Find the inverase of Matrix `A=[{:(1,3,-2),(-3,0,-5),(2,5,0):}]` by using elementary row transformation.

Answer

N/a

Explanation

we know that
`A=I.A`
`A=[{:(1,3,-2),(-3,0,-5),(2,5,0):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]A`
`implies [{:(1,3,-2),(0,9,-11),(0,-1,4):}]=[{:(1,0,0),(3,1,0),(-2,0,1):}]A`
`(R_(2)toR_(2)+3R_(1))`
`(R_(3)toR_(3)-2R_(1))`
`implies [{:(1,3,-2),(0,-1,4),(0,9,-11):}]=[{:(1,0,0),(-2,0,1),(3,1,0):}]A(R_(2)harrR_(3))`
`implies [{:(1,3,-2),(0,1,-4),(0,9,-11):}]=[{:(1,0,0),(2,0,-1),(3,1,0):}]A`
`[R_(2)toR_(2).(-1)]`
`implies[{:(1,0,10),(0,1,-4),(0,0,25):}]=[{:(-5,0,3),(2,0,-1),(-15,1,9):}]A`
`(R_(1)to R_(1)-3R_(2))`
`(R_(3)to R_(3)-9R_(2))`
`implies [{:(1,0,10),(0,1,-4),(0,0,1):}]=[{:(-5,0,3),(2,0,-1),(-(3)/(5),(1)/(25),(9)/(25)):}]A`
`(R_(3)to(1)/(25)R_(3))`
`implies [{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(1,-(2)/(5),-(3)/(5)),(-(2)/(5),(4)/(25),(11)/(25)),(-(3)/(5),(1)/(25),(9)/(25)):}]A`
`(R_(1)=R_(1)-10R_(3))`
`(R_(2)=R_(2)+4R_(3))`
`implies A^(-1)=[{:(1,-(2)/(5),(3)/(5)),(-(2)/(5),(4)/(25),(11)/(25)),(-(3)/(5),(1)/(25),(9)/(25)):}].`