I have to find a characteristic polynomial for the follwoing matrix:
A=(0−11−102120)
My goal to find the roots of the characteristic polynomial.
The characteristic polynomial is given by solving det(λI−A), and then we get:
det(λI−A)=det(t1−11t−2−1−2t)
The final polynomial I get is PA=t3−6t+4
While there is a way to solve this polynomial I believe there is a way to make this determinant more "comfortable" to solve.
Can I add the third column to the second column?
det(λI−A)=det(t0−11t−2−2−1t−2t)
And Then, can I do the following (and why?)
det(λI−A)=(t−2)∗det(t0−111−2−11t)
What kind of elementary operations, or other arithmetic ones I can do, to make the determinant easier to solve?
Thanks,
Alan
Answer
Personally, I prefer this method
|A−λI|=|−λ−11−1−λ212−λ|
=−λ|−λ22−λ|+|−121−λ|+|−1−λ12|
=−λ3+6λ−4=(2−λ)(λ2+2λ−2)
=(2−λ)(λ−√3+1)(λ+√3+1)
You can also triangularize the matrix first. This makes the determinant trivial to compute
|A−λI|=|−λ−11−1−λ212−λ|
=|−λ−1102−λ2−λ12−λ|
=|−λ−1102−λ2−λλ2λ−λ2|
=|−λ−1102−λ2−λ02λ−11−λ2|
=|−λ−1102−λ2−λ00−λ2−2λ+2|
=|1λ−1−λ−102−λ2−λ00λ2+2λ−2|
=(2−λ)(λ2+2λ−2)
=(2−λ)(λ−√3+1)(λ+√3+1)
In any case, I suggest you look up the properties of the determinant.
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