Show that
A = [ 2 -3
3 4 ]
satisfies the equation x^2 - 6x + 17 = 0. Hence find A^-1.

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Please find below the solution to the asked query:

We have : A = [\begin{matrix} 2 & - 3 \\ 3 & 4 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 2 & - 3 \\ 3 & 4 \end{matrix}] [\begin{matrix} 2 & - 3 \\ 3 & 4 \end{matrix}] = [\begin{matrix} 4 - 9 & - 6 - 12 \\ 6 + 12 & - 9 + 16 \end{matrix}] = [\begin{matrix} - 5 & - 18 \\ 18 & 7 \end{matrix}] L . H . S . = x^{2} - 6 x + 17 I, where I is indentity matrix . Put x = A \Rightarrow L . H . S . = A^{2} - 6 A + 17 I = [\begin{matrix} - 5 & - 18 \\ 18 & 7 \end{matrix}] - 6 [\begin{matrix} 2 & - 3 \\ 3 & 4 \end{matrix}] + 17 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} - 5 & - 18 \\ 18 & 7 \end{matrix}] + [\begin{matrix} - 12 & 18 \\ - 18 & - 24 \end{matrix}] + [\begin{matrix} 17 & 0 \\ 0 & 17 \end{matrix}] = [\begin{matrix} - 5 - 12 + 17 & - 18 + 18 + 0 \\ 18 - 18 + 0 & 7 - 24 + 17 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = R . H . S . Hence A satisfies given equation . Hence A^{2} - 6 A + 17 I = 0 \Rightarrow A (A - 6 I + 17 A^{- 1}) = 0 As A \neq 0, hence we get : A - 6 I + 17 A^{- 1} = 0 \Rightarrow 17 A^{- 1} = 6 I - A \Rightarrow 17 A^{- 1} = 6 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 2 & - 3 \\ 3 & 4 \end{matrix}] = [\begin{matrix} 6 & 0 \\ 0 & 6 \end{matrix}] - [\begin{matrix} 2 & - 3 \\ 3 & 4 \end{matrix}] \Rightarrow 17 A^{- 1} = [\begin{matrix} 4 & 3 \\ - 3 & 2 \end{matrix}] \Rightarrow A^{- 1} = [\begin{matrix} \frac{4}{17} & \frac{3}{17} \\ \frac{- 3}{17} & \frac{2}{17} \end{matrix}]

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Show that A = [ 2 -3 3 4 ] satisfies the equation x^2 - 6x + 17 = 0. Hence find A^-1.

Show that
A = [ 2 -3
3 4 ]
satisfies the equation x^2 - 6x + 17 = 0. Hence find A^-1.