The University of the West Indies St. Augustine Campus

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facultyreport The University of the West Indies St. Augustine Campus 07/08

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• Publié par: urmu
• Nombre de lectures: 29
• Langue: English

Extrait

Chapter

8

Determinants

The founder of the theory of determinants is usually taken to be Gottfried Wilhelm Leibniz (1646–1716), who also shares the credit for inventing calculus with Sir Isaac 1 Newton (1643–1727) . But the idea of 2×2 determinants goes back at least to the Chinese around 200bcword. The determinantitself was first used in its present sense in 1812 by Augustin-Louis Cauchy (1789–1857), and he developed much of the general theory we know today.

8.1 Inverses of2×2matrices, and determinants   a b LetA= be a 2×2 matrix. Recall that the determinant ofAis c d

a b =ad−bc=ad−cb.  c d We also denote this determinant by det(A) or detAor|A|. (Just as with other operators, such as cos, we avoid using the brackets except to resolve ambiguity. Thus we prefer T T cosθand detAin preference to cos(θ) and det(A), and detAmeans det(A).)   2 3 Examples.LetA= . Then detA=−2−6 =−8. Also, we have 2−1

1  3

2 =−2,  4

1 det I 2 =  0

0 = 1  1

and

2−1 = 4−4 = 0.   −4 2

T Lemma 8.1.LetAbe any 2×2 matrix. Then detA= detA.     a b a c T T T Proof.IfA= thenA= , and so detA=ad−cb= detA. c d b d

1 His dates by the Gregorian calendar are 4th January 1643 – 31st March 1727. But the Julian calendar was in use in England at the time, and his Julian dates are 25th December 1642 – 20th March 1727. At the time, the year in England started on 25th March, so that legally he died in 1726.

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 a Now letA= c

 b and let Δ := detA=ad−bcnote that if Δ. We 6= 0 and d   d/Δ−b/Δ B= −c/Δa/Δ

then     (ad−bc)/Δ (−ab+ba)/Δ 1 0 AB= =, (cd−dc)/Δ (−cb+da)/1Δ 0 and     (da−bc)/Δ (db−bd)/0Δ 1 BA= =, (−ca+ac)/Δ (−cb+ad)/Δ 0 1 −1 so thatAis invertible andA=B. Thus ifAis a 2×2 matrix and detA6= 0, then AWhat if detis invertible. A= 0? We shall show that in this caseAisnotinvertible, but first we show the following.

Theorem 8.2.IfAandBare 2×2 matrices then det(AB) = det(A) det(B).

Notation.Our conventions for bracketing the determinant operator are similar to those used for trigonometric functions. Thus detABmeans det(AB), even though (detA)Bmakes sense, and detAdetBmeans det(A) det(B) = (detA)(detB), though det(AdetB) makes less sense since we write the scalar first when notating scalar multi- plication of matrices. Similarly, ifλis a scalar then detλAmeans det(λAthat). Note while notations like det 3Aand detAdetBare often used, we shouldneverwrite things like det−Aor det−2Aor detA+B. Also, detABis seldom seen.     a b p q Proof.We prove this by direct computation. LetA= andB= . c d r s   ap+br aq+bs ThenAB= and cp+dr cq+ds

det(AB) = (ap+br)(cq+ds)−(cp+dr)(aq+bs) =apcq+apds+brcq+brds−cpaq−cpbs−draq−drbs =apds+brcq−cpbs−draq = (ad−cb)(ps−rq) = detAdetB.     1−2 2 3 Example.LetA= andB= . Then 1 3 4−1

−6 det(AB) =  14

5 = 0−70 =−70,  0

which is the same as detAdetB= (3−(−2))(−2−12) = 5×(−14) =−70.

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  a b Theorem 8.3.LetAa 2= be ×2-matrix. ThenAis invertible if and only c d   d/Δ−b/Δ −1 if detA6= 0, and when Δ := detA6= 0, thenA= . −c/Δa/Δ

Proof.We have already seen that when Δ = detA6= 0 thenAis invertible with inverse as above. It therefore just remains to show that when Δ = 0 thenAis not invertible. So suppose that detA= 0. Then det(AB) = detAdetB= 0 for every 2×2 matrixB. But since det I 2 = 1 this means there cannot exist a matrixBwithAB= I 2 . HenceA is not invertible.   2−1 Examples.1. LetAThen det= . A= 0, soAis not invertible. −4 2   1 2 2. LetAThen det= . A=−26= 0, soAis invertible, and 3 4     1 4 −2−2 1 −1 A= =. detA −33 1 /2−1/2

Let us check this:      1 2−2 1 1 −1 AA= = 3 4 3/2−1/2 0      −1 −2 1 1 2 1 A A= = 3/2−1/2 3 4 0

 0 = I 2 , 1  0 = I 2 . 1

We conclude this section by introducing theadjugateof a 2×2 matrix. This might not seem to be a terribly exciting or useful concept for 2×2 matrices, but is far more + useful for 3×3 matrices, and in general forn×nmatrices (n∈Nadjugate of). The a square matrixAis denoted adjA(or adj(A)), and for 2×2 matrices is defined by     a b d−b adj :=. c d−c a

