sol6 - SOLUTIONS TO HOMEWORK 6 Section 3.2 4. (a) We...

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SOLUTIONS TO HOMEWORK 6 Section 3.2 4. (a) We already have 1 in the top left corner. First turn the rest of the ±rst row into zeroes, by the elementary column operations C 2 m→ C 2 C 1 , C 3 m→ C 3 C 1 , and C 4 m→ C 4 C 1 , we arrive at the matrix 1 0 0 0 2 2 3 2 1 0 0 0 Next, use elementary row operations R 2 m→ R 2 2 R 1 and R 3 m→ R 3 R 1 to obtain 1 0 0 0 0 2 3 2 0 0 0 0 To obtain 1 in the (2 , 2) position, use C 2 m→ ( 1 / 2) C 2 . This yields 1 0 0 0 0 1 3 2 0 0 0 0 Now, “clear away” the rest of the non-zero entries via C 3 m→ C 3 +3 C 2 and C 4 m→ C 4 C 2 . As a result, obtain 1 0 0 0 0 1 0 0 0 0 0 0 Thus, the rank of our matrix equals 2 . 5. (d) A is invertible (rank A = 3). Compute A 1 using elementary row operations on the augmented matrix ( A | I 3 ): 0 2 4 1 0 0 1 1 1 0 1 0 2 4 5 0 0 1 −→ R 1 R 2 1 1 1 0 1 0 0 2 4 1 0 0 2 4 5 0 0 1 R 3 R 3 2 R 1 1 1 1 0 1 0 0 2 4 1 0 0 0 2 3 0 2 1 R 2 →− R 2 / 2 1 1 1 0 1 0 0 1 2 1 / 2 0 0 0 2 3 0 2 1 1
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2 SOLUTIONS TO HOMEWORK 6 −→ R 1 R 1 R 2 ,R 3 R 3 2 R 2 1 0 1 1 / 2 1 0 0 1 2 1 / 2 0 0 0 0 1 1 2 1 R 1 R 1 R 3 ,R 2 R 2 +2 R 3 1 0 0 1 / 2 3 1 0 1 0 3 / 2 4 2 0 0 1 1 2 1 Thus, A 1 = 1 / 2 3 1 3 / 2 4 2 1 2 1 (f) rank A = 2, hence A is not invertible. Once again, we apply elementary row operations to ( A | I 3 ). 1 2 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 R 2 R 2 R 1 ,R 3 R 3 R 1 1 2 1 1 0 0 0 2 0 1 1 0 0 1 0 1 0 1 R 2 R 3 1 2 1 1 0 0 0 1 0 1 0 1 0 2 0 1 1 0 R 2 →− R 2 / 2 1 2 1 1 0 0 0 1 0 1 0 1 0 2 0 1 1 0 R 3 R 3 2 R 2 1 2 1 1 0 0 0 1 0 1 0 1 0 0 0 3 1 2 We have arrived at an augmented matrix whose left side has a zero row. Thus, A cannot be invertible: its rank (the number of linearly independent rows) is less than 3 (recall that elementary transformations are rank-preserving). On the other hand, the two non- zero rows of the left side of the augmented matrix are linearly independent (they are not multiples of each other), hence rank A = 2. 6. (a) Let β = { p 0 ,p 1 2 } be the standard basis in P 2 ( R ) (here, p k ( x ) = x k ). Then T ( p 0 ) = p 0 , T ( p 1 ) = 2 p 0 p 1 , and T ( p 2 ) = 4 p 0 + 2 p 1 p 2 . Let A = [ T ] β = 1 2 4 0 1 2 0 0 1 To compute A 1 , we apply row transformations to the augmented matrix: 1 2 4 1 0 0 0 1 2 0 1 0 0 0 1 0 0 1 R 1 →− R 1 ,R 2 →− R 2 ,R 3 →− R 3 1 2 4 1 0 0 0 1 2 0 1 0 0 0 1 0 0 1 R 1 R 1 +2 R 2 1 0 8 1 2 0 0 1 2 0 1 0 0 0 1 0 0 1
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SOLUTIONS TO HOMEWORK 6 3 −→ R 1 R 1 +8 R 3 ,R 2 R 2 +2 R 3 1 0 0 1 2 8 0 1 0 0 1 2 0 0 1 0 0 1 Therefore, [ T 1 ] β = A 1 = 1 2 8 0 1 2 0 0
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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sol6 - SOLUTIONS TO HOMEWORK 6 Section 3.2 4. (a) We...

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