midterm 1 spring 2003 solutions

# Linear Algebra with Applications

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Math 21b Midterm I—Solutions 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 = 0 . 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

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Solution. This statement is false. Let A = 1 0 0 0 and B = 0 1 1 0 . Then x = 0 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 ) . Solution. This statement is false. Let A = 1 0 0 1 and B = - 1 0 0 - 1 . 2. (a) (5 points) Let A = 1 - 9 3 - 1 - 4 2 2 - 5 1 and y = 5 - 8 α For what value(s) of α , if any, will y be in the image of A .
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