102_solutions1

# 102_solutions1 - MATH 102 SOLUTIONS TO HW#1 Section 1.3...

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MATH 102 SOLUTIONS TO HW #1 Section 1.3, problem 4. The correct row operation is - c a × { Row 1 } + { Row 2 } ⇒ { Row 2 } . This pro- duces the equivalent linear system: ax + by = f , ( d - bc a ) y = g - fc a . If ( d - bc a ) = 0 then the formula for y is: y = ( d - bc a ) - 1 ( g - fc a ) . Of course the condition that the second pivot be zero (missing) is the same as saying that ( ad - bc ) = 0. Section 1.3, problem 18. The answer is no , it is not possible. a) If x = ( x, y, z ) and y = ( X, Y, Z ) are both solutions to Ax = b , then clearly so is z = 1 2 x + 1 2 y . b) Then they must also meet in the line joining these two points. Section 1.3, problem 30. First start with: u + v + w = 6 u + 2 v + 2 w = 11 2 u + 3 v - 4 w = 3 We apply, in order, the sequence of row operations: (1) - 1 × { Row 1 } + { Row 2 } ⇒ { Row 2 } (2) - 2 × { Row 1 } + { Row 3 } ⇒ { Row 3 } (3) - 1 × { Row 2 } + { Row 3 } ⇒ { Row 3 } to obtain the equivalent linear system: u + v + w = 6 v + w = 5 - 7 w = - 14 1

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2 One can now easily back-solve to find first that w = 2, then v = 3, and finally that u = 1. Applying the same elimination procedures to the second system we end up with: u + v + w = 7 v + w = 3 - 7 w = - 14
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