Menu Close

solve-using-matrix-method-x-y-4-2x-3y-5-

Tinku Tara 0 Comments
Pin




Question Number 37084 by Rio Mike last updated on 08/Jun/18
solve using matrix method x − y= 4 2x − 3y= 5
$$\:\:{solve}\:{using}\:{matrix}\:{method} \\ $$$$\:\:\:\:\:\:{x}\:−\:{y}=\:\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{2}{x}\:−\:\mathrm{3}{y}=\:\mathrm{5} \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
(s)⇔ (((1 −1)),((2 −3)) ) ((x),(y) ) = ((4),(5) ) ⇔ A. ((x),(y) ) = ((4),(5) ) det A=−1≠0 ⇒ ((x),(y) ) =A^(−1) . ((4),(5) ) A^(−1) = ((t(comA))/(det A)) with comA=(−1)^(i+j) A_(ij) = (((a_(11) a_(12) )),((a_(21) a_(22) )) ) = (((−3 −2)),(( 1 1)) ) and t(comA) = (((−3 1)),((−2 1)) ) ⇒A^(−1) = (((3 −1)),((2 −1)) ) ((x),(y) ) = (((3 −1)),((2 −1)) ) ((4),(5) ) = ((7),(3) ) ⇒x=7 and y=3
$$\left({s}\right)\Leftrightarrow\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:−\mathrm{3}}\end{pmatrix}\:\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:\:=\begin{pmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{pmatrix}\:\:\Leftrightarrow\:{A}.\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{pmatrix} \\ $$$${det}\:{A}=−\mathrm{1}\neq\mathrm{0}\:\Rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:\:={A}^{−\mathrm{1}} .\begin{pmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{pmatrix} \\ $$$${A}^{−\mathrm{1}} \:=\:\:\frac{{t}\left({comA}\right)}{{det}\:{A}}\:\:\:{with}\:{comA}=\left(−\mathrm{1}\right)^{{i}+{j}} \:{A}_{{ij}} \\ $$$$=\begin{pmatrix}{{a}_{\mathrm{11}} \:\:\:\:\:\:\:{a}_{\mathrm{12}} }\\{{a}_{\mathrm{21}} \:\:\:\:\:\:\:\:{a}_{\mathrm{22}} }\end{pmatrix}\:\:\:=\begin{pmatrix}{−\mathrm{3}\:\:\:\:\:\:\:\:\:−\mathrm{2}}\\{\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:{and} \\ $$$${t}\left({comA}\right)\:=\:\begin{pmatrix}{−\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\Rightarrow{A}^{−\mathrm{1}} \:=\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:\:\:=\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\:\:\begin{pmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{7}}\\{\mathrm{3}}\end{pmatrix}\:\:\Rightarrow{x}=\mathrm{7}\:{and}\:{y}=\mathrm{3} \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
this method is general for n unknown (x_i ) if the matrix A is inversible .
$${this}\:{method}\:{is}\:{general}\:{for}\:{n}\:{unknown}\:\left({x}_{{i}} \right)\:{if} \\ $$$${the}\:{matrix}\:{A}\:{is}\:{inversible}\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *