प्रारम्भिक पंक्ति रूपान्तरणों के प्रयोग से आव्यूह A `[{:(1,3,-2),(-3,0,-5),(2,5,0):}]` का व्युत्क्रम ज्ञात कीजिए ।

Explanation

हम जानते है कि -
`A=I*A`
`rArr[{:(1,3,-2),(-3,0,-5),(2,5,0):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]A`
`[{:(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)`
`rArr[{:(1,3,-2),(0,-1,4),(0,9,-11):}]=[{:(1,0,0),(-2,0,1),(3,1,0):}]A " " R_(2)harrR_(3)`
`rArr[{:(1,3,-2),(0,1,-4),(0,9,-11):}]=[{:(1,0,0),(2,0,-1),(3,1,0):}]A" " R_(2)toR_(2)*(-1)`
`rArr[{:(1,0,10),(0,1,-4),(0,0,25):}]=[{:(-5,0,3),(2,0,-1),(-15,1,9):}]A" " R_(1)toR_(1)-3R_(2)`
`R_(3)toR_(3)-9R_(2)`
`rArr[{:(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)`
`rArr[{:(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),):}`
`rArrA^(-1)=[{:(1,-(2)/(5),-(3)/(5)),(-(2)/(5),(4)/(25),(11)/(25)),(-(3)/(5),(1)/(25),(9)/(25)):}]`