Linear Algebra Solutions 14

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

Ask a homework question - tutors are online