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201y03mt1

# 201y03mt1 - B U Department of Mathematics Math 201 Matrix...

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Unformatted text preview: B U Department of Mathematics Math 201 Matrix Theory Summer 2003 First Midterm This archive is a property of Bo˘gazi¸ ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. Let A =   1 2 1- 1 3 4 1 2 1 1   , and let R be its row-reduced echelon form. a) Find all solutions of Ax = 0 by first finding R . Solution: A E 1 : r 1 + r 2 → r 2 ,E 2 :- r 1 + r 3 → r 3-→   1 2 1 2 4 4 1   E 3 : 1 2 r 2 → r 2-→   1 2 1 1 2 2 1   E 4 :- 2 r 2 + r 1 → r 1-→   1- 3- 4 1 2 2 1   E 5 :4 r 3 + r 1 → r 1 ,E 6 :- 2 r 3 + r 2 → r 2-→   1- 3 1 2 1   = R. Ax = 0 ⇔ Rx = 0, so x 1- 3 x 3 = x 2 + 2 x 3 = x 4 = . Setting x 3 to be the free parameter, the complete solutions are x =     x 1 x 2 x 3 x 4     =     +3 t- 2 t t     t ∈ R . b) Find a 3 x 3 matrix S such that SA = R . (Hint: Think of the elementary matrices corre- sponding to the operations performed in part (a).) Solution: The row operations applied above corresponds to the following matrices E i in 3 × 3 dimensions: R =   1 1- 2 1     1 4 1 1     1- 2 1 1     1 1 / 2 1     1 1- 1 1     1 1 1 1   = A, so E 6 E 5 E 4 E 3 E 2 E 1 = S ....
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201y03mt1 - B U Department of Mathematics Math 201 Matrix...

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