# solq6 - Solutions to quiz 6(October 14 1 Arranging vectors...

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Unformatted text preview: Solutions to quiz # 6 (October 14) 1. Arranging vectors as the columns of a matrix and applying elementary row operations, we obtain 1 1 2 1 2 5 a a- 1 a + 1 -→ 1 1 2 1 3- 1 1- a -→ 1 1 2 0 1 3 0 0 4- a . Hence the kernel of the matrix consists of ~ 0 if and only if a 6 = 4 (in other words, there are no free variables if and only if a 6 = 4). Answer: The vectors are linearly independent for a 6 = 4. 2. Matrix A defines a linear transformation R 3-→ R 5 , ~x 7-→ A~x whereas matrix B defines a linear transformation R 5-→ R 3 , ~x 7-→ B~x . Matrix AB defines a linear trans- formation R 5-→ R 5 , ~x 7-→ ( AB ) ~x = A ( B~x ). Since dimker A + dimim A = 3 , we conclude that dimim A ≤ 3. Alternatively, the image of A is spanned by the three columns of matrix A and hence dimim A ≤ 3. Since the kernel of A lies in R 3 , we have dimker A ≤ 3....
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solq6 - Solutions to quiz 6(October 14 1 Arranging vectors...

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