Solution Manual For A First Course in the Finite Element Method 6th Edition by Daryl L. Logan chapte

Appendix A A.1  4 0   2 0  6 0  (a) [A] + [B] =      1 8  2 4  3 12 (b) [A] + [C], Nonsense, [A] and [C] not same order

(c) [A] [C]T, Nonsense, columns [A]  rows [C]T  5 2 1  3    (d) [D] [E] =  2 10 0  2    1 0 5  1  5(3)  2(2)  1(1)   20      = 2(3)  10(2)  0(1)  =  26  1(3)  0(2)  5(1)   8     

(e) [D] [C], Nonsense, columns [D]  rows [C] 5 2 1   3 2 0   (f) [C] [D] =   2 10 0   1 0 2   1 0 5  3(5)  (2)(2)  0 6  20  0 3  0  0  =  2  0  0 1  0  10  5  0  2

 19 26 3 =    3 2 9

A.2 [A] =

1 0 1 4

[A]–1 =

[C ]T |[ A] |

C11 = (–1)1 + 1 (8) = 8,

C12 = (–1)1 + 2 (1) = -1

C21 = (–1)2 + 1 (0) = 0,

C22 = (–1)2 + 2 (4) = 4

 8 1 [C] =   0 4 

| [A] | = A11 C11 + A12 C12 = (4) (8) + (0) (-1) = 32  8 0 [C]T =    1 4 

 8 0  1 4   1   –1  [A] = =  4 32   1

32

0  1  8 629

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