InterviewSolution
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InterviewSolution
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यदि `A=[(2,-1,1),(-1,2,-1),(1-1,2)]`, तो सत्यापित कीजिए कि `A^(3)-6A^(2)+9A-4I=O` है तथा इसकी सहायता से `A^(-1)` ज्ञात कीजिए। |
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Answer» `A=[(2,-1,1),(-1,2,-1),(1,-1,2)]` `:.A^(2)=A.A[(2,-1,1),(-1,2,-1),(1,-1,2)][(2,-1,1),(-1,2,-1),(1,-1,2)]` `=[(4+1+1,-2-2-1,2+1+2),(-2-2-1,1+4+1,-1-2-2),(2+1+2,-1-2-2,1+1+4)]` `=[(6,-5,5),(-5,6,-5),(5,-5,6)]` `=A^(3)=A^(2).A=[(6,-5,5),(-5,6,-5),(5,-5,6)][(2,-1,1),(-1,2,-1),(1,-1,2)]` `[(12+5+5,-6-10-5,6+5+10),(-10-6-5,5-12+5,-5-6-10),(10+56,-5-10-6,5+5+12)]` `=[(22,-21,21),(-21,22,-21),(21,-21,22)]` अब L.H.S `=A^(3)-6A^(2)+9A-4I` `=[(22,-21,21),(-21,22,-21),(21,-21,22)]-6[(6,-5,5),(-5,6,-5),(5,-5,6)]` `+9[(2,-1,1),(-1,2,-1),(1,-1,2)]-4[(1,0,0),(0,1,0),(0,0,1)]` `=[(22,-21,21),(-21,22,-21),(21,-21,22)]+[(-39,30,-30),(30,-36,30),(-30,30,-36)]` `+[(18,-9,9),(-9,18,-9),(9,-9,18)]+[(-4,0,0),(0,-4,0),(0,0,-4)]` `=[(0,0,0),(0,0,0),(0,0,0)]=0=R.H.S` अब `|A|=|(2,-1,1),(-1,2,-1),(1,-1,2)|` `=2(4-1)-(-1)(-2+1)+1(1-2)` `=6-1-1=4!=0` `:.A^(-1)` का अस्तित्व है। हम सिद्ध कर चुके हैं कि `A^(3)-6A^(2)+9A-4I=0` `impliesA^(-1)(A^(3)-6A^(2)+9A-4I)=A^(-1)0` `impliesA^(2)-6A+9I-4A^(-1)=0` `implies4A^(-1)=A^(2)=6A+9I` `[(6,-5,5),(-5,6,-5),(5,-5,6)]-6[(2,-1,1),(-1,2,-1),(1,-1,2)]+9[(1,0,0),(0,1,0),(0,0,1)]` `=[(6,-5,5),(-5,6,-5),(5,-5,6)]+[(-12,6,-6),(6,-12,6),(-6,6,-12)]+[(9,0,0),(0,9,0),(0,0,9)]` `=[(3,1,-1),(1,3,1),(-1,1,3)]` `impliesA^(-1)=1/4[(3,1,-1),(1,3,1),(-1,1,3)]` `(adjA).A=|A|.I_(3)=|A|[(1,0,0),(0,1,0),(0,0,1)]` `=[(|A|,0,0),(0,|A|,0),(0,0,|A|)]` `implies|(adjA).A|=|(|A|,0,0),(0,|A|,0),(0,0,|A|)|=|A|^(3)` `implies |adjA|=|A|^(2)` |
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