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Unformatted text preview: Section 1.1 Chapter 1 Section 1.1 1.1. 1 bracketleftbigg x + 2 y = 1 2 x + 3 y = 1 bracketrightbigg 2 1st equation bracketleftbigg x + 2 y = 1 y = 1 bracketrightbigg ( 1) bracketleftbigg x + 2 y = 1 y = 1 bracketrightbigg 2 2nd equation bracketleftbigg x = 1 y = 1 bracketrightbigg , so that ( x,y ) = ( 1 , 1). 1.1. 3 bracketleftbigg 2 x + 4 y = 3 3 x + 6 y = 2 bracketrightbigg 2 bracketleftbigg x + 2 y = 3 2 3 x + 6 y = 2 bracketrightbigg 3 1st equation bracketleftbigg x + 2 y = 3 2 = 5 2 bracketrightbigg . So there is no solution. 1.1. 5 bracketleftbigg 2 x + 3 y = 0 4 x + 5 y = 0 bracketrightbigg 2 bracketleftbigg x + 3 2 y = 0 4 x + 5 y = 0 bracketrightbigg 4 1st equation bracketleftbigg x + 3 2 y = 0 y = 0 bracketrightbigg ( 1) bracketleftbigg x + 3 2 y = 0 y = 0 bracketrightbigg 3 2 2nd equation bracketleftbigg x = 0 y = 0 bracketrightbigg , so that ( x,y ) = (0 , 0). 1.1. 7 x + 2 y + 3 z = 1 x + 3 y + 4 z = 3 x + 4 y + 5 z = 4 I I x + 2 y + 3 z = 1 y + z = 2 2 y + 2 z = 3 2( II ) 2( II ) x + z = 3 y + z = 2 = 1 This system has no solution. 1.1. 9 x + 2 y + 3 z = 1 3 x + 2 y + z = 1 7 x + 2 y 3 z = 1 3( I ) 7( I ) x + 2 y + 3 z = 1 4 y 8 z = 2 12 y 24 z = 6 ( 4) x + 2 y + 3 z = 1 y + 2 z = 1 2 12 y 24 z = 6 2( II ) +12( II ) x z = 0 y + 2 z = 1 2 = 0 This system has infinitely many solutions: if we choose z = t , an arbitrary real number, then we get x = z = t and y = 1 2 2 z = 1 2 2 t . Therefore, the general solution is ( x,y,z ) = ( t, 1 2 2 t,t ) , where t is an arbitrary real number. 1.1. 11 bracketleftbigg x 2 y = 2 3 x + 5 y = 17 bracketrightbigg 3( I ) bracketleftbigg x 2 y = 2 11 y = 11 bracketrightbigg 11 bracketleftbigg x 2 y = 2 y = 1 bracketrightbigg +2( II ) bracketleftbigg x = 4 y = 1 bracketrightbigg , so that ( x,y ) = (4 , 1). See Figure 1.1. 1.1. 13 bracketleftbigg x 2 y = 3 2 x 4 y = 8 bracketrightbigg 2( I ) bracketleftbigg x 2 y = 3 = 2 bracketrightbigg , which has no solutions. (See Figure 1.2.) 1.1. 15 The system reduces to x = 0 y = 0 z = 0 so the unique solution is ( x,y,z ) = (0 , , 0). The three planes intersect at the origin. 1 Chapter 1 Figure 1.1: for Problem 1.1.11. Figure 1.2: for Problem 1.1.13. 1.1. 17 bracketleftbigg x + 2 y = a 3 x + 5 y = b bracketrightbigg 3( I ) bracketleftbigg x + 2 y = a y = 3 a + b bracketrightbigg ( 1) bracketleftbigg x + 2 y = a y = 3 a b bracketrightbigg 2( II ) bracketleftbigg x = 5 a + 2 b y = 3 a b bracketrightbigg , so that ( x,y ) = ( 5 a + 2 b, 3 a b )....
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 Spring '08
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