`[{:(2,0,-1),(5,1,0),(0,1,3):}]`

Explanation

मान `A=[{:(2,0,-1),(5,1,0),(0,1,3):}]`
अब `A*A^(-1)=I`
`rArr[{:(2,0,-1),(5,1,0),(0,1,3):}]A^(-1)=[{:(1,0,0),(0,1,0),(0,0,1):}]`
`rArr[{:(1,0,-(1)/(2)),(5,1," "0),(0,1," "3):}]A^(_1)=[{:((1)/(2),0,0),(0,1,0),(0,0,1):}] " " R_(1)to(1)/(2)R_(1)`
`rArr[{:(1,0,-(1)/(2)),(0,1," "(5)/(2)),(0,1," "3):}]A^(-1)=[{:((1)/(2),0,0),(-(5)/(2),1,0),(0,0,1):}]R_(2)toR_(2)-5R_(1)`
`rArr[{:(1,0,-(1)/(2)),(0,1," "(5)/(2)),(0,0," "(1)/(2)):}]A^(_1)=[{:((1)/(2),0,0),(-(5)/(2),1,0),((5)/(2),-1,1):}]R_(3)toR_(3)-R_(2)`
`rArr[{:(1,0,-(1)/(2)),(0,1," "(5)/(2)),(0,0," "1):}]A^(-1)=[{:((1)/(2),0,0),(-(5)/(2),1,0),(5,-2,2):}]" " R_(3)to2R_(3)`
`rArr[{:(1,0,0),(0,1,0),(0,0,1):}]A^(_1)=[{:(3,-1," "1),(-15," "6,-5),(5,-2," "2):}]`
`R_(1)toR_(1)+(1)/(2)R_(3)` और `R_(2)toR_(2)-(5)/(2)R_(3)`
`rArrA^(_1)=[{:(3,-1," "1),(-15," "6,-5),(5,-2," "2):}]`