In the given matrix, one of the eigenvalues is 1. the eigenvectors corresponding to the eigenvalue 1 are
⎡ 1 -1 2 ⎤ ⎢ 0 1 0 ⎥ ⎣ 1 2 1 ⎦
(A) A
(B) B
(C) C
(D) D
Answer: (B)
Explanation: Let z represents the eigenvalues.
And let the given matrix be A (square matrix of order 3 x3) The characteristic equation for this is : AX = zX ( X is the required eigenvector ) AX - zX = 0 [ A - z I ] [X] = 0 ( I is an identity matrix of order 3 ) put z = 1 ( because one of the eigenvalue is 1 ) [ A - 1 I ] [X] = 0 The resultant matrix is : [ 0 -1 2 ] [x1] [0] | 0 0 0 ] |x2] =|0| [ 1 2 0 ] |x3] [0] Multiplying thr above matrices and getting the equations as: -x2 + 2x3 = 0 ----------------(1) x1 + 2x2 = 0-----------------(2) now let x1 = k, then x2 and x3 will be -k/2 and -k/4 respectively. hence eigenvector X = { (k , -k/2, -k/4) } where k != 0 put k = -4c ( c is also a constant, not equal to zero ), we get X = { ( -4c, 2c, 1c ) }, i.e. { c ( -4, 2, 1 ) } Hence option B.