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2x+ 7y+z= 4x+ 4y-z= 9x+ 3y= 3We can express asAX=Bas follows27114-1130xyz=493.We have already seen thatA-1=-32-321121212-3212-12-12.139
Using inverse of matrix method to solveX=A-1B, i.e.xyz=-32-321121212-3212-12-12493=(-32)(4) + (-32)(9) + (112)(3)(12)(4) + (12) + (-32)(3)(12)(4) + (-12)(9) + (-12)(3)=-6-272+3322 +92-922-92-32=-32-4.Hencex=-3,y= 2 andz=-4.Cramer’s RuleThis method uses determinant to solve the system of linear equations.Consider the system of linear equationAX=Bwith unique solution. Then•detAexist, say|A|=D.•LetDx=|Ax|whereAxis a matrix obtained by replacing the column ofcoefficients ofxbyB.SimilarlyDy=|Ay|whereAyis a matrix obtained by replacing thecolumn of coefficients ofybyB;Dy=|Ay|in similar fashion, etc.•Then we can obtain unique solutionx=DxD;y=DyD;z=DzD;etc. whereD6= 0.140
Dy=|Ay|=24119-1130=2413130130= 1×31313= 3×3-13×1 = 9-13 =-4.Dz=|Az|=274149133=01-21490-1-6=-1×1-2-1-6=-[1×(-6)-(-1)×(-2)] =-(-6-2) = 8.Sox=DxD=6-2=-3,y=DyD=-4-2= 2,z=DzD=8-2=-4.5.63×3systems with parametersConsider the system of linear equations:142
Example 5.6.1.x+ 3y+ (2-3k)z= 32x+ 7y+ (4-7k)z= 72x+ky+ (k+ 2)z= 2k-1Determine the values ofk∈Rsuch that(a) The system has no solution(b) The system has infinitely many solutions(c) The system has unique (only one) solution (consistent).Solution:We perform Gauss-Jordan elimination of the augmented matrix to obtain:132-3k274-7k2kk+ 2372k-1∼132-3k01-k0k-67k-2312k-7R2-2R1R3-2R1∼10201-k00k2+k-201k-1R1-3R2R3-(k-6)R2∼10201-k00(k+ 2)(k-1)01k-1Then(a) The system has no solution if(k+ 2)(k-1) = 0 andk-16= 0, sok=-2 andk6= 1, since fork=-2,143
we obtain10201200001-3which is clearly impossible.(b) The system has infinitely many solutions if(k+ 2)(k-1) = 0 andk-1 = 0, thusk= 1.(c) The system has unique solution if(k+ 2)(k-1)6= 0 thusk6=-2 andk6= 1.Tutorials 184.108.40.206. Use Gauss-Jordan Elimination to find values ofx,yandzwhere possible and describe whether each type of solution is eitherunique, no solution or infinitely many solutions.(a)x+ 3y+ 2z= 4-2x-7y+z= 93x+ 10y+ 9z= 19(b)x+ 2y+z= 22x+ 3y+ 3z= 3-3x-4y-5z=-5(c)x+ 2y+ 3z= 32x+ 3y+ 8z= 45x+ 8y+ 19z= 11(d) From (a), (b) and (c) above, where you found a unique solution, usethe inverse of matrix method to confirm the solution.