Maharashtra State BoardHSC Commerce 12th Board Exam
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Find the inverse of the following matrices by the adjoint method [123024005]. - Mathematics and Statistics

Sum

Find the inverse of the following matrices by the adjoint method `[(1, 2, 3),(0, 2, 4),(0, 0, 5)]`.

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Solution

Let A = `[(1, 2, 3),(0, 2, 4),(0, 0, 5)]`

∴ |A| = `|(1, 2, 3),(0, 2, 4),(0, 0, 5)|`

= 1(10 – 0) –2(0 –0) + 3(0 –0) = 10 ≠ 0
∴ A–1 exists.
A11 = (– 1)1+1 M11 = `1|(2, 4),(0, 5)|` = 1(10 – 0) = 10

A12 = (– 1)1+2 M12 = `-1|(0, 4),(0, 5)|` = 1(0 – 0) = 0

A13 = (– 1)1+3 M13 = `1|(0, 2),(0, 0)|` = 1(0 – 0) = 0

A21 = (– 1)2+1 M21 = `-1|(2, 3),(0, 5)|` = – 1(10 – 0) = – 10

A22 = (– 1)2+2 M22 = `1|(1, 3),(0, 5)|` = 1(5 – 0) = 5

A23 = (– 1)2+3 M23 = `-1|(1, 2),(0, 0)|` = –1(0 – 0) = 0

A31 = (– 1)3+1 M31 = `1|(2, 3),(2, 4)|` = 1(8 – 6) = 2

A32 = (– 1)3+2 M32 = `-1|(1, 3),(0, 4)|` = –1(4 – 0) = – 4

A33 = (– 1)3+3 M33 = `1|(1, 2),(0, 2)|` = 1(2 – 0) = 2

∴ The matrix of the co-factor is

[Aij]3x3 = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)] = [(10, 0, 0),(-10, 5, 0),(2, -4, 2)]`

Now,adj A = `["A"_"ij"]_(3xx3)^"T" = [(10, -10, 2),(0, 5, -4),(0, 0, 2)]`

∴ A–1 = `(1)/|"A"| ("adj A") = (1)/(10)[(10, -10, 2),(0, 5, -4),(0, 0, 2)]`.

Concept: Inverse of Matrix
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