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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। |
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Answer» `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)]` |
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