Definition D1
For A = (a b ; c d), set det A = ad − bc.
Proof.
(1)det assigns one real number to one matrix. A 2×2 matrix with first row a b and second row c d is written (a b ; c d).convention
(2)det I = 1·1 − 0·0 = 1(1), A14, N
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Definition D2
If there is a matrix B with AB = BA = I, then A is called invertible, and that B is called the inverse of A, written A⁻¹.
Proof.
(1)I is the matrix satisfying AI = IA = A for every A.A7
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Definition D3
For A = (a b ; c d), set adjA = (d −b ; −c a).
Proof.
(1)The two diagonal entries are swapped and the signs of the other two are reversed.convention
(2)All four entries of adjA are 0 only when a = b = c = d = 0.(1), A12
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Lemma L1
For every A, A·adjA = adjA·A = (det A) I
Proof.
(1)(A·adjA)₁₁ = a·d + b·(−c) = ad − bcA1, D3, A12, A13
(2)(A·adjA)₁₂ = a·(−b) + b·a = −(ab) + ab = 0A1, D3, A12, A9, A11
(3)(A·adjA)₂₁ = c·d + d·(−c) = cd − dc = 0A1, D3, A12, A9, A11, A13
(4)(A·adjA)₂₂ = c·(−b) + d·a = −(bc) + ad = ad − bcA1, D3, A12, A9, A10, A13
(5)The entries of (det A)I are det A · 1 = det A on the diagonal and det A · 0 = 0 elsewhere.A5, A14
(6)All entries agree, so A·adjA = (det A)I(1), (2), (3), (4), (5), D1
(7)The entries of adjA·A are d·a + (−b)·c = ad − bc, d·b + (−b)·d = 0, (−c)·a + a·c = 0, (−c)·b + a·d = ad − bc.A1, D3, A12, A9, A10, A11, A13
(8)Therefore adjA·A = (det A)I holds as well.(5), (7), D1
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Theorem T1
If det A ≠ 0, then A is invertible and A⁻¹ = (1/det A)·adjA.
Proof.
(1)Setting t = det A, we have t ≠ 0, so t⁻¹ exists.assumption, D1, A15
(2)Set B = t⁻¹·adjA.(1), convention
(3)AB = A(t⁻¹·adjA) = t⁻¹(A·adjA)(2), A4
(4)By L1, AB = t⁻¹(tI) = (t⁻¹t)I = 1·I = I(1), (3), L1, A6, A15, A5, A14
(5)In the same way, BA = t⁻¹(adjA·A) = I(2), L1, A4, A6, A15, A5, A14
(6)Therefore B is the inverse of A, and A is invertible.(4), (5), D2
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Theorem T2
If A is invertible, then det A ≠ 0.
Proof.
(1)Suppose there is a B with AB = I.assumption, D2
(2)Taking det of both sides gives det(AB) = det I = 1(1), D1, E
(3)By multiplicativity, det A · det B = 1(2), A3
(4)If det A = 0, then det A · det B = 0A14, A9
(5)Since 0 ≠ 1, this contradicts (3).(3), (4), A16
(6)Therefore det A ≠ 0.(5)
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Theorem T3
A is invertible if and only if det A ≠ 0.
Proof.
(1)If det A ≠ 0, then A is invertible.T1
(2)If A is invertible, then det A ≠ 0.T2
(3)Therefore the two conditions are equivalent.(1), (2)
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Corollary C
If det A = 0 and A ≠ O, then adjA ≠ O and A·adjA = O.
Proof.
(1)By L1, A·adjA = (det A)I = 0·I = Oassumption, L1, A5, A14
(2)Since A ≠ O, the entries a, b, c, d are not all 0.assumption
(3)Therefore adjA ≠ O.(2), D3
(4)That is, neither A nor adjA is the zero matrix, yet their product is. Such an A is called a zero divisor.(1), (3), convention
(5)If A = (1 1 ; 1 1), then det A = 0, adjA = (1 −1 ; −1 1), and A·adjA = O.D1, D3, A1, N
(6)If A were invertible, then adjA = I·adjA = (A⁻¹A)adjA = A⁻¹(A·adjA) = A⁻¹O = O, contradicting (3).(1), (3), D2, A2, A7, A8
(7)Therefore an A with det A = 0 and A ≠ O is not invertible. The zero divisors of Proof 111 arise here.(4), (6)
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