Direct calculation shows that we haveA(adjA) = (adjA)A= (detA)I 2 for all 2×2 matricesAif det. Thus A6= 0 (precisely the conditions needed forAto be invertible) − 1 1 we haveA= (adjArelations generalise to). These n×nmatrices for alln>1. detA We emphasise that the adjugate of a matrix is not the same as its adjoint, and you must be careful not to confuse the two terms. (Unfortunately, what we now call the adjugate has in the past been termed the adjoint!) Theadjointof a matrix overC ∗ † is defined to the transpose of its complex conjugate, and is denotedAorA, though ∗ sometimesAsimply means the complex conjugate ofAthe adjoint of matrix. Thus overRis the same as its transpose. (If you have not yet met complex numbers and the complex conjugate, you will have done so by the time you take your first year exams.)

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8.2

Determinants of3×3secimrta

Consider the 3×3 matrix   a 1 b 1 c 1   A= a 2 b 2 c 2 . a 3 b 3 c 3 We define thedeterminantofAby: a 1 b 1 c 1 detA= a 2 b 2 c 2 :=a∙(b×c)   a 3 b 3 c 3 where       a 1 b 1 c 1       a= a 2 ,b= b 2 andc= c 2 . a 3 b 3 c 3 Thus       a 1 b 1 c 1       detA= a 2 ∙ b 2 × c 2 a 3 b 3 c 3     a 1 b 2 c 2 b 1 c 1 b 1 c 1   = a 2 ∙i−j+k     b 3 c 3 b 3 c 3 b 2 c 2 a 3 b 2 c 2 b 1 c 1 b 1 c 1 =a 1 −a 2 +a 3   b 3 c 3 b 3 c 3 b 2 c 2 =a 1 (b 2 c 3 −b 3 c 2 )−a 2 (b 1 c 3 −b 3 c 1 ) +a 3 (b 1 c 2 −b 2 c 1 ). Example.Let   3 2−1   A = 2 0−3 . −2 1 1 Then 0−3 2−1 2−1 detA= 3−2−2     1 1 1 1 0−3 = 3(3)−2(3)−2(−6) = 15. Remark.Notice that the right-hand side of our equation a∙(b×c) =a 1 (b 2 c 3 −b 3 c 2 )−a 2 (b 1 c 3 −b 3 c 1 ) +a 3 (b 1 c 2 −b 2 c 1 ) can be rearranged to givea 1 (b 2 c 3 −c 2 b 3 )−b 1 (a 2 c 3 −c 2 a 3 ) +c 1 (a 2 b 3 −b 2 a 3 we). Thus can also write: a 1 b 1 c 1 b 2 c 2 a 2 c 2 a 2 b 2 a 2 b 2 c 2 =a 1 −b 1 +c 1    b 3 c 3 a 3 c 3 a 3 b 3   a 3 b 3 c 3 =a 1 (b 2 c 3 −b 3 c 2 )−b 1 (a 2 c 3 −a 3 c 2 ) +c 1 (a 2 b 3 −a 3 b 2 ) =a 1 (b 2 c 3 −c 2 b 3 )−b 1 (a 2 c 3 −c 2 a 3 ) +c 1 (a 2 b 3 −b 2 a 3 ).

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This formula (“expanding along the first row”) is often used to define a 3×3 determinant (even by me!), and is more frequently seen than the triple scalar product definition (“expanding along the first column”) used in this course. The reason for our seemingly strange approach is that we can easily prove many properties of 3×3 determinants using results we have already established for triple scalar products (see Theorem 8.5). Note that it also follows from this rearrangement that

a 1 b 1  c 1

a 2 b 2 c 2

a 3 a 1 b 3 = a 2  c 3 a 3

b 1 b 2 b 3

c 1 c 2 .  c 3

In other words, the determinant of a 3×3 matrixAis equal to the determinant of its T transposeA, a result we record as Theorem 8.4. Another helpful rearrangement of the terms in detAis to group those terms with positive and negative coefficients together, to get

detA=a 1 b 2 c 3 +a 2 b 3 c 1 +a 3 b 1 c 2 −a 1 b 3 c 2 −a 2 b 1 c 3 −a 3 b 2 c 1 .

The products having positive coefficient correspond to diagonals going in the direction top-left to bottom-right in the diagram below.

c 2

b 3 c 3

a 1 b 1 c 1

a 2 b 2 c 2

a 3 b 3 c 3

b 1 c 1

c 2

The top-right to bottom-left ‘diagonals’ in the above diagram correspond to those prod- ucts with negative coefficient.

T Theorem 8.4.For all 3×3 matricesAwe have detA= detA.

Proof.See above.  a 1  Theorem 8.5.LetA= a 2 a 3

b 1 b 2 b 3

 c 1  c 2 . Then c 3

(i) Interchanging two columns ofAnegates detA. For example

b 1 b 2  b 3

a 1 a 2 a 3

c 1 a 1 c 2 =− a 2  c 3 a 3

b 1 b 2 b 3

c 1 c 2 =−detA.  c 3

(ii) Multiplying any column ofAbyλmultiplies detAbyλ.

