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Question
Find the adjoint of matrix A = `[(2, 0, -1),(3, 1, 2),(-1, 1, 2)]`
Solution
A11 = (–1)1+1 M11 = `1|(1, 2),(1, 2)|` = 1(2 – 2) = 0
A12 = (–1)1+2 M12 = `(-1)|(3, 2),(-1, 2)|` = (–1)(6 + 2) = –8
A13 = (–1)1+3 M13 = `1|(3, 1),(-1, 1)|` = 1(3 + 1) = 4
A21 = (–1)2+1 M21 = `(-1)|(0, -1),(1, 2)|` = (–1)(0 + 1) = –1
A22 = (–1)2+2 M22 = `1|(2, -1),(-1, 2)|` = 1(4 – 1) = 3
A23 = (–1)2+3 M23 = `(-1)|(2, 0),(-1, 1)|` = (–1)(2 – 0) = –2
A31 = (–1)3+1 M31 = `1|(0, -1),(1, 2)|` = 1(0 + 1) = 1
A32 = (–1)3+2 M32 = `(-1)|(2, -1),(3, 2)|` = (–1)(4 + 3) = –7
A33 = (–1)3+3 M33 = `1|(2, 0),(3, 1)|` = 1(2 – 0) = 2
∴ adj (A) = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)]^"T"`
= `[(0, -8, 4),(-1, 3, -2),(1, -7, 2)]^"T"`
= `[(0, -1, 1),(-8, 3, -7),(4, -2, 2)]`
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