Transcription of Matrices - solving two simultaneous equations
1 Matrices - solving two simultaneous equationssigma-matrices8-2009-1 One of the most important applications of Matrices is to the solution of linear simultaneous this leaflet we explain how this can be simultaneous equations in matrix formConsider the simultaneous equationsx+ 2y= 43x 5y= 1 Provided you understand how Matrices are multiplied together you will realise that these can bewritten in matrix form as(1 23 5)(xy)=(41)WritingA=(1 23 5),X=(xy),andB=(41)we haveAX=BThis is thematrix formof the simultaneous equations . Here the only unknown is the matrixX,sinceAandBare already called thematrix of the simultaneous equationsGivenAX=Bwe can multiply both sides by the inverse ofA, provided this exists, to giveA 1AX=A 1 BButA 1A=I, the identity matrix.
2 Furthermore,IX=X, because multiplying any matrix by anidentity matrix of the appropriate size leaves the matrix unaltered. SoX=A 1 BifAX=B,thenX=A 1 BThis result gives us a method for solving simultaneous equations . All we need do is write themin matrix form, calculate the inverse of the matrix of coefficients, and finally perform a mathcentre the simultaneous equationsx+ 2y= 43x 5y= have already seen these equations in matrix form:(1 23 5)(xy)=(41).We need to calculate the inverse ofA=(1 23 5).A 1=1(1)( 5) (2)(3)( 5 2 3 1)= 111( 5 2 3 1)ThenXis given byX=A 1B= 111( 5 2 3 1)(41)= 111( 22 11)=(21)Hencex= 2,y= 1is the solution of the simultaneous the simultaneous equations2x+ 4y= 2 3x+y= matrix form:(2 4 3 1)(xy)=(211).
3 We need to calculate the inverse ofA=(2 4 3 1).A 1=1(2)(1) (4)( 3)(1 43 2)=114(1 43 2)ThenXis given byX=A 1B=114(1 43 2)(211)=114( 4228)=( 32)Hencex= 3,y= 2is the solution of the simultaneous equations . You should check the solutionby substitutingx= 3andy= 2into both given equations , and verifying in each case that theleft-hand side is equal to the right-hand that a video tutorial covering the content of this leaflet is available mathcentre 2009
