Factorizing Quadratics

[jonnburton]

I have been working through this worked example (on finding the eigenvalues) for a matrix and can't understand a step in the working.

Expanding the determinant ends up in:

\(\displaystyle (1-\lambda)\{\lambda^2-\lambda - 3\}+1-\lambda = 0\)

\(\displaystyle 1 -\lambda = 0 \) or \(\displaystyle \lambda^2 -\lambda -2 = 0\)

But, surely the last statement should be \(\displaystyle \lambda^2 -2\lambda -2 = 0\)

If not, what has happened to the \(\displaystyle - \lambda\) at the end?

Any information would be much appreciated!

 

[jonnburton]

Oh, actually I see where I've been going wrong. That was a silly mistake which came about by trying to take shortcuts! After multiplying out and factorizing again, I can see how this works.

 

[Deleted member 4993]

I have been working through this worked example (on finding the eigenvalues) for a matrix and can't understand a step in the working.

Expanding the determinant ends up in:

\(\displaystyle (1-\lambda)\{\lambda^2-\lambda - 3\}+1-\lambda = 0\)

\(\displaystyle (1-\lambda)\{\lambda^2-\lambda - 3\} \ + \ 1*(1-\lambda) = 0\)

\(\displaystyle (1-\lambda)\{\lambda^2-\lambda - 3 + 1\} \ = 0\)

\(\displaystyle (1-\lambda)\{\lambda^2-\lambda - 2\} \ = 0\)

\(\displaystyle 1 -\lambda = 0 \) or \(\displaystyle \lambda^2 -\lambda -2 = 0\)

Just so that everybody knows the intermediate steps.