Chapter 1 Notes - ISM Linear Algebra Section 1.1 Chapter 1...

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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 0 = - 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 = 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 0 = - 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 , 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 = 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. 12. x - 2 y = 3 2 x - 4 y = 6 - 2( I ) x - 2 y = 3 0 = 0 2
ISM: Linear Algebra Section 1.1 Figure 1.1: for Problem 1.1.11. This system has infinitely many solutions: If we choose y = t , an arbitrary real number, then the equation x - 2 y = 3 gives us x = 3 + 2 y = 3 + 2 t . Therefore the general solution is ( x, y ) = (3 + 2 t, t ), where t is an arbitrary real number. (See Figure 1.2.) Figure 1.2: for Problem 1.1.12. 13. x - 2 y = 3 2 x - 4 y = 8 - 2( I ) x - 2 y = 3 0 = 2 , which has no solutions. (See Figure 1.3.) Figure 1.3: for Problem 1.1.13. 3
Chapter 1 ISM: Linear Algebra 14. The system reduces to x + 5 z = 0 y - z = 0 0 = 1 , so that there is no solution; no point in space belongs to all three planes. Compare with Figure 2b. 15. The system reduces to x = 0 y = 0 z = 0 so the unique solution is ( x, y, z ) = (0 , 0 , 0). The three planes intersect at the origin.

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