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İ S. Y

F ,

Surname, Name:

Q . §. T M E AX=B

Let u=

−1

1

−1

and A=

1 0 1

1 1 0

0 1 1

. Give two solutions answering the

following question:

Is uin the subset of R3spanned by the columns of A?

A .

SOLUTION 1. The vector uis in the subset spanned by the columns of Aif

and only if the system Ax=uhave a solution. Therefore, we need to decide

if Ax=uis consistent or not.

101−1

1 1 0 1

011−1

(−1)R1+R2↔R2

//

1 0 1 −1

0 1 −1 2

0 1 1 −1

(−1)R2+R3↔R3

//

1 0 1 −1

0 1 −1 2

0 0 2 −3

Note that we don’t have a row of the form [0 0 0 1]. The system above is

consistent by Theorem 2. As a result, the answer is "YES, uis in the in the

subset of R3spanned by the columns of A."

SOLUTION 2. The system Ax=bhave a solution for every bif and only if A

has a pivot position in every row by THEOREM 4. Therefore, we need to see

if Ahas a pivot position in every row:

1 0 1

1 1 0

0 1 1

(−1)R1+R2↔R2

//

1 0 1

0 1 −1

0 1 1

(−1)R2+R3↔R3

//

1 0 1

0 1 −1

0 0 2

(1/2)R3↔R3

(1)R3+R2↔R2

//

101

010

001

(−1)R3+R1↔R1

//

100

010

001

Note that Ahas pivot positions in EVERY row. As a result, the answer is "YES,

uis in the in the subset of R3spanned by the columns of A."

1

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