यदि A=`[{:(2,-3,3),(2," "2,3),(3,-2,2):}]`तो `A^-1` ज्ञात कीजिए

Explanation

मान `A=[{:(2,-3,3),(2," "2,3),(3,-2,2):}]`
अब , A=IA
`rArr[{:(2,-3,3),(2," "2,3),(3,-2,2):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]A`
`rArr[{:(1,1,4),(2,2,3),(3,-2,2):}]=[{:(1,1,-1),(0,1,0),(0,0,1):}]A " " R_(1)toR_(1)+R_(2)-R_(3)`
`rArr[{:(1," "1," "4),(0," "0,-5),(0,-5,-10):}]=[{:(1," "1,-1),(-2,-1," "2),(-3,-3," "4):}]A " " {:(R_(2)toR_(2)-2R_(1),),(R_(3)toR_(3)-3R_(1),):}`
`rArr[{:(1,1,4),(0,-5,-10),(0,0,-5):}]=[{:(1,1,-1),(-3,-3," "4),(-2,-1," "2):}]A " " R_(2)harrR_(3)`
`rArr[{:(1,1,4),(0,1,2),(0,0,1):}]=[{:(1,1,-1),((3)/(5),(3)/(5),-(4)/(5)),((2)/(5),(1)/(5),-(2)/(5)):}]A " " {:(R_(2)to-(1)/(5)R_(2),),(R_(3)to-(1)/(5)R_(3),):}`
`rArr[{:(1,0,2),(0,1,2),(0,0,1):}]=[{:((2)/(5),(2)/(5),-(1)/(5)),((3)/(5),(3)/(5),-(4)/(5)),((2)/(5),(1)/(5),-(2)/(5)):}]A" " R_(1)toR_(1)-R_(2)`
`rArr[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(-(2)/(5),0,(3)/(5)),(-(1)/(5),(1)/(5),0),((2)/(5),(1)/(5),-(2)/(5)):}]A " " R_(1)toR_(1)-2R_(3),R_(2)toR_(2)-2R_(3)`
`thereforeA^(-1)=[{:(-(2)/(5),0,(3)/(5)),(-(1)/(5),(1)/(5)," "0),((2)/(5),(1)/(5),-(2)/(5)):}]=-(1)/(5)[{:(2,0,-3),(1,-1," "0),(-2,-1," "2):}]`