Linear Algebra Solutions 23

Linear Algebra Solutions 23 - b 1 c-1 ‚ =-1 b c...

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Notice that A n 1 7 3 3 4 4 as n → ∞ . This problem is a special case of a more general result about Markov matrices. 20. Let A = a b c d be a matrix whose elements are non–negative real numbers satisfying a 0 , b 0 , c 0 , d 0 , a + c = 1 = b + d. Also let P = b 1 c - 1 and suppose that A 6 = I 2 . (i) det P = - b - c = - ( b + c ). Now b + c 0. Also if b + c = 0, then we would have b = c = 0 and hence d = a = 1, resulting in A = I 2 . Hence det P < 0 and P is non–singular. Next, P - 1 AP = - 1 b + c - 1 - 1 - c b ‚• a b c d ‚• b 1 c - 1 = - 1 b + c - a - c - b - d - ac + bc - cb + bd ‚• b 1 c - 1 = - 1 b + c - 1 - 1 - ac + bc - cb + bd ‚•
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Unformatted text preview: b 1 c-1 ‚ =-1 b + c •-b-c (-ac + bc ) b + (-cb + bd ) c-ac + bc + cb-bd ‚ . Now-acb + b 2 c-c 2 b + bdc =-cb ( a + c ) + bc ( b + d ) =-cb + bc = 0 . Also-( a + d-1)( b + c ) =-ab-ac-db-dc + b + c =-ac + b (1-a ) + c (1-d )-bd =-ac + bc + cb-bd. Hence P-1 AP =-1 b + c •-( b + c )-( a + d-1)( b + c ) ‚ = • 1 a + d-1 ‚ . 26...
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