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Just wondering what the author is doing here, given

##\displaystyle \left (\frac{2EI}{L^3} \begin {bmatrix} 6&3L\\3L&2L^2 \end {bmatrix} -\omega^2 \frac{\rho A L}{420} \begin {bmatrix} 156&22L\\22L&4L^2 \end{bmatrix} \begin {bmatrix} W\\ \theta \end {bmatrix} \right )=\begin {bmatrix} 0\\ 0 \end {bmatrix}##

He proceeds to set the determinant of the matrix above to 0 to obtain a quadratic characteristic equation in ##\omega^2##

The equation being

##15120-1224 \lambda + \lambda^2=0##, where ##\displaystyle \lambda=\frac{\rho A L^4}{EI} \omega^2##

Using wolfram http://www.wolframalpha.com/input/?i=determinant+of++(2Z/l^3{{6,3l},{3l,2l^2}}-((omega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I dont see how he arrives at this quadratic expression or how he determined ##\lambda## ...thanks