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math 235 hw3 - Homework 3 Due March 2 Adam Gamzon Related...

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Homework 3, Due March 2 Adam Gamzon Related reading: sections 2.4 - 3.2 from the text 1. Determine whether or not 1 0 0 , 1 2 0 , 1 2 3 are linearly independent. Solution: To do this, solve the system a 1 0 0 + b 1 2 0 + c 1 2 3 = 0 . Using Gauss-Jordan elimination, 1 1 1 | 0 0 2 2 | 0 0 0 3 | 0 1 1 1 | 0 0 1 1 | 0 0 0 3 | 0 1 0 0 | 0 0 1 1 | 0 0 0 3 | 0 1 0 0 | 0 0 1 1 | 0 0 0 1 | 0 1 0 0 | 0 0 1 0 | 0 0 0 1 | 0 . Hence the only solution is a = b = c = 0, so these vectors are linearly independent. 2. Find the inverse of the matrix 0 0 1 0 1 0 1 0 1 . Solution: 0 0 1 | 1 0 0 0 1 0 | 0 1 0 1 0 1 | 0 0 1 1 0 1 | 0 0 1 0 1 0 | 0 1 0 0 0 1 | 1 0 0 1 0 0 | - 1 0 1 0 1 0 | 0 1 0 0 0 1 | 1 0 0 1
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So the inverse is - 1 0 1 0 1 0 1 0 0 . 3. (a) Show that if A 2 is invertible then A is invertible. Solution: If A 2 is invertible then there is a matrix, ( A 2 ) - 1 , such that A 2 ( A 2 ) - 1 = I n . Since A 2 = AA , associativity of matrix multiplication gives I n = A 2 ( A 2 ) - 1 = A ( A ( A 2 ) - 1 ). That is, A is invertible and its inverse is A ( A 2 ) - 1 . (b) Let T : R n R m be a linear transformation. If vectors v and w are in ker( T ), is the vector v + 3 w in ker( T )? Why? What is T ( v + 3 w )? Solution: The vector v + 3 w is in ker( T ) because T ( v + 3 w ) = T ( v ) + 3 T ( w ) = 0 + 0 = 0 . (Alternatively, one could argue that the kernel is a subspace, so v + 3 w must be in ker( T ) because subspaces are closed under addition and scalar multiplication.) 4. (a) For the matrix A = 0 1 0 0 0 1 0 0 0
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