Using elementary transformations (operations), find the inverse of the following matrices, if it exists `[(2,-1,3),(-5,3,1),(-3,2,3)]`

Explanation

For getting the inverse of the given matrix A by row elementary operations we may write the given matrix as
A=IA
(i) `because [{:(2,-1,3),(-5,3,1),(-3,2,0):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]A`
`rArr [{:(2,-1,3),(-3, 2,4),(-3,2,3):}]=[{:(1,0,0),(1,1,0),(0,0,1):}]A [because R_(2)rArrR_(2)+R_(1)]`
`rArr[{:(2,-1,3):}]=[{:(1,0,0),(1,1,0),(0, 0,1):}]A[becauseR_(3)rArrR_(3)-R_(2)]`
`rArr [{:(-1,1,7),(-3,2,4),(0,0,-1):}]=[{:(2,1,0),(-5,-2,0),(-1,-1,1):}]A[because R_(2)rArrR_(2)-3R_(1)]`
`rArr[{:(-1,0,-10),(0,-1,-17),(0,0,1):}]=[{:(-3,-1,0),(-5,-2,0),(1,1,-1):}]A[because R_(1)rArrR_(1)+R_(2)"and" R_(3)rArr-1.R_(3)]`
`rArr[{:(1,0,0),(0,1,0),(0,0 ,1):}]=[{:(-7,-9,10),(1,1, -1):}]A[because R_(1)rArr-1R_(1) "and" R_(2)rArr-1R_(2)]`
So, the inverse of A is `=[{:(-7,-9,10),(-12,-15,17),(1,1,-1):}]`
(ii) `therefore [{:(2,3,-3),(-1,-2,2),(1,1,-1):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]A`
`rArr [{:(0,1,-1), (0,-1,1),(1,1,-1):}]=[{:(1,0,-2),(0,1,1),(0,0,1):}]A [because R_(2)rArrR_(2)+R_(3) "and" R_(1)rArrR_(1)-2R_(3)]`
`rArr[{:(0,1,-1),(0,0,0),(1,1,1):}]=[{,(1,0,-2),(2,1,-2),(0,0,1):}] [because R_(2)rArrR_(2)+R_(1)]`
Since, second row of the matrix A on LHS is containing all zeroes, so we can say that inverse of matrix A does not exist.
(iii) `therefore [{:(2,0,-1),(5,1,0),(0,1,3):}]=[{:(1,0,0),(0,1,0),(0,0, 1):}]A`
`rArr[{:(2,0,-1),(3,1,1),(0,1,3):}]=[{:(1,0,0),(-1,1,0),(0,0,1):}]A[because R_(2)rArrR_(2)-R_(1)]`
`rArr [{:(2,0,-1),(1,1,2),(2,1,2):}]=[{:(1,0,0),((-5)/(2),1,0),(2, 0,1):}]A[because R_(3)rArrR_(3)+R_(1)"and" R_(2)rArrR_(2)-(1)/(2)R_(1)]`
`rArr[{:(2,0,(-1),(0,1,(5)/(2)),(0,1,3):}]=[{:(1,0,0),((-5)/(2),1,0),(0,0,1):}][because R_(3)rArrR_(3)-2R_(1)]`
`rArr [{:(1,0,(-1)/(2)),(0,1,(5)/(2)),(0,0,1):}]=[{:(because R_(1)rArr(1)/(2)R_(1) "and" R_(3)rArr2R_(3)]`
`rArr [{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(3,-1,1),(-15,6,-5),(5,-2,2):}]A[{:(because R_(1)rArrR_(1)R_(1)+(1)/(2)R_(3) "and" R_(2)rArrR_(2)-(5)/(2)R_(3)]`
Hence, `[{:(3,-1,1),(-15,6,-5),(5,-2,2):}]` is the inverse of given matrix A.