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Applied Finite Mathematics HW Solutions 67

# Applied Finite Mathematics HW Solutions 67 - EXERCISES 3.6...

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Unformatted text preview: EXERCISES 3.6 65 3- 3 4 1 0 Rl—r—Rg+R1__) 1—1 1‘—1 01 _ 2 5 0 1 Rg—)—2R1+R2 _H 1—1 1 —1 H 1 —1 1 4.1314132“?1 0 7 —2 3 Jag—.132” 0 1 —2/7 3/7 3(1 21.511 :37) 7 -2 3 9- 1 3 1 0 H 2 0 0 1 Rg—r—QRl-i-Rz '0 —5 The inverse matrix is l ( 5 _4). 1 3 1 0 —-2 1 Rg—i—Rz/ﬁ a 13 1 0 R1—1—3Rg+R1__) 1 0 0 1/2 0 1 1/3 —1/6 0 1 1/3 —1/6 . . . 1 0 3 The Inverse matmc 15 E (2 _1). 10. 4 2 1 0 RIHR'Q __, 2 1 0 1 210 1 32—131 4 2 1 0 R2—1w2R1+R2 92101 001#2 There is no inverse matrix. 1 —5 10 10 The inverse matrix is m —-3 1 11 . 2593345 9223100 R1—1R3 1 412010 E4 1210011334131 _2 13. hill-IN . _ . A. _ Tr._:____:._. . -01 :-_'.;-|... D-..I...«-m Dual...- mmuhmlim in rum-I nr in ﬁll] in sfridlv ”011111?le 11- . 1 —3 1 0 - _} 1 —3 1 0 2 —6 0 1 R2 —1 “ZRI + R2 0 0 —2 1 There is no inverse matrix. 12. _ ' 1 2 4 1 0 0 1 2 4 1 0 0 3 —2 1 0 1 0 R2 —1 —3R1+R2 .__, 0 —8 —11 —3 1 0 R2 —) 3R3 + R2 0 3 1 0 0 1 ‘ 0 3 1 0 0 1 1 2 4 1 0 0 R1 —) —2R2 + R1 1 0 20 7 —2 —6 ——> 0 1 —8 —3 1 3 ——» 0 1 —8 m3 1 3 0 3 1 0 0 1 R3—r—3R2+Ra 0 0 25 9 —3 —8 R3—1R3/25 1 .0 20 7 —-2 —6 R1 —1 —-20R3 +R1 1 0 0 —1/5 ' 2/5 2/5 .__, 0 1 —-8 —3 1 I 3 R2 —1 833 + R2 ——? 0 1 0 —-3/25 1/25 11/25 0 0 1 9/25 HIS/25 —-8/25 0 0 1 9/25 —3/25 ~8/25 ) ...
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