Math 33A Answer KEy

Math 33A Answer KEy - ISM: Linear Algebra Section 1.1...

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Unformatted text preview: ISM: Linear Algebra Section 1.1 Chapter 1 1.1 1. x + 2 y = 1 2 x + 3 y = 1-2 1st equation x + 2 y = 1-y =-1 (-1) x + 2 y = 1 y = 1-2 2nd equation x =-1 y = 1 , so that ( x, y ) = (-1 , 1). 2. 4 x + 3 y = 2 7 x + 5 y = 3 4 x + 3 4 y = 1 2 7 x + 5 y = 3-7 1st equation x + 3 4 y = 1 2-1 4 y =-1 2 (-4) x + 3 4 y = 1 2 y = 2-3 4 2nd equation x =-1 y = 2 , so that ( x, y ) = (-1 , 2). 3. 2 x + 4 y = 3 3 x + 6 y = 2 2 x + 2 y = 3 2 3 x + 6 y = 2-3 1st equation x + 2 y = 3 2 =-5 2 So there is no solution. 4. 2 x + 4 y = 2 3 x + 6 y = 3 2 x + 2 y = 1 3 x + 6 y = 3-3 1st equation x + 2 y = 1 = 0 This system has infinitely many solutions: if we choose y = t , an arbitrary real number, then the equation x + 2 y = 1 gives us x = 1-2 y = 1-2 t . Therefore the general solution is ( x, y ) = (1-2 t, t ), where t is an arbitrary real number. 5. 2 x + 3 y = 0 4 x + 5 y = 0 2 x + 3 2 y = 0 4 x + 5 y = 0-4 1st equation x + 3 2 y = 0-y = 0 (-1) x + 3 2 y = 0 y = 0-3 2 2nd equation x = 0 y = 0 , so that ( x, y ) = (0 , 0). 6. x + 2 y + 3 z = 8 x + 3 y + 3 z = 10 x + 2 y + 4 z = 9-I-I x + 2 y + 3 z = 8 y = 2 z = 1-2( II ) x + 3 z = 4 y = 2 z = 1-3( III ) x = 1 y = 2 z = 1 , so that ( x, y, z ) = (1 , 2 , 1). 1 Chapter 1 ISM: Linear Algebra 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. 8. x + 2 y + 3 z = 0 4 x + 5 y + 6 z = 0 7 x + 8 y + 10 z = 0-4( I )-7( I ) x + 2 y + 3 z = 0-3 y-6 z = 0-6 y-11 z = 0 (-3) x + 2 y + 3 z = 0 y + 2 z = 0-6 y-11 z = 0-2( II ) +6( II ) x-z = 0 y + 2 z = 0 z = 0 + III-2( III ) x = 0 y = 0 z = 0 , so that ( x, y, z ) = (0 , , 0). 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. 10. x + 2 y + 3 z = 1 2 x + 4 y + 7 z = 2 3 x + 7 y + 11 z = 8-2( I )-3( I ) x + 2 y + 3 z = 1 z = 0 y + 2 z = 5 Swap : II III x + 2 y + 3 z = 1 y + 2 z = 5 z = 0-2( II ) x-z =-9 y + 2 z = 5 z = 0 + III-2( III ) x =-9 y = 5 z = 0 , so that ( x, y, z ) = (-9 , 5 , 0). 11. x-2 y = 2 3 x + 5 y = 17-3( I ) x-2 y = 2 11 y = 11 11 x-2 y = 2 y = 1 +2( II ) x = 4 y = 1 , so that ( x, y ) = (4 , 1). See Figure 1.1....
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Math 33A Answer KEy - ISM: Linear Algebra Section 1.1...

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