Linear Algebra Solutions 14

Linear Algebra Solutions 14 - Section 2.7 1. [A|I2 ] = 1 13...

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Unformatted text preview: Section 2.7 1. [A|I2 ] = 1 13 R2 R2 → 14 −3 1 14 01 10 01 R2 → R2 + 3R1 1 0 3/13 1/13 14 0 13 R1 → R1 − 4R2 Hence A is non–singular and A−1 = 10 31 10 01 1/13 −4/13 . 3/13 1/13 1/13 −4/13 . 3/13 1/13 Moreover E12 (−4)E2 (1/13)E21 (3)A = I2 , so A−1 = E12 (−4)E2 (1/13)E21 (3). Hence A = {E21 (3)}−1 {E2 (1/13)}−1 {E12 (−4)}−1 = E21 (−3)E2 (13)E12 (4). 2. Let D = [dij ] be an m × m diagonal matrix and let A = [ajk ] be an m × n matrix. Then n dij ajk = dii aik , (DA)ik = j =1 as dij = 0 if i = j . It follows that the ith row of DA is obtained by multiplying the ith row of A by dii . Similarly, post–multiplication of a matrix by a diagonal matrix D results in a matrix whose columns are those of A, multiplied by the respective diagonal elements of D. In particular, diag (a1 , . . . , an )diag (b1 , . . . , bn ) = diag (a1 b1 , . . . , an bn ), as the left–hand side can be regarded as pre–multiplication of the matrix diag (b1 , . . . , bn ) by the diagonal matrix diag (a1 , . . . , an ). Finally, suppose that each of a1 , . . . , an is non–zero. Then a−1 , . . . , a−1 n 1 all exist and we have diag (a1 , . . . , an )diag (a−1 , . . . , a−1 ) = diag (a1 a−1 , . . . , an a−1 ) n n 1 1 = diag (1, . . . , 1) = In . Hence diag (a1 , . . . , an ) is non–singular and its inverse is diag (a−1 , . . . , a−1 ). n 1 17 ...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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