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Question
Complete the following activity to find inverse of matrix using elementary column transformations and hence verify.
`[(2, 0, -1),(5, 1, 0),(0, 1, 3)]` B−1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
C1 → C1 + C3
`[("( )", 0, -1),("( )", 1, 0),("( )", 1, 3)]` B−1 = `[("( )", 0, 0),("( )", 1, 0),("( )", 0, 1)]`
C3 → C3 + C1
`[(1, 0, 0),("( )", 1, "( )"),(3, 1, "( )")]` B−1 = `[(1, 0, "( )"),(0, 1, 0),("( )", 0, "( )")]`
C1 → C1 – 5C2, C3 → C3 – 5C2
`[(1, "( )", 0),(0, 1, 0),("( )", 1, "( )")]` B−1 = `[(1, 0, "( )"),("( )", 1, -5),(1, "( )", 2)]`
C1 → C1 – 2C3, C2 → C2 – C3
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` B−1 = `[(3, -1, "( )"),("( )", 6, -5),(5, "( )", "( )")]`
B−1 = `[("( )", "( )", "( )"),("( )", "( )", "( )"),("( )", "( )", "( )")]`
`[(2, "( )", -1),("( )", 1, 0),(0, 1, "( )")] [(3, "( )", "( )"),("( )", 6, "( )"),("( )", -2, "( )")] = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Solution
`[(2, 0, -1),(5, 1, 0),(0, 1, 3)]` B−1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying C1 → C1 + C3, we get
`[(1, 0, -1),(5, 1, 0),(3, 1, 3)]` B−1 = `[(1, 0, 0),(0, 1, 0),(1, 0, 1)]`
Applying C3 → C3 + C1, we ge
`[(1, 0, 0),(5, 1, 5),(3, 1, 6)]` B−1 = `[(1, 0, 1),(0, 1, 0),(1, 0, 2)]`
Applying C1 → C1 – 5C2, C3 → C3 – 5C2, we get
`[(1, 0, 0),(0, 1, 0),(-2, 1, -1)]` B−1 = `[(1, 0, 1),(-5, 1, -5),(1, 0, 2)]`
Applying C1 → C1 – 2C3, C2 → C2 – C3, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` B−1 = `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`
B−1 = `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`
`[(2, 0, -1),(5, 1, 0),(0, 1, 3)] [(3, -1, 1),(-15, 6, -5),(5, -2, 2)] = [(6 - 0 - 5, -2 + 0 + 2, 2 - 0 - 2),(15 - 15 + 0, -5 + 6 - 0, 5 - 5 + 0),(0 - 15 + 15, 0 + 6 - 6, 0 - 5 + 6)]`
= `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
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