260S0708M2Sol - METU Math 260 Last Name Name ID Number 4...

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Instructor Time Duration Signature : : : : : : Last Name Name ID Number M E T U Department of Mathematics Math 260 Basic Linear Algebra Exam 2 01.08.2008 13: 00 100 minutes 4 QUESTIONS ON 4 PAGES TOTAL 30 POINTS ANSWER KEY 1 2 3 4 (1) (3 + 3 + 1 = 7 pts.) Let W = { ( x 1 ,x 2 ,x 3 ,x 4 ,x 5 ) R 5 : x 1 - 2 x 2 = 0 , x 3 + 4 x 4 = 0 } . (a) Show that W is a subspace of R 5 . The zero vector (0 , 0 , 0 , 0 , 0) is contained in W since its components satisfy the relations x 1 - 2 x 2 = 0 , x 3 + 4 x 4 = 0 . So, W 6 = . Let u = ( x 1 , ··· ,x 5 ) ,v = ( y 1 , ··· ,y 5 ) W, that is, x 1 - 2 x 2 = 0 , x 3 + 4 x 4 = 0 and y 1 - 2 y 2 = 0 , y 3 + 4 y 4 = 0 , and let c R . Now, u + v = ( x 1 + y 1 , ··· ,x 5 + y 5 ) , and cu = ( cx 1 , ··· ,cx 5 ) . Since ( x 1 + y 1 ) - 2( x 2 + y 2 ) = x 1 - 2 x 2 + y 1 - 2 y 2 = 0 and ( x 3 + y 3 ) + 4( x 4 + y 4 ) = x 3 +4 x 4 + y 3 +4 y 4 = 0 we have u + v W which means that W is closed under addition. Moreover, ( cx 1 ) - 2( cx 2 ) = c ( x 1 - 2 x 2 ) = 0 and ( cx 3 )+4( cx 4 ) = c ( x 3 +4 x 2 ) = 0 implies that cv W, that is, W is closed under scalar multiplication. Thus, W is a subspace of R 5 . (b) Find a basis for W. We should solve the homogeneous system x 1 - 2 x 2 = 0 , x 3 + 4 x 4 = 0 , which has the coefficient matrix " 1 - 2 0 0 0 0 0 1 4 0 # , from which it can be seen that x 2 ,x 4 and x 5 are free variables. Letting x 2 = 1 ,x 4 = x 5 = 0 we obtain x 1 = 2 ,x 3 = 0 and hence u 1 = (2 , 1 , 0 , 0 , 0) is a fundamental solution. For
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This note was uploaded on 05/26/2010 for the course MATHEMATIC 260 taught by Professor Uguz during the Spring '10 term at Middle East Technical University.

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260S0708M2Sol - METU Math 260 Last Name Name ID Number 4...

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