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