let A= [((a b)),((c d)) ]use the augmented matrix[A I] and elementary row operation to show A^(−1) = (1/(ad bc)) [((a b)),((c d)) ]and show that det(A^(−1) )=(1/(det(A)))


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Question Number 78670 by berket last updated on 19/Jan/20

$l e t A = [\begin{matrix} a b \\ c d \end{matrix}] u s e t h e a u g m e n t e d m a t r i x [A I] a n d e l e m e n t a r y r o w o p e r a t i o n t o s h o w A^{- 1} = \frac{1}{a d b c} [\begin{matrix} a b \\ c d \end{matrix}] a n d s h o w t h a t d e t (A^{- 1}) = \frac{1}{d e t (A)}$
Commented by abdomathmax last updated on 20/Jan/20

$P c (A) = d e t (A - x I) = \| \begin{matrix} a - x b \\ c d - x \end{matrix} \|$ $= (a - x) (d - x) - b c = a d - a x - d x + x^{2} - b c$ $= a d - b c - (a + d) x + x^{2}$ $c a y l e y h a m i t o n \Rightarrow A^{2} - (a + d) A + (a d - b c) I = 0 \Rightarrow$ $A^{2} - (a + d) A = (b c - a d) I i f b c - a d \neq 0 w e g e t$ $\frac{1}{b c - a d} A (A - (a + d) I) = I \Rightarrow$ $A^{- 1} = \frac{1}{b c - a d} (A - (a + d) I = \frac{1}{b c - a d} ((\begin{matrix} a b \\ c d \end{matrix}) - (\begin{matrix} a + d 0 \\ 0 a + d \end{matrix}))$ $= \frac{1}{b c - a d} (\begin{matrix} - d b \\ c - a \end{matrix})$
Commented by abdomathmax last updated on 20/Jan/20

$A^{- 1} = (\begin{matrix} \frac{d}{a d - b c} \frac{- b}{a d - b c} \\ \frac{- c}{a d - b c} \frac{a}{a d - b c} \end{matrix})$
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