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Find the inverse of the matrix $$\begin{pmatrix} -1 & 2& 0 \\ 1& 1 &0 \\ 2 & -1& 2 \end{pmatrix}$$ using the Cayley–Hamilton theorem.

Thanks!

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    $\begingroup$ Show your work. Compute the polynomial, show the matrix is invertible, and compute the inverse by knowing $p(A)=0$ for $p=\chi_A$. $\endgroup$ – Pedro Tamaroff May 3 '14 at 22:14
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The matrix $A$ is:

$A = \begin{bmatrix} -1 & 2 & 0 \\ 1 & 1 & 0 \\ 2 & -1 & 2 \end{bmatrix}, \tag{1}$

so the characteristic polynomial $p_A(\lambda)$ is

$p_A(\lambda) = \det(A - \lambda I) = \det \begin{bmatrix} -1 - \lambda & 2 & 0 \\ 1 & 1 - \lambda & 0 \\ 2 & -1 & 2 - \lambda \end{bmatrix}$ $= ( -1 - \lambda)(1 - \lambda) (2 - \lambda) - 2(2 - \lambda) = (\lambda^2 - 1)(2 - \lambda) - 4 + 2\lambda$ $= -\lambda^3 + 2\lambda^2 + 3\lambda - 6, \tag{2}$

and by Cayley-Hamilton we have

$0 = p_A(A) = -A^3 + 2A^2 + 3A - 6I; \tag{3}$

(3) may be written as

$A(-A^2 +2A + 3I) = 6I, \tag{4}$

or

$A(\dfrac{1}{6}(-A^2 +2A + 3I)) = I, \tag{5}$

which shows that

$A^{-1} = \dfrac{1}{6}(-A^2 +2A + 3I). \tag{6}$

I leave the explicit calculation of $A^{-1}$ from (6) to any interested readers; it is not difficult.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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  • $\begingroup$ @Amzoti: thanks, doc! Out in the garden, harvesting low-hanging fruit! $\endgroup$ – Robert Lewis May 3 '14 at 22:42
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Hint: The equation $$ A^n + a_{n-1}A^{n-1} + \cdots + a_1 A + a_0 I = 0 $$ can be rewritten as $$ A(A^{n-1} + a_{n-1}A^{n-2} + \cdots + a_1I) = -a_0I $$

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