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May2005Solutions

# May2005Solutions - MAT223 Final April-May 2005 Note Ill...

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MAT223 Final April-May 2005 Note: I’ll write column vectors horizontally and with commas sometimes, to save space. Part I 1. (2). If k = 2 the last equation disappears and the system has infinitely many solutions. For any other values of k , the last equation simplifies to ( k + 2) x 3 = 1 . This is inconsist with the second equation unless k + 2 = 1 2 , in which case again the system has infinitely many solutions. 2. (4). B is true because det A 2 = ( det A ) 2 . D is equivalent to dim ( row A ) = 0 , i.e., A = 0 . C is true by a theorem and for A you can easily find counterexamples, for instance A = diag ( 1 , 1 , 1 , 1 , , 1) . 3. (2). C is not given by a homegenous linear condition, A and B are. 4. (2). The two equations are clearly independent, so the dimension is 4 2=2 . 5. (4). The second column of A is 2 times the first, so it is unnecessary. Perform Gram- Schmidt on the other two. 6. (3). W = span { [1 , 0 , 0] , [0 , 1 , 1] } and that basis is orthogonal, so the required distance is the norm of X proj W X = X 3[1 , 0 , 0] 4[0 , 1 , 1]=[0 , 2 , 2] , that is, 2 2 . 7. (1). det ( 2 A 3 adjA ) =( 2) 3 ( det A ) 3 det ( adj A ) = 2 3 2 3 det ( adj A ) = det ( adj A ) = det (( detA ) A 1 )= ( det A ) 3 ( det A ) 1 = 4 .

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May2005Solutions - MAT223 Final April-May 2005 Note Ill...

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