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# solution5 - M341 Homework Solution 5 Section 2.2 11(a Let...

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M341 Homework Solution 5 Section 2.2: 11(a) : Let X 1 = [ u 1 , u 2 , · · · , u n ] and let X 2 = [ v 1 , v 2 , · · · , v n ] . Since AX = O we have for each 1 i m we have a i 1 u 1 + a i 1 u 2 + · · · , a in u n = 0 and a i 1 v 1 + a i 1 v 2 + · · · , a in v n = 0 . Therefore a i 1 ( u 1 + v 1 )+ a i 1 ( u 2 + v 2 )+ · · · , a in ( u n + v n ) = a i 1 u 1 + a i 1 u 2 + · · · , a in u n + a i 1 v 1 + a i 1 v 2 + · · · , a in v n = 0 . Therefore X 1 + X 2 is also a solution. Furthermore a i 1 cu 1 + a i 1 cu 2 + · · · , a in cu n = c ( a i 1 u 1 + a i 1 u 2 + · · · , a in u n ) = c 0 = 0 . Therefore c X 1 is also a solution. A simple proof is following: A ( X 1 + X 2 ) = AX 1 + AX 2 = O + O = O since AX 1 = AX 2 = O and A ( c X 1 ) = c AX 1 = c O = O Hence X 1 + X 2 and c X 1 are also solutions. 11(b) : Consider the linear system x 1 + x 2 + x 3 = 1 x 1 + x 2 x 3 = 1 has two solutions X 1 = [1 , 0 , 0] and X 2 = [0 , 1 , 0] . But X 1 + X 2 = [1 , 1 , 0] is not a solution since 1 + 1 + 0 ̸ = 1 . 11(c) : Using the notation from part (a) we have a i 1 u 1 + a i 1 u 2 + · · · , a in u n = 0 and a i 1 v 1 + a i 1 v 2 + · · · , a in v n = b i . Therefore a i 1 ( u 1 + v 1 ) + a i 1 ( u 2 + v 2 ) + · · · , a in ( u n + v n ) = a i 1 u 1 + a i 1 u 2 + · · · , a in u n + a i 1 v 1 + a i 1 v 2 + · · · , a in v n = 0 + b i = b i . Therefore X 1 + X 2 is a solution of AX = B . A simple proof is following: A ( X 1 + X 2 ) = AX 1 + AX 2 = B + O = B since AX 1 = B and AX 2 = O . Hence X 1 + X 2 is also a solution. 11(d) : We will prove by contradiction. Suppose AX = 0 has a nontrivial solution Y and AX = B with B ̸ = O has a unique solutions X 1 by the assumption. Then by part(c), X 2 = X 1 + Y is also a solution and X 2 ̸ = X 1 since Y ̸ = 0 . It contradicts to the fact AX = B has only one solution. 12

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solution5 - M341 Homework Solution 5 Section 2.2 11(a Let...

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