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MIT18_06S10_pset1_s10_soln

# MIT18_06S10_pset1_s10_soln - 18.06 Problem Set 1 Solutions...

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Unformatted text preview: 18.06 Problem Set 1 Solutions Total: 100 points Section 1.2. Problem 23: The figure shows that cos( α ) = v 1 / v and sin( α ) = v 2 / v . Similarly cos( β ) is and sin( β ) is . The angle θ is β − α . Substitute into the trigonometry formula cos( α ) cos( β )+sin( β ) sin( α ) for cos( β − α ) to find cos( θ ) = v w/ v w . · Solution (4 points) First blank: w 1 / w . Second blank: w 2 / w . Substituting into the trigonome- try formula yields cos( β − α ) = ( w 1 / w )( v 1 / v ) + ( w 2 / w )( v 2 / v ) = v w/ v w . · Section 1.2. Problem 28: Can three vectors in the xy plane have u v < and · v w < and u w < 0? · · Solution (12 points) 1 √ 3 1 √ 3 Yes. For instance take u = (1 , 0), v = ( − 2 , 2 ), w = ( − 2 , − 2 ). Notice 1 u v = v w = u w = − 2 . · · · Section 1.3. Problem 4: Find a combination x 1 w 1 + x 2 w 2 + x 3 w 3 that gives the zero vector: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 4 7 w 1 = ⎣ 2 ⎦ ; w 2 = ⎣ 5 ⎦ ; w 3 = ⎣ 8 ⎦ . 3 6 9 Those vectors are (independent)(dependent). The three vectors lie in a . The matrix W with those columns is not invertible . Solution (4 points) Observe w 1 − 2 w 2 + w 3 = 0. The vectors are dependent . They lie in a plane . Section 1.3. Problem 13: The very last words say that the 5 by 5 centered difference matrix is not invertible. Write down the 5 equations Cx = b . Find a 1 combination of left sides that gives zero. What combination of b 1 ,b 2 ,b 3 ,b 4 ,b 5 must be zero? Solution (12 points) The 5 by 5 centered difference matrix is ⎞ ⎛ 1 C = ⎜ ⎜ ⎜ ⎜ ⎝ − 1 1 − 1 1 − 1 1 ⎟ ⎟ ⎟ ⎟ ⎠ . − 1 The five equations Cx = b are x 2 = b 1 , − x 1 + x 3 = b 2 , − x 2 + x 4 = b 3 , − x 3 + x 5 = b 4 , − x 4 = b 5 ....
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MIT18_06S10_pset1_s10_soln - 18.06 Problem Set 1 Solutions...

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