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Inverse of a Matrix

If A =[ 1 0...

Question

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦$ , $I = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ and
$A^{- 1} = \frac{1}{6} (A^{2} + α A + β I)$ , find $β / 11$

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Solution

$I f A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦, I = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ a n d A^{- 1} = \frac{1}{6} (A^{2} + α A + β I) \Rightarrow A^{- 1} A = \frac{1}{6} (A^{2} A + α A A + β I A) \Rightarrow 6 I = A^{3} + α A^{2} + β A - - - - - - - 1 A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & - 1 & 5 0 & - 10 & 14 \end{matrix} ⎤ ⎥ ⎦ A^{3} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & - 1 & 5 0 & - 10 & 14 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & - 11 & 19 0 & - 38 & 46 \end{matrix} ⎤ ⎥ ⎦ f r o m e q u a t i o n - - - - 1 ⎡ ⎢ ⎣ \begin{matrix} 6 & 0 & 0 0 & 6 & 0 0 & 0 & 6 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & - 11 & 19 0 & - 38 & 46 \end{matrix} ⎤ ⎥ ⎦ + α ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & - 1 & 5 0 & - 10 & 14 \end{matrix} ⎤ ⎥ ⎦ + β ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦ c o m p a r i n g e l e m e n t s o n b o t h s i d e s g i v e s α + β = 5 a n d - α + β = 17 H e n c e, β = 11$

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A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦, I = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

A^{- 1} = [\frac{1}{6} (A^{2} + c A + d I)]

(c, d)

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦, I = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

A^{- 1} = [\frac{1}{6} (A^{2} + c A + d I)]

, then the value of

c

d

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 1 0 & - 2 & 4 \end{matrix} ⎤ ⎥ ⎦, 6 A^{- 1} = A^{2} + c A + d I

(c, d)

A = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}]

A^{3} - 6 A^{2} + 9 A - 4 I = O

and hence find A⁻¹.

If $A = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 1 & 1 - 1 & 2 & - 1 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$ verify that $A^{3} - 6 A^{2} + 9 A - 4 I = 0$ and hence, find $A^{- 1}$

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