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Verify Cayley Hamilton theorem for the matrix A and hence ,find $A^{-1}$ & $A^4$ ,where $A = \begin{bmatrix} \ 1 & 2 & -2 \\ \ -1 & 3 & 0 \\ \ 0 & -2 & 1 \\ \end{bmatrix}$
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To verify C- H theorem characteristic eqn |A- λI| =0

i.e. $ \begin{vmatrix} 1-λ&2&-2 \\ -1&3-λ&0 \\ 0&-2&1-λ \end{vmatrix} = 0 $

Expanding we get a cubic equation in λ as $λ^3 - 5λ^2 + 9λ - 1 =0$

To verify C-H theorem we have to show that $A^3 - 5A^2 + 9A - I =0$ ………………..(1)

Taking L.H.S. of eqn (1)

$A^3 - 5A^2 + 9A – I = \begin{bmatrix} -13&42&-2 \\ -11&9& 10 \\ 10&-22& -3 \end{bmatrix} -5 \begin{bmatrix} -1&12&-4 \\ -4&7& 2 \\ 2&-8& 1 \end{bmatrix} +9 \begin{bmatrix} 1&2&-2 \\ -1&3& 0 \\ 0&-2& 1 \end{bmatrix} - \begin{bmatrix} 1& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix} 0&0& 0 \\ 0&0& 0 \\ 0&0& 0 \end{bmatrix} $

Therefore C-H theorem verified

To find $A^{-1}$ , multiply eqn (1) by $A^{-1}$ we get $A^2 – 5A + 9I - A-1 =0 \,\,\,\,\,\,\,\,\,\,$ Since ( $A A^{-1} =I$)

i.e.

$A^{-1} = A^2 – 5A + 9I = \begin{bmatrix} -1&12&-4 \\ -4&7& 2 \\ 2&-8& 1 \end{bmatrix} - 5 \begin{bmatrix} 1&2&-2 \\ -1&3& 0 \\ 0&-2& 1 \end{bmatrix} +9 \begin{bmatrix} 1& 0& 0 \\ 0& 1& 0 \\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix} 3& 2& 6 \\ 1& 1& 2 \\ 2& 2& 5 \end{bmatrix} $

To find $A^4$ , multiply eqn (1) by A we get $A^4 – 5A^3 + 9A^2 - A = 0$ i.e.

$ A^4 = 5A^3 - 9A^2 + A = 5 \begin{bmatrix} -13&42&-2 \\ -11&9& 10 \\ 10&-22& -3 \end{bmatrix} -9 \begin{bmatrix} -1&12&-4 \\ -4&7& 2 \\ 2&-8& 1 \end{bmatrix} + \begin{bmatrix} 1&2&-2 \\ -1&3& 0 \\ 0&-2& 1 \end{bmatrix} = \begin{bmatrix} -55& 104& 24 \\ -20&-15& 32 \\ 32&-42& 13 \end{bmatrix} $

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