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Problem of the Week #56 - April 22nd, 2013

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Chris L T521

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Jan 26, 2012
995
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $A$ be an $n\times n$ matrix whose characteristic polynomial is
\[p(\lambda)=\lambda^n+a_1\lambda^{n-1}+\ldots+ a_{n-1}\lambda+ a_n.\]
If $A$ is nonsingular, show that
\[A^{-1}=-\frac{1}{a_n}\left( A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n\right).\]

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Hint:
Use the Cayley-Hamilton theorem.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
995
This week's problem was correctly answered by Sudharaka. You can find his solution below.

Using the Cayley-Hamilton theorem we get,

\[p(A)=A^{n} + a_1A^{n-1} + \ldots + a_{n-1}A+a_{n}I_n=0_n\]

where \(0_n\) is the \(n\times n\) zero matrix.

Since \(A\) is non-singular multiplying by \(A^{-1}\) we get,

\[A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n+a_n A^{-1}=0_n\]

\[\therefore A^{-1}=-\frac{1}{a_n}\left( A^{n-1} + a_1A^{n-2} + \ldots + a_{n-2}A+a_{n-1}I_n\right)\]
 
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