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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 =adbc=adcb. 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=26 =8. Also, we have 21
1 3
2 =2, 4
1 det I2= 0
0 = 1 1
and
21 = 44 = 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=adcb= 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=adbcnote that if Δ. We 6= 0 and d   d/Δb/Δ B= c/Δa/Δ
then     (adbc)/Δ (ab+ba)/Δ 1 0 AB= =, (cddc)/Δ (cb+da)/1Δ 0 and     (dabc)/Δ (dbbd)/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 detAor det2Aor 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+brdscpaqcpbsdraqdrbs =apds+brcqcpbsdraq = (adcb)(psrq) = detAdetB.     12 2 3 Example.LetA= andB= . Then 1 3 41
6 det(AB) = 14
5 = 070 =70, 0
which is the same as detAdetB= (3(2))(212) = 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/Δb/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 I2= 1 this means there cannot exist a matrixBwithAB= I2. HenceA is not invertible.   21 Examples.1. LetAThen det= . A= 0, soAis not invertible. 4 2   1 2 2. LetAThen det= . A=26= 0, soAis invertible, and 3 4     1 422 1 1 A= =. detA33 1 /21/2
Let us check this:     1 22 1 1 1 AA= = 3 4 3/21/2 0     12 1 1 2 1 A A= = 3/21/2 3 4 0
0 = I2, 1 0 = I2. 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 (nNadjugate of). The a square matrixAis denoted adjA(or adj(A)), and for 2×2 matrices is deﬁned by     a b db adj :=. c dc a
Direct calculation shows that we haveA(adjA) = (adjA)A= (detA)I2for 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   a1b1c1   A=a2b2c2. a3b3c3 We deﬁne thedeterminantofAby: a1b1c1 detA=a2b2c2:=a(b×c)   a3b3c3 where       a1b1c1       a=a2,b=b2andc=c2. a3b3c3 Thus       a1b1c1       detA=a2b2×c2 a3b3c3     a1 b2c2b1c1b1c1   =a2ij+k    b3c3b3c3b2c2 a3 b2c2b1c1b1c1 =a1a2+a3  b3c3b3c3b2c2 =a1(b2c3b3c2)a2(b1c3b3c1) +a3(b1c2b2c1). Example.Let   3 21   A= 2 03. 2 1 1 Then 03 21 21 detA= 322     1 1 1 1 03 = 3(3)2(3)2(6) = 15. Remark.Notice that the right-hand side of our equation a(b×c) =a1(b2c3b3c2)a2(b1c3b3c1) +a3(b1c2b2c1) can be rearranged to givea1(b2c3c2b3)b1(a2c3c2a3) +c1(a2b3b2a3we). Thus can also write: a1b1c1 b2c2a2c2a2b2 a2b2c2=a1b1+c1    b3c3a3c3a3b3   a3b3c3 =a1(b2c3b3c2)b1(a2c3a3c2) +c1(a2b3a3b2) =a1(b2c3c2b3)b1(a2c3c2a3) +c1(a2b3b2a3).
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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
a1 b1 c1
a2 b2 c2
a3a1 b3=a2 c3a3
b1 b2 b3
c1 c2. c3
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=a1b2c3+a2b3c1+a3b1c2a1b3c2a2b1c3a3b2c1.
The products having positive coeﬃcient correspond to diagonals going in the direction top-left to bottom-right in the diagram below.
c2
b3 c3
a1 b1 c1
a2 b2 c2
a3 b3 c3
b1 c1
c2
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. a1 Theorem 8.5.LetA=a2 a3
b1 b2 b3
c1 c2. Then c3
(i) Interchanging two columns ofAnegates detA. For example
b1 b2 b3
a1 a2 a3
c1a1 c2=a2 c3a3
b1 b2 b3
c1 c2=detA. c3
(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         120212212 012 0   2 0 1 1 1 1 = 10 0210(0(0)1 = 2(2)1(2)) =20.    2 02 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 I3, 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 3It turns out that= 1. Ais invertible whenever detA6But= 0.
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ﬁrst, we deﬁne the adjugate, adjA, of a 3×3 matrixA. LetAijbe 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   detA11detA21detA31   adjA:=detA12detA22detA32. detA13detA23detA33 Theorem 8.7.For all 3×3 matricesAwe haveA(adjA) = (adjA)A= (detA)I3. 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 innunknownsx1, x2, . . . , xn as a matrix equation Ax=d(8.1) whereAis anm×nmatrix,xis ann×1 matrix anddis anm×1 matrix, with the     d1x1     entries ofA= (a) anddknown and the entries of= being x ij m×n.=.dmxn being the unknowns. The linear equations areaj1x1+∙ ∙ ∙+ajnxn=djfor 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 11 A Ax=Ad. 1 SinceA Ax= Inx=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,Aijwill denote the (m1)×(n1) matrix obtained fromm×nmatrixAby deleting thei-th row andj-column. The notationAijonly ˜ makes sense whenm, n>1. The (i, j)-entries of the matricesA,AandBshall be denotedaij, ˜aijandbijrespectively.
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