let-A-1-1-2-3-1-find-A-1-2-calculate-A-n-3-determine-e-A-and-e-2A-

Question Number 60500 by prof Abdo imad last updated on 21/May/19

l e t A = (\begin{matrix} 1 1 \\ - 2 3 \end{matrix})

1) f i n d A^{- 1}

2) c a l c u l a t e A^{n}

3) d e t e r m i n e e^{A} a n d e^{- 2 A} .

Commented by maxmathsup by imad last updated on 22/May/19

2) w e h a v e P_{c} (A) = d e t (A - x I) = | \begin{matrix} 1 - x 1 \\ - 2 3 - x \end{matrix} | = (1 - x) (3 - x) + 2 = 3 - x - 3 x + x^{2} + 2

= x^{2} - 4 x + 5 P_{c} (A) = 0 \Rightarrow x^{2} - 4 x + 5 = 0 \to Δ^{^{'}} = 4 - 5 = - 1 = i^{2} \Rightarrow

x_{1} = 2 + i a n d x_{2} = 2 - i

v (x_{1}) = K e r (A - x_{1} I) = {u / (A - x_{1}) u = 0} l e t u (\begin{matrix} x \\ y \end{matrix})

(A - x_{1}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow (\begin{matrix} - 1 - i 1 \\ - 2 1 - i \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \Rightarrow {_{- 2 + (1 - i) y = 0}^{(- 1 - i) x + y = o}

Δ_{s} = | \begin{matrix} - 1 - i 1 \\ - 2 1 - i \end{matrix} | = - (1 + i) (1 - i) + 2 = - (2) + 2 = 0 \Rightarrow

y = (1 + i) x \Rightarrow (x, y) = (x, (1 + i) x) = {x e}_{1} w i t h e_{1} (\begin{matrix} 1 \\ 1 + i \end{matrix})

v (x_{2}) = K e r (A - x_{2} I) = {u / (A - x_{2} I) u = 0} l e t u (\begin{matrix} x \\ y \end{matrix})

(A - x_{2} I) u = 0 \Rightarrow (\begin{matrix} - 1 + i 1 \\ - 2 1 + i \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \Rightarrow {_{- 2 x + (1 + i) y = 0}^{(- 1 + i) x + y = 0} \Rightarrow

Δ_{s} = | \begin{matrix} - 1 + i 1 \\ - 2 1 + i \end{matrix} | = - (1 - i) (1 + i) + 2 = 0 \Rightarrow y = (1 - i) x \Rightarrow

(x, y) = (x, (1 - i) x) = x (1, 1 - i) \Rightarrow e_{2} (\begin{matrix} 1 \\ 1 - i \end{matrix}) \Rightarrow M_{p} = (\begin{matrix} 1 1 \\ 1 + i 1 - i \end{matrix}) = P

a n d D = (\begin{matrix} 2 + i 0 \\ 0 2 - i \end{matrix})

w e h a v e A = P {D P}^{- 1} \Rightarrow A^{n} = {P D}^{n} P^{- 1}

P^{- 1} = \frac{t (c o m (P))}{d e t P} d e t P = 1 - i - (1 + i) = 1 - i - 1 - i = - 2 i

c o m (P) = {(- 1)}^{i + j} A_{i j} (\begin{matrix} α_{11} α_{12} \\ α_{2} 1 α_{22} \end{matrix})

α_{11} = 1 - i, α_{21} = - (1), α_{12} = - (1 + i), α_{22} = 1 \Rightarrow

c o m (P) = (\begin{matrix} 1 - i - 1 - i \\ - 1 1 \end{matrix}) \Rightarrow t (c o m (P)) = (\begin{matrix} 1 - i - 1 \\ - 1 - i 1 \end{matrix}) \Rightarrow

P^{- 1} = - \frac{1}{2 i} (\begin{matrix} 1 - i - 1 \\ - 1 - i 1 \end{matrix}) = (\begin{matrix} \frac{- 1 + i}{2 i} \frac{1}{2 i} \\ \frac{1 + i}{2 i} \frac{- 1}{2 i} \end{matrix})

= (\begin{matrix} \frac{- i - 1}{- 2} \frac{i}{- 2} \\ \frac{i - 1}{- 2} \frac{- i}{- 2} \end{matrix}) = (\begin{matrix} \frac{1 + i}{2} \frac{- i}{2} \\ \frac{1 - i}{2} \frac{i}{2} \end{matrix})

A^{n} = \frac{1}{2} (\begin{matrix} 1 1 \\ 1 + i 1 - i \end{matrix}) (\begin{matrix} {(2 + i)}^{n} 0 \\ 0 {(2 - i)}^{n} \end{matrix}) (\begin{matrix} 1 + i - i \\ 1 - i i \end{matrix})

r e s t t o f i n i s h t h e c a l c u l u s \dots .

Commented by maxmathsup by imad last updated on 22/May/19

1) w e h a v e P_{c} (X) = x^{2} - 4 x + 5 c a y l e y h a m i l t o n t h e o r e m g i v e

P_{c} (A) = 0 \Rightarrow A^{2} - 4 A + 5 I = 0 \Rightarrow A^{2} = 4 A - 5 I \Rightarrow A = 4 I - 5 A^{- 1} \Rightarrow

5 A^{- 1} = - A + 4 I = - (\begin{matrix} 1 1 \\ - 2 3 \end{matrix}) + (\begin{matrix} 4 0 \\ 0 4 \end{matrix}) = (\begin{matrix} 3 - 1 \\ 2 1 \end{matrix})

\Rightarrow A^{- 1} = (\begin{matrix} \frac{3}{5} \frac{- 1}{5} \\ \frac{2}{5} \frac{1}{5} \end{matrix}) b u t f i r s t w d m u s t v e r i f y t h a t d e t A \neq 0!