Calculate Sum of Products (x,y,z)

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SUMMARY

The discussion focuses on calculating the sum of products using arrays and substitution methods. Given the arrays x=(-3,2,5) and y=(2,4,-5), the inner summation is computed as (-3)(2) + (2)(4) = -6 + 8 = 2. The outer product is then calculated as 3(2) = 6. The conversation also addresses potential misunderstandings regarding the formulas for product and summation, clarifying the correct application of these operations to arrive at the final results of 24 and 144 for the respective calculations.

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I don't quite understand the method to solve this type of question.

Let x=(-3,2,5), y=(2,4,-5), and z=(1,6,7). Calculate:
 

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I view such qns playing with 'arrays' and 'susbstituion'.

Generally, i will view it this way:-

x is an array of (-3,2,5)
y is an array of (2,4,-5)

first part is u do the summation first - i call it inner.

Inner: (-3)(2) + (2)(4)

Then you do the Products - i call it outer.

But is your question complete? there's no 'j' in your formulaes pasted.
 
tstuddud said:
I don't quite understand the method to solve this type of question.

Let x=(-3,2,5), y=(2,4,-5), and z=(1,6,7). Calculate:

What you have written,
[tex]\prod_{j= 1}^3\sum_{i=1}^2 x_iy_i[/tex] and
[tex]\sum_{j=1}^3\prod_{i=1}^2 x_iy_i[/tex]
are just
[tex]\prod_{j=1}^3(x_1y_1+ x_2y_2+ x_3y_3)= \prod_{j=1}^3((-3)(2)+ (2)(4)+ (5)(-5))= \prod_{j=1}^3(-6+ 8- 10)= 3(8)= 24[/tex]
and
[tex]\sum{j= 1}^3((x_1y_1)(x_2y_2))= \sum_{j=1}^3 (-3)(2)(2)(4)= \sum_{j=1}^3 48= 3(48)= 144[/tex]

But I suspect you meant
[tex]\prod_{j=1}^3\sum_{i= 1}^2 x_iy_j[/tex] and
[tex]\sum{j=1}^3\Pi_{i=1}^2 x_iy_j[/tex]

The first of those is
[tex]\prod_{j=1}^3(x_1+ x_2)y_j= (x_1+ x_2)\prod_{j=1}^3y_i= (x_1+ x_2)(y_1y_2y_3)[/tex]
surely you can do that arithmetic yourself.
 

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