midterm 1 spring 2003 solutions

Linear Algebra with Applications

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Unformatted text preview: Math 21b Midterm ISolutions Thursday, March 6, 2003 1. (12 points) True or False. No justification is necessary, simply circle T or F for each statement. T F (a) If { v 1 , v 2 , v 3 } is a linearly independent set in R n , then { v 1 + v 2 , v 2 + v 3 , v 1 + v 3 } is also a linearly independent set in R n . Solution. This statement is true. Suppose that a ( v 1 + v 2 )+ b ( v 2 + v 3 )+ c ( v 1 + v 3 ) = ( a + c ) v 1 +( a + b ) v 2 +( b + c ) v 3 = . Then a + c = 0 a + b = 0 b + c = 0 , since { v 1 , v 2 , v 3 } is a linearly independent set. If we solve this system, then we see that a = b = c = 0. Therefore, the vectors v 1 + v 2 , v 2 + v 3 , v 1 + v 3 must be linearly independent. T F (b) It is possible to have a 5 3 matrix A such that the dimension of the kernel of A is four. Solution. This statement is false. Since rank( A ) + nullity( A ) = 3, the dimension of the kernel of A must be less than or equal to 3. T F (c) If A 2 + 2 A- 5 I 3 = 0 for a 3 3 matrix A , then A is invertible. Solution. This statement is true. Solving for I 3 , I 3 = A 1 5 A + 2 5 I 3 . Therefore, A- 1 = 1 5 A + 2 5 I 3 . T F (d) If A and B are n n matrices and x is in the kernel of A , then x must also be in the kernel of AB . 1 Solution. This statement is false. Let A = 1 0 0 0 and B = 0 1 1 0 . Then x = 1 is in the kernel of A but is not in the kernel of AB . T F (e) Row operations on an m n matrix A can change the kernel of A . Solution. This statement is false, since it statement contradicts Gauss-Jordan elimination. T F (f) If A and B are m n matrices, then rank( A + B ) = rank( A ) + rank( B ) ....
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This homework help was uploaded on 01/23/2008 for the course MATH 21B taught by Professor Judson during the Spring '03 term at Harvard.

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midterm 1 spring 2003 solutions - Math 21b Midterm...

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