Find the eigenvalues of a given matrix

[blouqu6]

1. The 3x3 Matrix A=[33, -12, -70; 0, 1, 0; 14, -6, -30] has three distinct eigenvalues, λ₁<λ₂<λ₃.
Find each eigenvalue.2. det(A-λI)=0 where I denotes the appropriate identity matrix (3x3 in this case)3. Here's my attempt:

--> det([33, -12, -70; 0, 1, 0; 14, -6, -30]-λ[1, 0, 0; 0, 1, 0; 0, 0, 1])=0

--> det([33-λ, -12, -70; 0, 1-λ, 0; 14, -6, -30-λ]=0

--> -λ³+4λ²+7λ-10=0

And this is where I'm not exactly sure what to do. I don't believe I can effectively use grouping to solve for λ. Any help here would be appreciated.

 

[Dick]

1. The 3x3 Matrix A=[33, -12, -70; 0, 1, 0; 14, -6, -30] has three distinct eigenvalues, λ₁<λ₂<λ₃.
Find each eigenvalue.2. det(A-λI)=0 where I denotes the appropriate identity matrix (3x3 in this case)3. Here's my attempt:

--> det([33, -12, -70; 0, 1, 0; 14, -6, -30]-λ[1, 0, 0; 0, 1, 0; 0, 0, 1])=0

--> det([33-λ, -12, -70; 0, 1-λ, 0; 14, -6, -30-λ]=0

--> -λ³+4λ²+7λ-10=0

And this is where I'm not exactly sure what to do. I don't believe I can effectively use grouping to solve for λ. Any help here would be appreciated.

The rational roots theorem. Any possible rational root must divide 10. Can you guess one? Once you find a root r, divide by λ-r. Now you have a quadratic.

 

[blouqu6]

Ahh yes

And that did it. Thanks man, I ended up with λ₁=-2,λ₂=1, and λ₃=5, which is the correct answer. Really appreciate the help.

 

[Dick]

And that did it. Thanks man, I ended up with λ₁=-2,λ₂=1, and λ₃=5, which is the correct answer. Really appreciate the help.

You are welcome. If they give you a cubic to solve without using a computing device, it will likely have one easy root. If you find that you are home free.