मान लीजिए `A=[(1,-2,1),(-2,3,1),(1,1,5)]` हो तो सत्यापित कीजिए कि
`((A')')=A`

Explanation

`A=[(1,-2,1),(-2,3,1),(1,1,5)]`
`:.|A|=|(1,-2,1),(-2,3,1),(1,1,5)|`
`=1(15-1)-(-2)(-10-1)+1(-2-3)`
`=14-22-5=-13!=0`
`A_(11)=14, A_(12)=11,A_(13)=-5`
`A_(21)=11,A_(22)=4, A_(23)=-3`
`A_(31)=-5,A_(32)=-3,A_(33)=-1`
`:.adjA=[(14,11,-5),(11,4,-3),(-5,-3,-1)]=[(14,11,-5),(11,4,-3),(-5,-3,-1)]`
और `A^(-1)=1/(|A|).adjA=-1/13[(14,11,-5),(11,4,-3),(-5,-3,-1)]`
`=[(-14/13,-11/13,5/12),(-11/13,-4/13,3/13),(5/13,3/13,1/13)]` ltbr (i) माना `B=adjA=[(14,11,-5),(11,4,-3),(-5,-3,-1)]`
`implies|B|=|(14,11,-5),(11,4,-3),(-5,-3,-1)|`
`=14(-4-9)-11(-11-15)-5(-33+20)`
`=-182+286+65=169!=0`
`B_(11)=-13, B_(12)=26,B_(13)=-13`
`B_(21)=26,B_(22)=-39, B_(23)=-13`
`B_(31)=-13, B_(32)=-13,B_(33)=-65`
`:.adjB=[(-13,26,-13),(26,-39,-13),(-13,-13,65)]=[(-13,26,-13),(29,-39,-13),(-13,-13,-65)]`
और `B^(-1)=1/(|B|)adjB=1/169[(-13,26,-13),(26,-39,-13),(-13,-13,-65)]`

`=1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`
माना `C=A^(-1)=[(-14/13, -11/13, 5/13),(-11/13,-4/13,3/13),(5/13,3/13,1/13)]`
`C_(11)=-13/169=-1/13,C_(12)=26/169=2/13`
`C_(13)=-13/169=-1/13`
`C_(21)=26/169=2/13,C_(22)=39/169=-3/13`
`C_(23)=-13/169=-1/13`
`C_(31)=-13/169=-1/13,C_(32)=-13/169=-1/13`
`C_(33)=-65/169=-5/13`
`:.adjC=adjA^(-1)=[(-1/13,2/13,-1/13),(2/13,-3/13,-1/13),(-1/13,-1/13,-5/13)]`
`=[(-1/13,2/13,-1/13),(2/13,-3/13,-1/13),(-1/13,-1/13,-5/13)]`
`=1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]=B^(-1)-(adjA)^(-1) `
(ii) `|A^(-1)|=-14/13(-4/169-9/169)`
`+11/13(-11/169-15/169)`
`+5/13(-33/169+20/169)`
`=-1/13`
`:.(A^(-1))^(-1)=1/(|A^(-1)|).adjA^(-1)`
`=1/(-1/13) . 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`
`=[(1,-2,1),(2,3,1),(-1,1,5)]=A`