fjnd-inverse-of-matrix-2-5-1-3-

Question Number 141400 by Raffaqet last updated on 18/May/21

f j n d i n v e r s e o f m a t r i x [\begin{matrix} 2 & - 5 \\ 1 & 3 \end{matrix}]

Answered by iloveisrael last updated on 18/May/21

A^{- 1} = \frac{1}{6 - (- 5)} [\begin{matrix} 3 5 \\ - 1 2 \end{matrix}]

A^{- 1} = [\begin{matrix} 3 / 11 5 / 11 \\ - 1 / 11 2 / 11 \end{matrix}]

Answered by EDWIN88 last updated on 18/May/21

Cayley - Hamilton theorem

∣ λ I - A ∣ = 0 \Rightarrow p (λ) = λ^{2} - tr . A + \det (A) = 0

p (A) = A^{2} - 5 A + 11 I = 0

\Rightarrow A - 5 I = - {11 A}^{- 1}

\Rightarrow A^{- 1} = \frac{1}{11} {(\begin{matrix} 5 0 \\ 0 5 \end{matrix}) - (\begin{matrix} 2 - 5 \\ 1 3 \end{matrix})}

\Rightarrow A^{- 1} = \frac{1}{11} (\begin{matrix} 3 5 \\ - 1 2 \end{matrix}) ⋇

Answered by mathmax by abdo last updated on 19/May/21

p_{c} (A) = \det (A - xI) = | \begin{matrix} 2 - x - 5 \\ 1 3 - x \end{matrix} | = (2 - x) (3 - x) + 5

= (x - 2) (x - 3) + 5 = x^{2} - 5 x + 6 + 5 = x^{2} - 5 x + 11

cayley hamilton \Rightarrow A^{2} - 5 A + 11 I = 0 \Rightarrow A^{2} - 5 A = - 11 I \Rightarrow

- \frac{1}{11} (A^{2} - 5 A) = I \Rightarrow - \frac{1}{11} A \times (A - 5 I) = I \Rightarrow

A^{- 1} = - \frac{1}{11} (A - 5 I) = \frac{1}{11} (5 I - A) = \frac{5}{11} (\begin{matrix} 1 0 \\ 0 1 \end{matrix}) - \frac{1}{11} (\begin{matrix} 2 - 5 \\ 1 3 \end{matrix})

= (\begin{matrix} \frac{3}{11} \frac{5}{11} \\ - \frac{1}{11} \frac{2}{11} \end{matrix})