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Find the inverse of matrix B = [315278125] by using adjoint method - Mathematics and Statistics

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Question

Find the inverse of matrix B = `[(3,1, 5),(2, 7, 8),(1, 2, 5)]` by using adjoint method

Sum

Solution

B = `[(3,1, 5),(2, 7, 8),(1, 2, 5)]` 

∴ |B| = `|(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

∴ B–1 exists.

Here,

b11 = 3

∴ M11 = `|(7, 8),(2, 5)|`

= 35 – 16

= 19

and B11 = (–1)1+1 M11 = 19

b12 = 1

∴ M12 = `|(2, 8),(1, 5)|`

= 10 – 8

= 2

and B12 = (–1)1+2 M12 = – 2

b13 = 5

∴ M13 = `|(2, 7),(1, 2)|`

= 4 – 7

= – 3

and B13 = (–1)1+3 M13 = – 3

b21 = 2

∴ M21 = `|(1, 5),(2, 5)|`

= 5 – 10

= – 5

and B21 = (–1)2+1 M21 = 5

b22 = 7

∴ M22 = `|(3, 5),(1, 5)|`

= 15 – 5

= 10

and B22 = ( –1)2+2 M22 = 10

b23 = 8

∴ M23 = `|(3, 1),(1, 2)|`

= 6 – 1

= 5

and B23 = ( –1)2+3 M23 = – 5

b31 = 1

∴ M31 = `|(1, 5),(7, 8)|`

= 8 – 35

= – 27

and B31 = ( –1)3+1 M31 = – 27

b32 = 2

∴ M32 = `|(3, 5),(2, 8)|`

= 24 – 10

= 14

and B32 = ( –1)3+2 M32 = – 14

b33 = 5

∴ M33 = `|(3, 1),(2, 7)|`

= 21 – 2

= 19

and B33 = ( –1)3+3 M33 = 19

∴ The matrix of the co-factors is

[Bij]3×3 = `[("B"_11, "B"_12, "B"_13),("B"_21, "B"_22, "B"_23),("B"_31, "B"_32, "B"_33)]`

= `[(19, -2, -3),(5, 10, -5),(-27, -4, 19)]`

Now, adj B = `["B"_"ij"]_(3 xx 3)^"T"`

= `[(19, 5, -27),(-2, 10, -14),(-3, -5, 9)]`

∴ B–1 = `1/|"B"|` (adj B)

= `1/40 [(19, 5, -27),(-2, 10, -14),(-3, -5, 9)]`

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Chapter 1.2: Matrices - Q.4

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