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Unformatted text preview: Math 152, Spring 2010 Solutions assignment #9 Notes: Each question is worth 5 marks. Due in class: Monday, March 22 for MWF sections; Tuesday, March 23 for TTh sections. Solutions will be posted Tuesday, March 23 in the afternoon. No late assignments will be accepted. 1. Find all the values of for which the matrix A = 2 1  1 1  2 is not invertible. Solution: The matrix A is not invertible if and only if det( A ) = 0. In this case we have det( A ) = ( (  2) 1)+2(1) = 3 2 2 +2 = (  2)(  1)( +1) . This shows that det( A ) = 0 if and only if (  2)(  1)( + 1) = 0. Therefore the values of for which A is not invertible are = 1 , 1 , 2. 2. Consider the following linear system x + y = 1 3 x + 2 y + 1 z = 1 x 2 y + z = 5 (a) Write this system in the form A x = b for a matrix A and a vector b . Solution: The system can be written in the form A x = b , where the matrix A and the vector b can be read from the system. In this case we get A = 1 1 3 2 1 1 2 1 and b = 1 1 5 (b) Compute det( A ). Is the matrix A invertible? Please explain....
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 Spring '08
 Caddmen
 Math

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