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Applied Finite Mathematics HW Solutions 74

Applied Finite Mathematics HW Solutions 74 - 72 EXERCISES...

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Unformatted text preview: 72 ' EXERCISES 3.7 1 2 —3 13 —> 0 H3 2 ~13 -—31 0 0 0 'When we convert the last row to an equation, We obtain 0 = —31, an impossibiiity. The system has no solutions; it is inconsistent. l .11 4 11. Consider finding the inverse of the coefficient matrix A = (1 3 45) . 1 1 -—1 1 —1 4 1 O 0 1 w1 4 1 0 0 ' 1 3 —6 U 1 0 R; —> —R1 + R; -—r 0 4 —1D —1 1 0 R2 —> —2R3 —I— R; 1 1 -1 0 0 1 R3 —¥ —R1 + R3 0 2 -5 -1 0- 1 1—14 —>000 02—5 1 0 0 1 I —2 —1 0 1 Since the inverse matrix does not exist, we consider the augmented matrix. 1 —1 4 o 1 —1 4 e 1 3 —s 0 Rg—r—R1+R2 —. 0 4 —10 0 R2—¥R3 1 1 _1 0 Rg—r—R1+Ra e 2 —5 e R3—>R2/2 . 1 H1 4 0 1 —1 4 e —: 0 2 ~5 0 —: 0 2 —5 0 Rz—FR2/2 0 2 “5 D)R3—>—~R2+R3 (o 0 u o) _ I -—-1 4 —) D 1 —5/2 D U U o 1 0 3/2 0 R1—>R2+R1 —» o 1 —5/2 0 o 0‘ o 0 D D We now convert this augmented matrix to an equivalent system of equations, 32: $+-§'—-U, 52: ””3"” The system therefore has an infinity of solutions, - 33 52: . chm—f, y—-2--, zarbitrary. 2 2 —3 12. Consider finding the inverse of the coefficient matrix A = 2 1 —1 . 3 —2 1 22—3100 22—3100R1—rR3 2 1 -'1 D 1 D —1 2 I —1 0 1 U .3 H2 1 D 01 Rs—rfiR1+Rs 1 *4 4 -1 01 R3—rR1 —1_01 21—2 ' 1 —4 4 —} 2 l —1 3 0 —2 Ra—r—Rg-FR3 410 1 1 —4 4 0 1 o Rg—r—2R1+R2—b o 9 —9 2 2 —3 1 o o Rafi—231+R3 o 10 —1'1 _'_A I____L__4.I_ _ YY_:_.__:I.. -fll--!J-L- “4“...4--- I7...nL_-__..A.._K:.u. 2.. find nu L. a." :. “OJ—DI" ML:L:‘AA I! "fl—“H‘s ...
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