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Unformatted text preview: Spring 2010 (331) Solutions for EXAM 2 Dr. A. Agarwal 1. Let B be the following matrix B = 0 1 0 0 0 0 1 0 0 0 0 1 a b c d (a) For what values of a (and no restrictions on b,c,d ) will B be invertible? Explain. ANS: Observe that the first three rows of B are linearly independent no matter what a,b,c,d are. So we consider two cases: • If a = 0 , then the fourth row can be expressed as a linear combination R 4 = bR 1 + cR 2 + dR 3 of the first three rows, hence the rows will be linearly dependent, hence by the fundamental theorem of invertible matrices, B is not invertible. • If a negationslash = 0 , then the fourth row cannot be expressed as a linear combination of the first three rows because all of them have a zero as their first entry. Thus the rows are linearly independent, hence B is invertible. (b) Find B 1 when it exists. ANS: Using rowreduction (start by making the last row as the first row) it can be easily shown that B 1 = − b/a − c/a − d/a 1 /a 1 1 1 2. Suppose { v 1 , v 2 , v 3 , v 4 } is a set of linearly independent vectors in R n . Determine, with proper justification, which of the following sets are linearly independent and which are dependent. (a) S = { v 1 + v 2 , v 2 + v 3 , v 3 + v 4 , v 4 − v 1 } . ANS: To check the independence we start with the definition and set up a ( v 1 + v 2 ) + b ( v 2 + v 3 ) + c ( v 3 + v 4 ) + d ( v 4 − v 1 ) = If we can show that the ONLY solution is a = b = c = d = 0, then the set S is independent, otherwise it is dependent. Observe that a ( v 1 + v 2 ) + b ( v 2 + v 3 ) + c ( v 3 + v 4 ) + d ( v 4 − v 1 ) = v 1 ( a − d ) + v 2 ( a + b ) + v 3 ( b + c ) + v 4 ( c + d ) = Now we use the fact that { v 1 , v 2 , v 3 , v 4 } is a set of linearly independent vectors. This means the only combination of these vectors which can add up to is the zero combination. Hence we get a − d = 0 a + b = 0 b + c = 0 c + d = 0 Solving this system gives that the only solution is a = b = c = d = 0. Therefore S is linearly independent . (b) T = { v 1 + v 2 , v 2 + v 3 , v 3 − v 4 , v 4 − v 1 } . ANS: Following the same idea as in the previous question we start with a ( v 1 + v 2 ) + b ( v 2 + v 3 ) + c ( v 3 − v 4 ) + d ( v 4 − v 1 ) = This leads to a − d = 0 a + b = 0 b + c = 0 d − c = 0 This has NONTRIVIAL solutions. For example: a = 1 ,b = − 1 ,c = 1 ,d = 1. Hence T is linearly dependent . (c) U = { v 1 , v 1 + v 2 , v 1 + v 2 + v 3 , v 1 + v 2 + v 3 + v 4 } . ANS: Similarly we can show that U is also independent set. 3. For the matrix A given below A = 1 − 2 3 2 − 5 − 3 − 2 6 5 15 10 2 6 18 8 6 write your row −−−−−−−−−−−→ reduced matrix here R = 1 0 0 − 2 3 1 − 1 0 0 0 1 1 0 0 0 (a) Find TWO different bases for the row space row( A ) ....
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This note was uploaded on 05/20/2010 for the course MTH 1016.331 taught by Professor Dr.anuragagarwal during the Spring '10 term at RIT.
 Spring '10
 Dr.AnuragAgarwal
 Linear Algebra, Algebra

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