The characteristic polynomial is det(A - λI) = (2-λ)(2-λ) - 1 = λ^2 - 4λ + 3 = 0, giving eigenvalues 1 and 3.

The inverse of A is (1/det(A)) * adj(A) = (1/-2) * [[4, -2], [-3, 1]] = [[-2, 1]; [1.5, -0.5]].

AB = [[1*2 + 0*4, 1*3 + 0*5], [0*2 + 1*4, 0*3 + 1*5]] = [[2, 3], [4, 5]].

A^n = I for any integer n, where I is the identity matrix.

A + B = [[1+5, 2+6], [3+7, 4+8]] = [[6, 8], [10, 12]].

A^2 = A * A = [[1*1 + 2*3, 1*2 + 2*4], [3*1 + 4*3, 3*2 + 4*4]] = [[7, 10], [15, 22]].

The determinant of A is (1*4) - (2*3) = 4 - 6 = -2.

A^2 = A * A = [[1*1 + 2*3, 1*2 + 2*4], [3*1 + 4*3, 3*2 + 4*4]] = [[7, 10], [15, 22]].

The adjoint of A is [[4, -2], [-3, 1]].

The determinant of A is (1*4) - (2*3) = 4 - 6 = -2.

The eigenvalues are found from the characteristic polynomial λ^2 - 5λ + 2 = 0, which gives λ = 5.

The inverse of A is (1/det(A)) * adj(A) = (1/(-2)) * [[4, -2], [-3, 1]] = [[-2, 1], [1.5, -0.5]].

The trace of A is the sum of the diagonal elements: 1 + 4 = 5.

The eigenvalues of A are the diagonal elements: 2 and 3.

Since A^2 = I, A is invertible because the inverse of A is A itself.

The determinant of a matrix is the product of its eigenvalues. Thus, det(A) = 1 * (-1) = -1.

The trace of a matrix is the sum of its eigenvalues. Therefore, trace(A) = 3 + 5 = 8.

The trace of a matrix is the sum of its eigenvalues. Thus, trace(A) = 2 + 3 + 4 = 9.

The characteristic polynomial is det(A - λI) = det([[1-λ, 2], [3, 4-λ]]) = (1-λ)(4-λ) - 6 = λ^2 - 5λ + 2.

The characteristic polynomial is given by det(A - λI) = det([[2-λ, 1], [1, 2-λ]]) = (2-λ)(2-λ) - 1 = λ^2 - 3λ + 3.

The determinant is calculated as (0*0) - (1*1) = 0 - 1 = -1.

The inverse of A is A itself since A is an involutory matrix.

The inverse of A is (1/det(A)) * adj(A) = (1/-2) * [[4, -2], [-3, 1]] = [[-2, 1], [1.5, -0.5]].

The rows of A are linearly dependent, hence the rank is 2.

The rank of the matrix is 2, as the rows are linearly dependent.

The trace of A is the sum of the diagonal elements: 1 + 5 + 9 = 15.