Matrices & Determinants
Orthogonal Matrices
Grade Class 12
Question:
If A and B are two orthogonal matrices of order 3, then -<br>(A) A and B both will be invertible matrices<br>(B) matrix ABA will also be orthogonal<br>(C) matrix A^2B^2 will also be orthogonal<br>(D) maximum value of det\left(\frac{A}{2} adj(2B)\right) is 8.
(A) A and B both will be invertible matrices
(B) matrix ABA will also be orthogonal
(C) matrix A^2B^2 will also be orthogonal
(D) maximum value of det\left(\frac{A}{2} adj(2B)\right) is 8.
Step-by-Step Solution
Key Concept: An orthogonal matrix A satisfies A^T A = I, which implies det(A) = \pm1, so it is always invertible. The product of orthogonal matrices is orthogonal. For (D), use properties of determinant and adjoint: det(kA) = k^n det(A) and det(adj(M)) = (det(M))^(n-1).
Since A and B are orthogonal, A^T A = I and B^T B = I. Thus, det(A)^2 = 1 and det(B)^2 = 1, so det(A) = \pm1 and det(B) = \pm1. Hence, A and B are invertible. (A) is correct. For (B), (ABA)^T (ABA) = A^T B^T A^T A B A = A^T B^T I B A = A^T (B^T B) A = A^T I A = A^T A = I. So ABA is orthogonal. (B) is correct. For (C), (A^2 B^2)^T (A^2 B^2) = (B^2)^T (A^2)^T A^2 B^2 = (B^T)^2 (A^T)^2 A^2 B^2 = (B^T B)^2 (A^T A)^2 = I^2 I^2 = I. So A^2 B^2 is orthogonal. (C) is correct. For (D), det(A/2 adj(2B)) = (1/2)^3 det(A) det(adj(2B)) = (1/8) det(A) (det(2B))^(3-1) = (1/8) det(A) (2^3 det(B))^2 = (1/8) det(A) (8 det(B))^2 = (1/8) det(A) (64 det(B)^2) = 8 det(A) det(B)^2. Since det(B)^2 = 1, this is 8 det(A). The maximum value is 8 when det(A) = 1. (D) is correct.
Correct Answer: A,B,C,D