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By using elementary row operations, find the inverese of the martrix A=[{:(1,3,-2),(-3,0,-5),(2,5,0):}]. |
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Answer» Solution :We have `[{:(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""[{:(1,0,10),(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)):}]` `implies""[{:(1,0,10),(0,1,-4),(0,0,25):}]=[{:(-5,0,3),(" "2,0,-1),(-15,1,9):}].A""[{:(R_(2)tp(-1).R_(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)toR_(1)-10R_(3)),(R_(2)toR_(2)+4R_(3)):}].` Hence, `A^(-1)=[{:(1,(-2)/(5),(-3)/(5)),((-2)/(5),(4)/(25),(11)/(25)),((-3)/(5),(1)/(25),(9)/(25)):}].` |
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