

Using property 2 again, we get:

det(A) * det(A(transpose)) = 1

Since A = A(transpose), we have:

det(A) * det(A) = 1

Taking the square root of both sides, we find:

det(A) = +/- 1

Therefore, if A(transpose) = A(inverse), the determinant of A is also either +1 or -1.