A is similar to B I There exists an invertible matrix P such that A = Pa#39;lBP ...(1) B is similar to C I There exists an invertible matrix Q such that B = Qa#39;lCQ ...(2) Substituting for B from (2) in (1), we get A = Pa#39;1Q_1CQP = (QP)_1 C (QP) = Ra#39;lCR whereanbsp;...
|Title||:||A Textbook of Engineering Mathematics (M.D.U, K.U., G.J.U, Haryana) Sem-II|
|Author||:||N. P. Bali|
|Publisher||:||Laxmi Publications - 2011-12-01|