- #26

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All I have to do is delete the last two lines in the first part and I will x + y.

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- Thread starter Dustinsfl
- Start date

- #26

- 699

- 5

All I have to do is delete the last two lines in the first part and I will x + y.

- #27

vela

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Let [itex]A, B \in S[/itex]. Let C=A+B. Then

[tex]A \in S \Rightarrow a=\sum_{i=1}^3 a_{i1} = \sum_{i=1}^3 a_{i2} = \sum_{i=1}^3 a_{i3} = \sum_{i=1}^3 a_{1i} = \sum_{i=1}^3 a_{2i} = \sum_{i=1}^3 a_{3i} = \sum_{i=1}^3 a_{ii}=\sum_{i=1}^3 a_{4-i,i}[/tex]

[tex]B \in S \Rightarrow b=\sum_{i=1}^3 b_{i1} = \sum_{i=1}^3 b_{i2} = \sum_{i=1}^3 b_{i3} = \sum_{i=1}^3 b_{1i} = \sum_{i=1}^3 b_{2i} = \sum_{i=1}^3 b_{3i} = \sum_{i=1}^3 b_{ii}=\sum_{i=1}^3 b_{4-i,i}[/tex]

The sum of the elements of the i-th row of C is

[tex]\sum_{j=1}^3 c_{ij} = \sum_{j=1}^3 (a_{ij}+b_{ij}) = \sum_{j=1}^3 a_{ij} + \sum_{j=1}^3 b_{ij} = a+b[/tex]

The first equality is from the definition of matrix addition; the second equality is from what you know about rewriting sums; and the third equality holds because A and B are in S. From this, you can see that all the rows sum to the same number, a+b.

I'll leave it to you to fill in the rest. Eventually, you'll be able to conclude the sums of the rows, columns, and diagonals are all equal to each other; therefore, C is an element of S, which is what you wanted to show.

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