Recitation2Sol - ORIE 3300/5300 - Fall 2010 Solutions for...

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ORIE 3300/5300 - Fall 2010 Solutions for Recitation 2 1. Suppose y and z are two solutions of Ax = b , that is, Ay = b and Az = b . Hence, we get A ( y - z ) = 0. Let u := y - z . By substitution, we have Au = 0. So, if A 1 ,A 2 ,...A n are the columns of A , then n j =1 u j A j = 0. The linear independence of the columns of A implies that u j = 0 j = 1 ,...n . This implies that y = z . Hence, there is only one solution of Ax = b . 2. (a) The column is linearly independent. The column does not span (not enough columns to span). There is no basis. The matrix is not invertible (only a square matrix can be invertible). (b) The columns are linearly independent. The columns span. There are two bases: [1,2] and [2,1] The matrix is invertible. (c) The columns are not linearly independent: we have 2 ± 1 - 3 ² + ± - 2 6 ² = ± 0 0 ² The columns do not span. There is no basis. The matrix is not invertible (because the columns are not linearly independent). (d) The columns are linearly dependent (there are too many columns to be linearly
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This note was uploaded on 01/25/2011 for the course ORIE 5300 taught by Professor Todd during the Fall '08 term at Cornell University (Engineering School).

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Recitation2Sol - ORIE 3300/5300 - Fall 2010 Solutions for...

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