Question 475047
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Any row in a matrix, *[tex \Large R_i\ =\ \alpha_{i,1}\ \alpha_{i,2}\ \cdots\ \alpha_{i,j}\ \cdots\ \alpha_{i,n\,-\,1}\ \alpha_{i,n}] can be replaced by the sum of a constant times any other row plus row <i>i</i>.  That is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\,\cdot\,R_j\ +\ R_i\ \Rightarrow\ \ R_i]


Actually, in more general terms, you can multiply both rows by constants if it makes sense to do so, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\,\cdot\,R_j\ +\  b\,\cdot\,R_i\ \Rightarrow\ \ R_i]



1.  Multiply Row 2 by 4:  12 32 4 0 4 0 | 40


2.  Add the result of Step 1 to Row 3:


12 + (-4) = 8


32 + (-32) = 0


4 + 1 = 5


0 + 0 = 0


4 + 0 = 4


0 + 1 = 1


40 + 0 = 40


Hence: 8 0 5 0 0 4 1 | 40


3.  Replace Row 3 with the results of Step 2.


Note: Just because you multiplied Row 2 by 4 in order to drive the 3rd Row 2nd Column element to zero, doesn't mean you have to replace Row 2 with the results of the multiplication.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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