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Question
Find the inverse of the matrices (if it exists).
`[(1,-1,2),(0,2,-3),(3,-2,4)]`
Solution
A = `[(1,-1,2),(0,2,-3),(3,-2,4)]`
So, adj A `= [(A_11,A_21,A_31),(A_12,A_22,A_32),(A_13,A_23,A_33)]`
`= [(2,0,-1),(-9,-2,3),(-6,-1,2)]`
`abs A= 1(8 - 6) + 1(0 + 9) + 2 (0 - 6)`
`= -1 ne 0 -> "A"^-1` exists.
`C_11 = (-1)^(1+1) |(2,-3), (-2,4)| = 8 - 6 = 2`
`C12 = (-1)^(1+2) |(0,-3), (3,4)| = -(0 + 9) = -9`
`C_13 = (-1)^(1+3)|(0,2),(3,-2)| = 0 - 6 = -6`
`C_21 = (-1)^(2+1) |(-1,2), (-2,4)| = -(-4 + 4) = 0`
`C_22 = (-1)^(2+2) |(1,2), (3,4)| = 4 - 6 = -2`
`C_23 = (-1)^(2+3) |(1,-1), (3,-2)| = -(2+3) = -1`
`C_31 = (-1)^(3+1) |(-1,2), (2,-3)| = 3 - 4 = -1`
`C_32 = (-1)^(3+2) |(1,2), (0,-3)| = -(-3 - 0) = 3`
`C_33 = (-1)^(3+3)|(1,-1), (0,2)| = 2 + 0 = 2`
`A^-1 = 1/abs A (adjA) = 1/abs A [(A_11,A_21,A_31),(A_12,A_22,A_32),(A_13,A_23,A_33)]`
`= 1/-1[(2,0,-1),(-9,-2,3),(-6,-1,2)]`
`= [(-2,0,1),(9,2,-3),(6,1,-2)]`
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