A = [4,0,1; -2,1,0; -2,0,1] (3x3 Matrix) What are the eigenvalues of this matrix? Please show all working. A = [4,0,1; -2,1,0; -2,0,1] (3x3 Matrix) What are the eigenvalues of this matrix? Please show all working. @Mathematics
\[A = \left[\begin{matrix}4 & 0 & 1 \\-2 &1 &0\\ -2 & 0 & 1\end{matrix}\right]\] \[|A- \lambda I|= det\left[\begin{matrix}4- \lambda & 0 & 1 \\-2 &1- \lambda &0\\ -2 & 0 & 1- \lambda\end{matrix}\right]\]
Unkle I get lost at the characteristic equation, sorting out the cubic equation to find the roots is causing me difficulties.
\[=(4- \lambda) \times (({1- \lambda)}^2 -0) - 0 (0-(-2 \times (1- \lambda))) + 1 (0-(-2 \times (1-\lambda)) \]
\[=(4- \lambda )(1- \lambda)^2+2(1- \lambda)\]
\[=(4-\lambda)(1-2 \lambda+ \lambda^2)+2- \lambda\]\[=(4-8\lambda+4 \lambda^2 - \lambda +2 \lambda^2-\lambda^3)+2- \lambda\]\[=6-10 \lambda +6 \lambda^2 -\lambda^3 \]
i am using matlab the eigenvalues it gives is lambda = 1 / 3 / 2
maybe i have made a mistake then
EDU>> a = [4,0,1;-2,1,0;-2,0,1] a = 4 0 1 -2 1 0 -2 0 1 EDU>> eig (a) ans = 1 3 2
just cant get all the working to it though.
the cubic equation is \[\lambda^3-6\lambda^2+11\lambda-6=0\]. You can see, that \[\lambda=1\] satisfies the equation, so you can divide \[\lambda^3-6\lambda^2+11\lambda-6=0\] by \[\lambda-1\]. Then for the rest of 2 values you have to solve the quadratic equatin \[\lambda^2-5\lambda+6=0\] and then you got other values: 2 and 3
i made a mistake o my second last line should read \[ 0=(4−8λ+4λ^2−λ+2λ^2−λ^3)+2−2λ\]\[0=6−11λ+6λ^2−λ^3\]
unkle are you able to produce the roots by hand calcs?
matmagician how do i divide it by lambda - 1 ? What does that do?
hope this helps
thanks mate, i will keep going if possible and only if you have got the time could you do a full solution and post, got an exam on this next friday and I am struggling.
thankyou very much, mathmagician and unklerhuk sorry couldnt give both of yous medals. thanks again.
Join our real-time social learning platform and learn together with your friends!