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Using elementarytransformations, find the inverse of the matrix :`[[2, 0,-1],[ 5, 1, 0],[ 0, 1, 3]]` |
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Answer» We have `[{:(2,0,-1),(5,1,0),(0,1,3):}]=[{:(1,0,0),(0,1,0),(0,0,1):}].A` `implies" "[{:(2,0,-1),(1,1,2),(0,1,3):}]=[{:(1,0,0),(-2,1,0),(0,0,1):}].A" "[R_(2)toR_(2)-2R_(1)]` `implies" "[{:(1,1,2),(2,0,-1),(0,1,3):}]=[{:(-2,1,0),(1,0,0),(0,0,1):}].A" "[R_(1)harrR_(2)]` `implies" "[{:(1,1,2),(0,-2,-5),(0,1,3):}]=[{:(-2,1,0),(5,-2,0),(0,0,1):}].A" "[R_(2)toR_(2)-2R_(1)]` `implies" "[{:(1," "1," "2),(0," "1," "3),(0,-2,-5):}]=[{:(-2," "1,0),(0," "0,1),(5,-2,0):}].A" "[R_(2)harrR_(3)]` `implies" "[{:(1,0,-1),(0,1,3),(0,0,1):}]=[{:(-2," "1,-1),(0," "0," "1),(5,-2," "2):}].A" "[{:(R_(1)toR_(1)-R_(2)),(R_(3)toR_(3)+2R_(2)):}]` `implies" "[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(" "3,-1," "1),(-15," "6,-5),(" "5,-2," "2):}].A" "[{:(R_(1)toR_(1)+R_(3)),(R_(2)toR_(2)-3R_(3)):}].` Hence, `A^(-1)=[{:(3,-1," "1),(-15," "6,-5),(5,-2," "2):}].` |
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