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Math254Mid3Sols - MIDTERM 3 SOLUTIONS MATH 254 SUMMER 2002...

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MIDTERM 3 - SOLUTIONS MATH 254 - SUMMER 2002 - KUNIYUKI CHAPTERS 6, 7 GRADED OUT OF 75 POINTS ¥ 2 = 150 POINTS TOTAL 1) The linear transformation T R R : 2 3 Æ is such that T 1 1 3 4 1 , , , ( ) ( ) = - and T 0 1 1 5 3 , , , ( ) ( ) = . Find T 4 5 , ( ) . Hint: Remember the definition of a linear transformation. (5 points) Express (4,5) as a linear combination of the given basis vectors for R 2 , (1,1) and (0,1). Observe that (4,5) = (4,4) + (0,1) = 4(1,1) + (0,1). T T T T 4 5 4 1 1 0 1 4 1 1 0 1 4 3 4 1 1 5 3 12 16 4 1 5 3 11 21 7 , , , , , , , , , , , , , , , ( ) = ( ) + ( ) ( ) = ( ) + ( ) = - ( ) + ( ) = - ( ) + ( ) = - ( ) 2) The linear transformation T R R : 2 2 Æ is such that T v v v v v v 1 2 1 2 1 2 2 2 3 , , ( ) ( ) = - + . Find the preimage of 17 5 , ( ) . (7 points) We want to solve the system (maybe by using Gauss-Jordan elimination) 2 17 2 3 5 1 2 1 2 v v v v - = + = Ï Ì Ó 2 1 2 3 17 5 - È Î Í ˘ ˚ ˙ Subtract Row 1 from Row 2. R R R 2 1 2 1 + - ( ) Æ . 2 1 0 4 17 12 - - È Î Í ˘ ˚ ˙ Divide Row 2 through by 4. 2 1 0 1 17 3 - - È Î Í ˘ ˚ ˙ Add Row 2 to Row 1. R R R 1 2 1 + Æ .
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2 0 0 1 14 3 - È Î Í ˘ ˚ ˙ Divide Row 1 through by 2. 1 0 0 1 7 3 - È Î Í ˘ ˚ ˙ The preimage is 7 3 , - ( ) { } . 3) The linear transformation T R R : 5 7 Æ is such that dim Ker T ( ) ( ) = 3 . (11 points total) a) What is the domain of T ? (2 points) R 5 b) What is nullity T ( ) ? (2 points) 3, since nullity( T ) = dim(Ker( T )). c) What is rank T ( ) ? (3 points) 2, since rank( T ) + nullity( T ) = dimension of domain = 5. d) True or False: Range T ( ) is a subspace of R 7 . Circle one: (2 points) True False Note that Range( T ) = Col( A ), where A is a 7 5 ¥ matrix.
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