Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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 .

$${this}\:{method}\:{is}\:{general}\:{for}\:{n}\:{unknown}\:\left({x}_{{i}} \right)\:{if} \\ $$$${the}\:{matrix}\:{A}\:{is}\:{inversible}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com