Undergrad Normal Mode calculation steps

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The calculation of X1:X2 after inputting the lower value of W^2 in equation 9.9 involves substituting known matrices A and B, along with the scalar omega, into the matrix equation Cx=0. This results in two equations with two unknowns, represented by the vector x. However, since any scalar multiple of a solution x is also a solution, the equations are not independent, leading to infinite solutions. Consequently, the relationship between the variables can only be expressed as a ratio. This process highlights the dependence of the variables rather than yielding unique solutions.
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Can someone please explain me the steps of calculation of X1:X2 after putting in the lower value of W^2 in equation 9.9 in "Riley, Hobson, Bence - Mathematical Methods for Physics and Engineering 2006 - pg 319"? I have attached the page as a PDF file.
Thank you.
 

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A and B are matrices you know, and ##\,omega## is a number that you know, so you can plug them all in and you get like like ##Cx=0## for some matrix ##C## that you know. You might at first guess that gives you two equations and two unknown (##x## is a vector of 2 dimensions, the 0 on the right side is also a vector of 2 dimensions so you get two equations) but if ##x## is a solution so is ##\alpha x## for any scalar ##\alpha##, which means the two equations are not independent, and there are infinite solutions. So all you can do is express one variable in terms of the other which is the same as computing their ratio
 
Office_Shredder said:
A and B are matrices you know, and ##\,omega## is a number that you know, so you can plug them all in and you get like like ##Cx=0## for some matrix ##C## that you know. You might at first guess that gives you two equations and two unknown (##x## is a vector of 2 dimensions, the 0 on the right side is also a vector of 2 dimensions so you get two equations) but if ##x## is a solution so is ##\alpha x## for any scalar ##\alpha##, which means the two equations are not independent, and there are infinite solutions. So all you can do is express one variable in terms of the other which is the same as computing their ratio
Thank you very much.
 

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