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Unformatted text preview: Section 2.7
1. [AI2 ] =
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.
 Fall '10
 Dreibelbis

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