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Find the inverse of [315278125] by adjoint method. - Mathematics and Statistics

Sum

Find the inverse of `[(3, 1, 5),(2, 7, 8),(1, 2, 5)]` by adjoint method.

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Solution

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

[Aij]3x3 = `[("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 = `["A"_"ij"]_(3xx3)^"T" = [(19, 5, -27),(-2, 10, -14),(-3, -5, 19)]`

∴ A–1 = `(1)/|"A"| ("adj A") = (1)/(40)[(19, 5, -27),(-2, 10, -14),(-3, -5, 19)]`.

Concept: Inverse of Matrix
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 2 Matrices
Miscellaneous Exercise 2 | Q 4.17 | Page 85
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