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Unformatted text preview: . Math 331 – Wylie – Midterm 2 – Solutions – November 3, 2011 1. True of False: (a) If A and B are invertible n × n matrices then AB = BA . False (b) If an n × n matrix A has linearly independent columns and the columns of another n × n matrix B span R n , then the columns of the matrix AB are linearly independent. True (c) If the rank of a 3 × 5 matrix is two, then Nul( A ) is a three dimensional subspace of R 5 . True (d) There is an onto linear transformation T : R 2 → R 3 . False (e) If B can be obtained from A by a sequence of row operations and A is invertible, then so is B . True 2. Let T ( x 1 ,x 2 ,x 3 ) = ( x 1 + 2 x 2 + 3 x 3 ,x 2 + 2 x 3 ,x 2 + 3 x 3 ,x 1 + 2 x 2 + 3 x 3 ) (a) Show that T is a linear mapping by finding a matrix (the standard matrix) which im plements the mapping. Solution: T x 1 x 2 x 3 = x 1 + 2 x 2 + 3 x 3 x 2 + 2 x 3 x 2 + 3 x 3 x 1 + 2 x 2 + 3 x 3 = x 1 1 1 + x 2 2 1 1 2 + x 3 3 2 3 3 = 1 2 3 0 1 2 0 1 3 1 2 3 x 1 x 2 x 3 So T ( x ) = A x where A = 1 2 3 0 1 2 0 1 3 1 2 3 , so T is a linear mapping with standard matrix A . (b) Determine whether T is one to one. Solution: T is one to one if there are no nontrivial solutions to T ( x ) = , which is equivalent to showing there is only one solution to A x = . We put A in echelon form. 1 2 3 0 1 2 0 1 3 1 2 3 IV I→ 1 2 3 0 1 2 0 1 3 0 0 0 III II→ 1 2 3 0 1 2 0 0 1 0 0 0 Since there is a pivot in every column, there is only the trivial solution to A x = , so T is one to one. 3. Let A = 1 3 1 , B = 1 0 2 2 0 1 , C = 1 1 1 1 . One of the two combinations, ( AB ) T C , C ( AB ) T is well defined. Compute the well defined combination. Solution: AB = 1 3 1 1 0 2 2 0 1 = 1 0 2 1 0 7 ( AB ) T = 1 1 0 0 2 7 ....
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 Fall '09
 Linear Algebra, Matrices, Det

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