
Transpose of a matrix
Ex 3.3, 12 If A = [■8(cos 𝛼&〖−sin〗𝛼@sin𝛼&cos𝛼 )] , then A + A’ = I , if the value of α is A. 𝛑/𝟔 B. 𝛑/𝟑 C. π D. 𝟑𝛑/𝟐 A = [■8(cos𝛼&〖−sin〗𝛼@sin𝛼&cos𝛼 )] A’= [■8(cos𝛼&sin𝛼@〖−sin〗𝛼&cos𝛼 )] and I = [■8(1&0@0&1)] Given A + A’ = I [■8(cos𝛼&〖−sin〗𝛼@sin𝛼&cos𝛼 )] + [■8(cos𝛼&sin𝛼@〖−sin〗𝛼&cos𝛼 )] = [■8(1&0@0&1)] [■8(cos〖𝛼+cos𝛼 〗&〖−sin〗〖𝛼+sin𝛼 〗@sin𝛼 〖−sin〗𝛼&cos〖𝛼+cos𝛼 〗 )]= [■8(1&0@0&1)] [■8(2cos𝛼&0@0&2cos𝛼 )]= [■8(1&0@0&1)] Since the matrices are equal, corresponding elements are equal 2cos 𝛼 = 1 cos𝛼 = 1/2 cos𝛼 = cos〖60°〗 Comparing angles 𝛼 = 60° 𝛼 = 60° × 𝜋/(180°) 𝛼 = 𝜋/3 So the correct answer is (B) (As 𝑐𝑜𝑠〖60°〗= 1/2)