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# ch3 - CHAPTER 3 LINEAR SYSTEMS AND MATRICES SECTION 3.1...

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132 Chapter 3 CHAPTER 3 LINEAR SYSTEMS AND MATRICES SECTION 3.1 INTRODUCTION TO LINEAR SYSTEMS This initial section takes account of the fact that some students remember only hazily the method of elimination for 2 2 × and 3 3 × systems. Moreover, high school algebra courses generally emphasize only the case in which a unique solution exists. Here we treat on an equal footing the other two cases — in which either no solution exists or infinitely many solutions exist. 1. Subtraction of twice the first equation from the second equation gives 5 10, y = − so y = 2, and it follows that x = 3. 2. Subtraction of three times the second equation from the first equation gives 5 15, y = − so y = –3, and it follows that x = 5. 3. Subtraction of 3/2 times the first equation from the second equation gives 1 3 , 2 2 y = so y = 3, and it follows that x = –4. 4. Subtraction of 6/5 times the first equation from the second equation gives 11 44 , 5 5 y = so y = 4, and it follows that x = 5. 5. Subtraction of twice the first equation from the second equation gives 0 1, = so no solution exists. 6. Subtraction of 3/2 times the first equation from the second equation gives 0 1, = so no solution exists. 7. The second equation is –2 times the first equation, so we can choose y = t arbitrarily. The first equation then gives 10 4 . x t = − + 8. The second equation is 2/3 times the first equation, so we can choose y = t arbitrarily. The first equation then gives 4 2 . x t = + 9. Subtraction of twice the first equation from the second equation gives 9 4 3. y z = − Subtraction of the first equation from the third equation gives 2 1. y z + = Solution of

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Section 3.1 133 these latter two equations gives 1, 3. y z = − = Finally substitution in the first equation gives x = 4. 10. Subtraction of twice the first equation from the second equation gives 3 5. y z + = − Subtraction of twice the first equation from the third equation gives 2 3. y z = Solution of these latter two equations gives 1, 2. y z = = − Finally substitution in the first equation gives x = 3. 11. First we interchange the first and second equations. Then subtraction of twice the new first equation from the new second equation gives 7, y z = and subtraction of three times the new first equation from the third equation gives 2 3 18. y z + = − Solution of these latter two equations gives 3, 4. y z = = − Finally substitution in the (new) first equation gives x = 1. 12. First we interchange the first and third equations. Then subtraction of twice the new first equation from the second equation gives 7 3 36, y z = − and subtraction of twice the new first equation from the new third equation gives 16 7 83. y z = − Solution of these latter two equations gives 3, 5. y z = = Finally substitution in the (new) first equation gives x = 1.
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ch3 - CHAPTER 3 LINEAR SYSTEMS AND MATRICES SECTION 3.1...

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