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MAS4105Su10c

# MAS4105Su10c - MAS 4105 Test 3 1(8 pts Indicate whether the...

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MAS 4105 Test 3 1. (8 pts) Indicate whether the following are true or false. i. If A, B M n × n ( R ) and A is invertible, then det( AB ) = det( B ). T F ii. If A M n × n ( R ) , then rank( A 2 ) < rank( A ). T F iii. If A M m × n ( R ) , then rank( A ) m . T F iv. If A M n × n ( R ) and 5 n then det( A ) = n i =1 ( - 1) i +5 A i 5 det( ˜ A i 5 ). T F v. If A M m × n ( R ) and n < m , then det( A ) = 0. T F vi. An elementary matrix is always invertible. T F vii. Every invertible matrix is the product of elementary matrices. T F viii. Let A M n × n ( R ); the equation Ax = b is consistent if the reduced echelon form of ( A | b ) has n pivots. T F

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2. (10 pts) Based on the following proofs, answer the questions below: Theorem 3.8. Let Ax = 0 be a homogeneous system of m linear equations in n unknowns over a field F . Let K denote the set of all solutions to Ax = 0. Then K = N ( L A ); hence K is a subspace of F n of dimension n - rank( L A ) = n - rank( A ). Proof. Clearly, K = { s F n : As = 0 } = N ( L A ) . The second part now follows from the dimension theorem. Corollary. If m < n , the system Ax = 0 has a nonzero solution. Proof. Suppose that m < n . Then rank( A ) = rank( L A ) m . Hence, dim( K ) = n - rank( L A ) n - m > 0 , where K = N ( L A ) . Since dim( K ) > 0 , K = { 0 } . Thus there exists a nonzero vector s K ; so s is a nonzero solution to Ax = 0 . i. Justify the equation: dim( K ) = n - rank( L A ).
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