आव्यूह `A=[(1,1,1),(1,2,-3),(2,-1,3)]` के लिए दर्शाइए कि `A^(3)-6A^(2)+5A+11I=O` इसकी सहायता से `A^(-1)` ज्ञात कीजिए।

Explanation

`A^(2)=A.A`
`=[(1,1,1),(1,2,-3),(2,-1,3)][(1,1,1),(1,2,-3),(2,-1,-3)]`
`=[(1+1+2,1+2-1,1-3+3),(1+2-6,1+4+3,1-6-9),(2-1+6,2-2-3,2+3+9)]`
`=[(4,2,1),(-3,8,-14),(7,-3,14)]`
और `A^(3)=A^(2).A`
` =[(4,2,1),(-3,8,-14),(7,-3,14)][(1,1,1),(1,2,-3),(2,-1,3)]`
`=[(4+2+2,4+4-1,4-6+3),(-3+8-28,-3+16+14,-3-24-42), (7-3+28,7-6-14,7+9+42)]`
`=[(8,7,1),(-23,27,-69),(32,-13,58)]`
अब L.H.S `=A^(3)-6A(3)+5A+11I`
`=[(8,7,1),(-23,27,-690),(32,-13,58)]-6[(4,2,1),(-3,8,14),(7,-3,14)]`
`+5[(1,1,1),(1,2,-3),(2,-1,3)]+11[(1,0,0),(0,1,0),(0,0,1)]`
`=[(8-24+5+11,7-12+5+0,1-6+5+0),(-23+18+5+0,27-48+10+11,-69+84-15+0),(32-42+10+0,-13+18-5+0,58-84+15+11)]`
`[(0,0,0),(0,0,0),(0,0,0)]=O=R.H.S`
अब `|A|=|(1,1,1),(1,2,-3),(2,-1,3)|`
`=1(6-3)-1(3+6)+1(-1-4)`
`=3-9-5=-11!=0`
`:.A^(-1)` का अस्‍त‌ित्‍व है।
अब `A^(3)-6A^(2)+5A+11I=O`
`impliesA^(-1)(A^(3)-6A^(2)+5A+11I)=A^(-1).O`
`impliesA^(2)-6A+5I+11A^(-1)=O`
`implies-11A^(-1)=A^(2)-6A+5I`
`=[(4,2,1),(-3,8,-14),(7,-3,14)]-6[(1,1,1),(1,2,-3),(2,-1,3)]+5[(1,0,0),(0,1,0),(0,0,1)]`
`=[(4-6+5,2-6+0,1-6+0),(-3-6+0,8-12+5,-14-18+0),(7-12+0,-3+6+0,14-18+5)]`
`=[(3,-4,-5),(-9,1,4),(-5,3,1)]`
`=A^(-1)=-1/11[(3,-4,-5),(-9,1,4),(-5,3,1)]`