2.5 Inverse Matrices - MIT Mathematics

1 Inverse Inverse MatricesSupposeAis a square matrix . We look for an Inverse matrix A 1of the same size, suchthatA 1timesAequalsI. WhateverAdoes,A 1undoes. Their product is the identitymatrix which does nothing to a vector, soA 1might not a matrix mostly does is to multiply a vectorx. MultiplyingAxDbbyA 1givesA 1 AxDA isxDA 1b. The productA 1 Ais like multiplying bya number and then dividing by that number. A number has an Inverse if it is not zero Matrices are more complicated and more interesting. The matrixA 1is called Ainverse. DEFINITIONThe matrixAisinvertibleif there exists a matrixA 1such thatA 1 ADIandAA 1DI:(1)Not all Matrices have inverses. This is the first question we ask about a square matrix :IsAinvertible?

2 We don t mean that we immediately calculateA 1. In most problemswe never compute it! Here are six notes aboutA 1 The Inverse exists if and only if elimination producesnpivots(row exchangesare allowed). Elimination solvesAxDbwithout explicitly using the matrixA 2 The matrixAcannot have two different inverses. SupposeBADIand alsoACDI. ThenBDC, according to this proof by parentheses :(2)This shows that aleft-inverseB(multiplying from the left) and aright-inverseC(multi-plyingAfrom the right to giveACDI) must be thesame 3 IfAis invertible, the one and only solution toAxDbisxDA 1b:MultiplyAxDbbyA 1:ThenxDA 1 AxDA 1b:Note 4(Important)Suppose there is a nonzero vectorxsuch have an matrix can bring0back invertible, thenAxD0can only have the zero solutionxDA 5A 2 by 2 matrix is invertible if and only ifad bcis not zero:2by2 Inverse : abcd 1D1ad bc d b ca :(3)This numberad bcis thedeterminantofA.

3 A matrix is invertible if its determinant is notzero (Chapter 5). The test fornpivots is usually decided before the determinant 2. Solving Linear EquationsNote 6A diagonal matrix has an Inverse provided no diagonal entries are zero:IfAD264d1:::dn375thenA 1D2641=d1:::1=dn375:Example 1 The 2 by 2 matrixAD 1212 is not invertible. It fails the test in Note 5,becausead bcequals2 2D0. It fails the test in Note 3, ; 1/. It fails to have two pivots as required by Note turns the second row of this matrixAinto a zero Inverse of a ProductABFor two nonzero numbersaandb, the sumaCbmight or might not be invertible. ThenumbersaD3andbD 3have inverses13and 13. Their sumaCbD0has no the productabD 9does have an Inverse , which is13times two matricesAandB, the situation is similar.

4 It is hard to say much about theinvertibility ofACB. But theproductABhas an Inverse , if and only if the two factorsAandBare separately invertible (and the same size). The important point is thatA 1andB 1come inreverse order:IfAandBare invertible then so isAB. The Inverse of a 1DB 1A 1:(4)To see why the order is reversed, multiplyABtimesB 1A 1. Inside that isBB 1DI: Inverse ofAB .AB/.B 1A 1/DAIA 1 DAA 1DI:We moved parentheses to multiplyBB 1first. SimilarlyB 1A 1timesABequalsI. Thisillustrates a basic rule of Mathematics : Inverses come in reverse order. It is also commonsense: If you put on socks and then shoes, the first to be taken off are the. The samereverse order applies to three or more Matrices :Reverse / 1DC 1B 1A 1:(5)Example 2 Inverse of an elimination 5 times row 1 from row 2,thenE 1adds5 times row 1 to row 2:ED24100 51000135andE 1D2410051000135:MultiplyEE 1to get the identity matrixI.

5 Also multiplyE 1 Eto getI. We are addingand subtracting the same 5 times row 1. Whether we add and then subtract (this isEE 1/or subtract and then add (this isE 1E/, we are back at the Inverse Matrices83 For square Matrices , an Inverse on one side is automatically an Inverse on the other automaticallyBADI. In that caseBisA 1. This is very useful to knowbut we are not ready to prove 3 SupposeFsubtracts 4 times row 2 from row 3, andF 1adds it back:FD241000100 4135andF 1D2410001004135:Now multiplyFby the matrixEin Example 2 to findFE. Also multiplyE 1timesF 1to 1. Notice the ordersFEandE 1F 1!FED2410 0 51 020 4135is inverted byE 1F 1D2410051004135:(6)The result is beautiful and correct. The productFEcontains 20 but its Inverse doesn 5 times row 1 from row 2.))

