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Applied Finite Mathematics HW Solutions 44

Applied Finite Mathematics HW Solutions 44 - 15 16 42...

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Unformatted text preview: 15. 16. 42 . ' EXERCISES 3.4. 1—2 3 '3 —'> 0?.——10 0 _ 1 m2 0 ——} 0 ’i‘ —-10. 0 Rg—I—Rg+R3 0 0 0 '1—2 3 0 ——1 o 1 “10/70 0 0 0 0 We now convert this augmented matrix to an equivalent system of equations, 07—10 0 0)R2 aRz/T 0 m - 21: + 3:: = 0, y — 1—33 = 0. 10 When we solve the second equation for y, we get 11— = ?z When we substitute this into the first equation, we obtain 10 z 2 = = — —. ac— (—- ’i‘ z) + 3z 0 => :1: 7, Thus, there is an infinity of solutions for the system i z 102: . a: -— —§, y_— 7, z arbltrary. We present a sequence of elementary row operations that lead to a. row echelon form for the augmented matrix of the given system. You may, however, use a difi‘erent sequence of operations, resulting in a different row echelon form. The solution of the system, however, must he the same. 1 —-1 1 2 3 2 6 ——1 6 4 I 1 -1 1 5 Rz—i—ZRl—I—Rg --—¥ 0 5 0 R3—1—6R1+R3 0 5 0 1 —1 1 4 —1 0 5 0 —2 --2 000 When we convert the last row to an equation, we get 0- —- —2, an impossibility. The system has no solutions; it is inconsistent. 4) —4 RgewRHRa We present a sequence of elementary row operations that lead to a row echelon form for the augmented matrix of the given system. You may, however, use a difierent sequence of operations, resulting in a. different row echelon form. The solution of the system, however, must be the same. 1 Rl—rRs 1 —-3 —-3 —3 —1 2 -—1 4 2 3 —4 1 2 —1 4 1 m3 —3 133—1121 3 —4 1 . 1—3—3 —1 0510 2 —3 R2 —! 4-2131 + R2 R3 -—| ~3R1+Ra 2 1 —3 —3 2 —? ——» o 5 1o —7 o 5 1o —_5 Rs—f—Ih-i-Rs o o 0 2 I When we convert the last row to an equation, we get 0 2 2, an impossibility. The system has no solutions; it is inconsistent. ___ 1.. __ 1 :_ _ _____u__ _. ...:.L --..._-..L. I-...¢...u.,. n.,:..m=h..ru“:e,.l... Dmbdm final... mcm:m :n ma n.- :n fiull Eu utridlvmhihiml ...
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