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प्रश्न
Find the inverse of `[(3, 1, 5),(2, 7, 8),(1, 2, 5)]` by adjoint method.
उत्तर
Let A = `[(3, 1, 5),(2, 7, 8),(1, 2, 5)]`
|A| = `|(3, 1, 5),(2, 7, 8),(1, 2, 5)|`
= 3(35 – 16) – 1(10 – 8) + 5(4 – 7)
= 3(19) – 1(2) + 5(– 3)
= 57 – 2 – 15
= 40 ≠ 0
∴ A–1 exists
A11 = (–1)1+1 × M11
= `1|(7, 8),(2, 5)|`
= 1(35 – 16)
= 19
A12 = (–1)1+2 M12
= `-1|(2, 8),(1, 5)|`
= 1(10 – 8)
= –2
A13 = (–1)1+3 × M13
`1|(2, 7),(1, 2)|`
= 1(4 – 7)
= –3
A21 = (–1)2+1 × M21
= `-1|(1, 5),(2, 5)|`
= (–1)(5 – 10)
= 5
A22 = (– 1)2+2 × M22
= `1|(3, 5),(1, 5)|`
= (1)(15 – 5)
= 10
A23 = (–1)2+3 × M23
= `(-1)|(3, 1),(1, 2)|`
= (–1)(6 – 1)
= –5
A31 = (–1)3+1 × M31
= `(1)|(1, 5),(7, 8)|`
= (1)(8 – 35)
= –27
A32 = (–1)3+2 M32
= `(-1)|(3, 5),(2, 8)|`
= (–1)(24 – 10)
= –14
A33 = (–1)3+3 M33
= `(+1)|(3, 1),(2, 7)|`
= 1(21 – 2)
= 19
∴ The matrix of the co-factors is
A = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)] = [(19, -2, -3),(5, 10, -5),(-27, -14, 19)]`
Now, adj A = `[(19, 5, -27),(-2, 10, -14),(-3, -5, 19)]`
∴ A–1 = `(1)/|"A"| ("adj A")`
∴ A–1 = `(1)/(40)[(19, 5, -27),(-2, 10, -14),(-3, -5, 19)]`
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