Calculate Sum of Products (x,y,z)

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The discussion centers on calculating the sum of products using the arrays x=(-3,2,5), y=(2,4,-5), and z=(1,6,7). The method involves performing an inner summation of products followed by an outer product operation. Clarifications are made regarding the formulas presented, with emphasis on ensuring the correct interpretation of the summation and product indices. The calculations yield results of 24 and 144 for the two different interpretations of the formulas. Understanding the correct application of these operations is crucial for solving such problems effectively.
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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? Theres 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,
\prod_{j= 1}^3\sum_{i=1}^2 x_iy_i and
\sum_{j=1}^3\prod_{i=1}^2 x_iy_i
are just
\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
and
\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

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

The first of those is
\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)
surely you can do that arithmetic yourself.
 
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