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Question Number 204218 by mnjuly1970 last updated on 08/Feb/24

	Answered by deleteduser1 last updated on 08/Feb/24

	$(\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ $\Rightarrow (\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}) (\begin{matrix} a x + b y \\ c x + d y \end{matrix}) = (\begin{matrix} (3 a + 2 c) x + (3 b + 2 d) y \\ (4 a + c) x + (4 b + d) y \end{matrix})$ $\Rightarrow (\begin{matrix} 3 a + 2 c & 3 b + 2 d \\ 4 a + c & 4 b + d \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \Rightarrow (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} \frac{- 1}{5} & \frac{2}{5} \\ \frac{4}{5} & \frac{- 3}{5} \end{matrix})$ $\Rightarrow A^{- 1} = 5^{- 1} (\begin{matrix} - 1 & 2 \\ 4 & - 3 \end{matrix}) = (\begin{matrix} - 3 & 6 \\ 12 & - 9 \end{matrix}) = (\begin{matrix} 4 & 6 \\ 5 & 5 \end{matrix})$
	Answered by Rasheed.Sindhi last updated on 08/Feb/24
	$let A^(−1) = ((w,x),(y,z) ) ((3,2),(4,1) ) ((w,x),(y,z) ) ≡ ((1,0),(0,1) ) (mod 7) (((3w+2y),(3x+2z)),((4w+y),(4x+z)) )≡ ((1,0),(0,1) ) (mod 7) Mod 7: { ((3w+2y≡1)),((4w+y≡0)) :} & { ((3x+2z≡0)),((4x+z≡1)) :} { ((3w+2y≡1)),((8w+2y≡0)) :} & { ((3x+2z≡0)),((8x+2z≡2)) :} 5w≡−1⇒5w≡6⇒w≡4 y≡−4w≡−4(4)≡−16≡5 & 5x≡2⇒x≡6 z≡1−4x=1−4(6)=−23≡5 A^(−1) ≡ ((w,x),(y,z) ) ≡ ((4,6),(5,5) ) Verification: ((3,2),(4,1) ) ((4,6),(5,5) ) = (((12+10),(18+10)),((16+5),(24+5)) ) = (((22),(28)),((21),(29)) )≡ ((1,0),(0,1) )$
	$l e t A^{- 1} = (\begin{matrix} w & x \\ y & z \end{matrix})$ $(\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}) (\begin{matrix} w & x \\ y & z \end{matrix}) \equiv (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (m o d 7)$ $(\begin{matrix} 3 w + 2 y & 3 x + 2 z \\ 4 w + y & 4 x + z \end{matrix}) \equiv (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (m o d 7)$ $Mod 7 :$ ${\begin{cases} 3 w + 2 y \equiv 1 \\ 4 w + y \equiv 0 \end{cases} & {\begin{cases} 3 x + 2 z \equiv 0 \\ 4 x + z \equiv 1 \end{cases}$ ${\begin{cases} 3 w + 2 y \equiv 1 \\ 8 w + 2 y \equiv 0 \end{cases} & {\begin{cases} 3 x + 2 z \equiv 0 \\ 8 x + 2 z \equiv 2 \end{cases}$ $5 w \equiv - 1 \Rightarrow 5 w \equiv 6 \Rightarrow w \equiv 4$ $y \equiv - 4 w \equiv - 4 (4) \equiv - 16 \equiv 5$ $&$ $5 x \equiv 2 \Rightarrow x \equiv 6$ $z \equiv 1 - 4 x = 1 - 4 (6) = - 23 \equiv 5$ $A^{- 1} \equiv (\begin{matrix} w & x \\ y & z \end{matrix}) \equiv (\begin{matrix} 4 & 6 \\ 5 & 5 \end{matrix})$ $V e r i f i c a t i o n :$ $(\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}) (\begin{matrix} 4 & 6 \\ 5 & 5 \end{matrix}) = (\begin{matrix} 12 + 10 & 18 + 10 \\ 16 + 5 & 24 + 5 \end{matrix})$ $= (\begin{matrix} 22 & 28 \\ 21 & 29 \end{matrix}) \equiv (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$
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