(iii) Adding a scalar multiple of any column ofAto any other column does not change the determinant.

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Proof.These all follow very easily from properties we have already proved for the scalar triple product. (i): This follows at once from Theorem 6.6, which gives detA=a∙(b×c) =b∙(c×a) = c∙(a×b) =−a∙(c×b) =−b∙(a×c) =−c∙(b×a). (ii): We haveλdetA=λ(a∙(b×c)) = (λa)∙(b×c) =a∙((λb)×c) =a∙(b×(λc)), by Theorem 6.3 and Section 3.2. (iii): LetBbe the matrix obtained by addingλtimes the second column to the first column ofA. Then standard results on the dot and cross product, such asb∙(b×c) = 0, give us detB= (a+λb)∙(b×c) =a∙(b×c) +λb∙(b×c) =a∙(b×c) = detA. The other cases are similar.

Example.We can use these properties to compute determinants quickly.

6 10 13 6 1 13 6 1 1 0 1 1 2 20 4 = 10 2 2 4 = 10 2 2 0 = 10 2 2 0         −1−20−2−1−2−2−1−2 0−1−2 0   2 0 1 1 1 1 = 10 0−2−10(0(0)1 = −2(2)−1(−2)) =−20.     −2 0−2 02 0

Here to get the first equality we have applied Part (ii) of the theorem to take out a factor of 10 from the second column, to get the second equality we have applied Part (iii) of the theorem to subtract twice the first column from the third column, and for the 3rd equality we have applied Part (iii) to subtract 6 times the third column from the first column. (Note that we can use the theorem to simplify the matrix even more, so that we end up computing det I 3 , which is 1.)

T Remark.Since detA= detAfor all 3×3 matricesA, Theorem 8.5 holds with the word ‘column’ replaced by ‘row’ throughout. Thus we can just as well use row operations as column operations when simplifying determinants.

The rest of this section is presented without proof. In particular, the next theorem is rather difficult to prove. (The one after that is not too hard.)

Theorem 8.6.For all 3×3 matricesAandBwe have det(AB) = detAdetB.

3 Proof.In theory, we can expand det(AB), which has 3!×63 = ×27 = 162 terms, 2 2 perform all the cancellations to be left with (3!) = 6 = 36 terms, and observe that this is detAdetB. This is not very satisfactory, and new ideas are needed, which will also extend to generaln×nthis is beyond the scope of this course,determinants. However, though I may write up the proof as an appendix in the hope that some of you might understand it after having completed MTH4104: Introduction to Algebra.

In view of the above theorem, it is evident that the 3×3 matrixAis not invertible if detAI= 0, since det 3 It turns out that= 1. Ais invertible whenever detA6But= 0.

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first, we define the adjugate, adjA, of a 3×3 matrixA. LetA ij be the 2×2 matrix obtained by removing thei-th row andj-th column fromA. (Note that this notation conflicts with what we used earlier.) Then   detA 11 −detA 21 detA 31   adjA:= −detA 12 detA 22 −detA 32 . detA 13 −detA 23 detA 33 Theorem 8.7.For all 3×3 matricesAwe haveA(adjA) = (adjA)A= (detA)I 3 . Thus − 1 1 if detA6= 0 thenAis invertible, and we haveA= (adjA). detA

8.3

Systems of linear equations as matrix equations

We can write any system ofmsimultaneous linear equations innunknownsx 1 , x 2 , . . . , x n as a matrix equation Ax=d(8.1) whereAis anm×nmatrix,xis ann×1 matrix anddis anm×1 matrix, with the     d 1 x 1     entries ofA= (a) anddknown and the entries of= being x ij m×n  .  =  .  d m x n being the unknowns. The linear equations area j1 x 1 +∙ ∙ ∙+a jn x n =d j for 16j6m. I commented previously that properly echelon form is a property of matrices. We define the matrixAto beechelon formif the corresponding systemAx=dof linear equations is in echelon form. SoAis in echelon form (Geometry I version) if each nonzero row ofA(This forces allcommences with strictly fewer 0s than those below it. zero rows to occur at the ‘bottom’ ofAdo not insist (for Geometry I) that the.) We first nonzero entry in a nonzero row be 1. IfAis a squaren×nmatrix (so nowm=n) such that detA6= 0 then the matrixA −1 has a (unique) inverseA, and by multiplying both sides of the matrix equation (8.1) −1 byAwe see that −1−1 A Ax=Ad. −1 SinceA Ax= I n x=xwe deduce that −1 x=Ad is a solution of the simultaneous equations, and indeed that it is theuniquesolution.

8.4

Determinants and inverses ofn×nmatrices

Notation.Throughout this section,A ij will denote the (m−1)×(n−1) matrix obtained fromm×nmatrixAby deleting thei-th row andj-column. The notationA ij only ˜ makes sense whenm, n>1. The (i, j)-entries of the matricesA,AandBshall be denoteda ij , ˜a ij andb ij respectively.

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