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Given,$$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{pmatrix}$$ and $$6A^{-1}=A^2+cA+dI$$ then $(c,d)=?$

I have no other clue than putting the values of $A$ and $A^{-1}$ and explicitly solving $c$ and $d$. Of course thats not too decent. Is there a better way out?

Thanks!!

Soham
  • 9,990

4 Answers4

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The characteristic polynomial must be $$f(x)=x^3+cx^2+dx-6.$$ By inspection $1$ is an eigenvalue of $A$. Therefore $f(1)=0$ and so $$c+d=5.$$ The trace of $A$ is $6$, so $c=-6$ and therefore $d=11$.

Angina Seng
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Note that $$\rho_A(x)=(1-x)(x-3)(x-2)$$ so by Cayley-Hamilton $$(I-A)(A-3I)(A-2I)=0=A^3-6A^2+11A-6I$$ and so $$6I=A^3-6A^2+11A$$ so $$6A^{-1}=A^2-6A+11I$$ since $A$ is invertble!

Chinnapparaj R
  • 11,589
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Hint :

Pre multiplying both sides by $A$ we get $$A^3+cA^2+dA-6I=0$$

That means $A$ satisfies the equation $$x^3+cx^2+dx+6=0$$

So by Cayley Hamilton theorem you just need to find the characteristic equation of $A$ and compare it with the cubic above to get $c, d$

Rohan Shinde
  • 9,737
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The characteristic polynomial: $$p(\lambda)=\det(A-\lambda I_3)=\begin{vmatrix}1-\lambda & 0 & 0\\ 0&1-\lambda & 1\\ 0&-2&4-\lambda\end{vmatrix}=0 \Rightarrow \\ p(\lambda)=-\lambda ^3+6\lambda ^2-11\lambda +6=0 \Rightarrow \\ p(A)=-A^3+6A^2-11A+6I=0 \Rightarrow \\ 6I=A^3-6A^2+11A \stackrel{\text{multiply by } A^{-1}}\Rightarrow \\ 6A^{-1}=A^2-6A+11I \Rightarrow c=-6, d=11.$$

farruhota
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