6 ThenFsubtracts 4 times thenewrow 2 (changedby row 1) from row this orderFE, row 3 feels an effect from row the orderE 1F 1, that effect does not happen. FirstF 1adds 4 times row 2 torow 3. After that,E 1adds 5 times row 1 to row 2. There is no 20, because row 3 doesn tchange this orderE 1F 1,row3feels no effect from row elimination orderFfollowsE. In reverse orderE 1followsF 1F 1is quick. The multipliers5,4fall into place below the diagonal of1 special multiplicationE 1F 1andE 1F 1G 1will be useful in the next sec-tion. We will explain it again, more completely. In this section our job isA 1, and weexpect some serious work to compute it. Here is a way to organize that 1by Gauss-Jordan EliminationI hinted thatA 1might not be explicitly needed.

7 The equationAxDbis solved byxDA 1b. But it is not necessary or efficient to computeA 1and multiply it goes directly tox. Elimination is also the way to calculateA 1,aswenowshow. The Gauss-Jordan idea is to solveAA 1DI,finding each column ofA the first column ofA 1(call thatx1/to give the first column ofI(callthate1/. This is our ;0;0/. There will be two more of the columnsx1,x2,x3ofA 1is multiplied byAto produce a column ofI:3columns ofA 1AA 1DA x1x2x3 D e1e2e3 DI:(7)To invert a 3 by 3 matrixA, we have to solve three systems of ;1;0 ;0;1/. Gauss-Jordan findsA 1this 2. Solving Linear EquationsTheGauss-Jordan methodcomputesA 1by solvingallnequations the augmented matrix Ab has one extra columnb. Now we have threeright sidese1;e2;e3(whenAis 3 by 3).))

8 They are the columns ofI, so the augmentedmatrix is really the block matrix A I . I take this chance to invert my favorite matrixK,with 2 s on the main diagonal and 1 s next to the 2 s: Ke1e2e3 D2642 101 0 0 12 10 1 00 120 0 1375 Start Gauss-Jordan onK!242 10100032 112100 !242 10100032 are halfway toK 1. The matrix in the first three columns isU(upper triangular). Thepivots2;32;43are on its diagonal . Gauss would finish by back substitution. The contributionof Jordan isto continue with elimination! He goes all the way to the reduced echelonform . Rows are added to rows above them, to producezeros above the pivots: Zero abovethird pivot !242 Zero abovesecond pivot ! last Gauss-Jordan step is to divide each row by its pivot.

9 The new pivots are 1. Wehave reachedIin the first half of the matrix , becauseKis three columnsofK 1are in the second half of I K 1 :(divide by2/(divide by32/(divide by43/26666410034121401012112001141234377 775D Ix1x2x3 D IK 1 :Starting from the 3 by 6 matrix K I , we ended with I K 1 . Here is the wholeGauss-Jordan process on one line for any invertible matrixA:Gauss-JordanMultiply AI byA 1to get IA 1 Inverse Matrices85 The elimination steps create the Inverse matrix while changingAtoI. For large Matrices ,we probably don t wantA 1at all. But for small Matrices , it can be very worthwhile toknow the Inverse . We add three observations about this particularK 1because it is animportant example. We introduce the wordssymmetric,tridiagonal, its main diagonal .)))

10 So isK (only three nonzero diagonals). ButK 1is a dense matrix withno zeros. That is another reason we don t often compute Inverse Matrices . Theinverse of a band matrix is generally a dense of This number 4 is 1involves division by the determinantK 1D142432124212335.(8)This is why an invertible matrix cannot have a zero 4 FindA 1by Gauss-Jordan elimination starting fromAD 2347 . There aretwo row operations and then a division to put1 s in the pivots: AI D 23104701 ! 231001 21 this is UL 1 ! 207 301 21 ! 1072 3201 21 this is IA 1 :ThatA 1involves division by the determinantad bcD2 7 3 4D2. The code userref, the row reduced echelon form from Chapter 3 Define thenbynidentity matrixRDrref. A I /I% Eliminate on the augmented matrix A I ;nC1 WnCn/% PickA 1from the lastncolumns ofRAmust be invertible, or elimination cannot reduce it toI(in the left half ofR).