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प्रश्न
If A = `[(1, 0, 0),(3, 3, 0),(5, 2, -1)]`, find A−1 by the adjoint method
उत्तर
Let A = `[(1,0,0),(3,3,0),(5,2,-1)]`
∴ |A| = `|(1,0,0),(3,3,0),(5,2,-1)|`
= 1(− 3 − 0) − 0 + 0
= − 3 ≠ 0
∴ A−1 exist
A11 = (−1)1+1 M11 = `|(3,0),(2,-1)|` = −3 − 0 = −3
A12 = (−1)1+2 M12 = −`|(3,0),(5,-1)|` = −(−3 − 0) = 3
A13 = (−1)1+3 M13 = `|(3,3),(5,2)|` = 6 − 15 = −9
A21 = (−1)2+1 M21 = −`|(0,0),(2,-1)|` = −(0 − 0) = 0
A22 = (−1)2+2 M22 = `|(1, 0),(5, -1)|` = − 1 − 0 = −1
A23 = (−1)2+3 M23 = −`|(1,0),(5,2)|` = −(2 − 0) = −2
A31 =(−1)3+1 M31 = `|(0,0),(3,0)|` = 0 − 0 = 0
A32 = (−1)3+2 M32 = −`|(1,0),(3,0)|` = −(0 − 0) = 0
A33 = (−1)3+3 M33 = `|(1,0),(3,3)|` = 3 − 0 = 3
Hence, matrix of the co-factors is
`["A"_"ij"]_(3 xx 3) = [("A"_11,"A"_12,"A"_13),("A"_21,"A"_22,"A"_23),("A"_31,"A"_32,"A"_33)]`
= `[(-3,3,-9),(0,-1,-2),(0,0,3)]`
Now, adj A = `["A"_"ij"]_(3 xx 3)^"T"`
= `[(-3,0,0),(3,-1,0),(-9,-2,3)]`
∴ A−1 = `1/|"A"|` (adj A)
= `1/(-3)[(-3,0,0),(3,-1,0),(-9,-2,3)]`
= `1/3[(3,0,0),(-3,1,0),(9,2,-3)]`
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