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Question-204218

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Question Number 204218 by mnjuly1970 last updated on 08/Feb/24
Answered by AST last updated on 08/Feb/24
((3,2),(4,1) ) ((a,b),(c,d) ) ((( x)),(( y)) )= ((1,0),(0,1) ) ⇒ ((3,2),(4,1) ) (((ax+by)),((cx+dy)) )= ((((3a+2c)x+(3b+2d)y)),(((4a+c)x+(4b+d)y)) ) ⇒ (((3a+2c),(3b+2d)),((4a+c),(4b+d)) )= ((1,0),(0,1) )⇒ ((a,b),(c,d) )= ((((−1)/5),(2/5)),((4/5),((−3)/5)) ) ⇒A^(−1) =5^(−1) (((−1),2),(4,(−3)) )= (((−3),6),((12),(−9)) )= ((4,6),(5,5) )
$$\begin{pmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\begin{pmatrix}{\:{x}}\\{\:{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{ax}+{by}}\\{{cx}+{dy}}\end{pmatrix}=\begin{pmatrix}{\left(\mathrm{3}{a}+\mathrm{2}{c}\right){x}+\left(\mathrm{3}{b}+\mathrm{2}{d}\right){y}}\\{\left(\mathrm{4}{a}+{c}\right){x}+\left(\mathrm{4}{b}+{d}\right){y}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{3}{a}+\mathrm{2}{c}}&{\mathrm{3}{b}+\mathrm{2}{d}}\\{\mathrm{4}{a}+{c}}&{\mathrm{4}{b}+{d}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\Rightarrow\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}=\begin{pmatrix}{\frac{−\mathrm{1}}{\mathrm{5}}}&{\frac{\mathrm{2}}{\mathrm{5}}}\\{\frac{\mathrm{4}}{\mathrm{5}}}&{\frac{−\mathrm{3}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\Rightarrow{A}^{−\mathrm{1}} =\mathrm{5}^{−\mathrm{1}} \begin{pmatrix}{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{−\mathrm{3}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{3}}&{\mathrm{6}}\\{\mathrm{12}}&{−\mathrm{9}}\end{pmatrix}=\begin{pmatrix}{\mathrm{4}}&{\mathrm{6}}\\{\mathrm{5}}&{\mathrm{5}}\end{pmatrix} \\ $$
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) )
$${let}\:{A}^{−\mathrm{1}} =\begin{pmatrix}{{w}}&{{x}}\\{{y}}&{{z}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{w}}&{{x}}\\{{y}}&{{z}}\end{pmatrix}\:\equiv\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\:\left({mod}\:\mathrm{7}\right) \\ $$$$\begin{pmatrix}{\mathrm{3}{w}+\mathrm{2}{y}}&{\mathrm{3}{x}+\mathrm{2}{z}}\\{\mathrm{4}{w}+{y}}&{\mathrm{4}{x}+{z}}\end{pmatrix}\equiv\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\:\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{Mod}\:\mathrm{7}: \\ $$$$\begin{cases}{\mathrm{3}{w}+\mathrm{2}{y}\equiv\mathrm{1}}\\{\mathrm{4}{w}+{y}\equiv\mathrm{0}}\end{cases}\:\&\:\begin{cases}{\mathrm{3}{x}+\mathrm{2}{z}\equiv\mathrm{0}}\\{\mathrm{4}{x}+{z}\equiv\mathrm{1}}\end{cases}\: \\ $$$$\begin{cases}{\mathrm{3}{w}+\mathrm{2}{y}\equiv\mathrm{1}}\\{\mathrm{8}{w}+\mathrm{2}{y}\equiv\mathrm{0}}\end{cases}\:\&\:\begin{cases}{\mathrm{3}{x}+\mathrm{2}{z}\equiv\mathrm{0}}\\{\mathrm{8}{x}+\mathrm{2}{z}\equiv\mathrm{2}}\end{cases}\: \\ $$$$\mathrm{5}{w}\equiv−\mathrm{1}\Rightarrow\mathrm{5}{w}\equiv\mathrm{6}\Rightarrow{w}\equiv\mathrm{4} \\ $$$${y}\equiv−\mathrm{4}{w}\equiv−\mathrm{4}\left(\mathrm{4}\right)\equiv−\mathrm{16}\equiv\mathrm{5} \\ $$$$\& \\ $$$$\mathrm{5}{x}\equiv\mathrm{2}\Rightarrow{x}\equiv\mathrm{6} \\ $$$${z}\equiv\mathrm{1}−\mathrm{4}{x}=\mathrm{1}−\mathrm{4}\left(\mathrm{6}\right)=−\mathrm{23}\equiv\mathrm{5} \\ $$$${A}^{−\mathrm{1}} \equiv\begin{pmatrix}{{w}}&{{x}}\\{{y}}&{{z}}\end{pmatrix}\:\equiv\begin{pmatrix}{\mathrm{4}}&{\mathrm{6}}\\{\mathrm{5}}&{\mathrm{5}}\end{pmatrix} \\ $$$${Verification}:\: \\ $$$$\begin{pmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{4}}&{\mathrm{6}}\\{\mathrm{5}}&{\mathrm{5}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{12}+\mathrm{10}}&{\mathrm{18}+\mathrm{10}}\\{\mathrm{16}+\mathrm{5}}&{\mathrm{24}+\mathrm{5}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\begin{pmatrix}{\mathrm{22}}&{\mathrm{28}}\\{\mathrm{21}}&{\mathrm{29}}\end{pmatrix}\equiv\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$

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