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I was thinking about the following problem:

Let $A$ be a $3\times 3$ matrix with complex entries whose eigenvalues are $1,\pm 2i.$ Suppose that for some $a,b,c\in \mathbb C$, $aA^{-1}=A^{2}+bA+cI,$ where $I$ is the $3\times 3$ identity matrix.Then $(a,b,c)$ equals to which of the following options:

(A)$(-1,-4,4),$

(B)$(4,-1,4),$

(C)$(-1,4,-2),$

(D)$(-1,-2,4).$

My attempts: From $aA^{-1}=A^{2}+bA+cI,$ we get $aI=A^{3}+bA^{2}+cA $ and so trace$(aI)$=trace$(A^{3}+bA^{2}+cA)$=$(\operatorname{trace}(A^{3})+b \operatorname{trace}(A^{2})+ c \operatorname{trace}(A))$.....$(1)$ Now if we take a $3\times 3$ diagonal matrix with diagonal entries $1,\pm 2i $ respectively then,from the above relation $(1)$ we get,$3a=1+b+c$. But from the given options, we see none of them satisfies $3a=1+b+c$.I am sure that i have done some mistake. Please provide a better approach to tackle the problem. Thanks in advance for your time.

3 Answers 3

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The characteristic polynomial of $A$ is $(\lambda-1)(\lambda-2i)(\lambda+2i)=(\lambda-1)(\lambda^2+4)=\lambda^3-\lambda^2+4\lambda-4,$ and a matrix always satisfies its characteristic polynomial. In this case, then, we have $A^3-A^2+4A-4I=0.$ From this, we can calculate a formula for $A^{-1}$ in terms of $I,A,A^2$, and then solve for $a,b,c$.

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Hint: This is equivalent to $A^3 +bA^2+cA-aI = 0$. Also, we know that any matrix satisfies it's own characteristic equation by the Cayley-Hamilton theorem.

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The characteristic equation is $(\lambda-1)(\lambda^2+4) = 0 \implies \lambda^3 - \lambda^2 + 4 \lambda - 4 = 0$ By Cayley Hamilton theorem, we have $A^3 - A^2 + 4A - 4I = 0 \implies 4A^{-1} = A^2 - A + 4I$