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If `A=[[2,3,4],[1,-1,0],[0,1,2]]`, find `A^(-1)`. Hence, solve the system of equations x-y=3, 2x+3y+4z=17, y+2z=7 |
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Answer» `** **A = [{:(,2,3,4),(,1,-1,0),(,0,1,2):}]` `|A| = 2 (-2) =- 3(2-0) + 4( 1-0) = - 6 ne 0` `therefore A^(-1) ` exists Cofactors `{:(,A_(11) = -2 ,A_(12)=- 2 , A_(13) = 1),(,A_(21) - 2 ,A_(22)=4,A_(23) = -2),(,A_(31) = 4,A_(34) = 4 ,A_(33) = - 5):}` `Adj A= [{:(,-2,-2,1),(,-2,4,-2),(,4,4,-5):}]` `Adj A= [{:(,-2,-2,4),(,-2,4,4),(,1,-2,-5):}]` `A^(-1) = (AdjA)/(|A|) = (1)/(-6) [{:(,-2,-2,4),(,-2,4,4),(,1,-2,-5):}]` System of equation can be written as AX = B `"Where A"=[{:(,2,3,4),(,1,-1,0),(,0,1,2):}],X =[{:(,x),(,y),(,z):}] ,B = [{:(,17),(,3),(,7):}]` Now AX = B `rArr X = A^(-1)B` `rArr =(1)/(-6)[{:(,-2,-2,4),(,-2,4,4),(,1,-2,-5):}][{:(,17),(,3),(,7):}]` `rArr X =(1)/(-6) [{:(,-34,-6,+28),(,-34,+12,+28),(,17,-6,35):}]` `rArr X = (1)/(-6)[{:(,-12),(,6),(,-24):}]` `X = [{:(,x),(,y),(,z):}] =[{:(,2),(,-1),(,4):}]` `rArr x= 2, y = - 1, z = 4` |
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