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।

Explanation

`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)`