Review• Elementary matrices performing row operations:1 0 0−2 1 00 0 1a b cd e fg h i=a b cd − 2a e − 2b f − 2cg h i• Gaussian elimination on A gives an LU decomposition A = LU:2 1 14 −6 0−2 7 2=12 1−1 −1 12 1 1−8 −21Uis the e chelo n form, and L records the inverse row operations we did.• LU decomposition allows us to solve Ax = b f or many b.•1 0 0a 1 00 0 1−1=1 0 0−a 1 00 0 1• Already not so clear:1 0 0a 1 00 b 1−1=1 0 0−a 1 0ab −b 1Goal for today : invert these and any other matrices (if possible)Armin [email protected] inverse of a matrixExample 1. The inverse of a real number a is denoted as a−1. For instance, 7−1=17and7 · 7−1= 7−1· 7 = 1.In the context of n × n matrix multiplicati on, the role of 1 is taken by the n × n identitymatrixIn=111.Definition 2. An n × n matrix A is invertible if the r e is a matrix B such thatAB = BA = In.In that case, B is the inverse of A and we wri te A−1= B.Example 3. We already saw that elem entary matrices are invertible.•1 0 02 1 00 0 1−1=1−2 11Armin [email protected].•The i n verse of a matrix is unique. Why? So A−1is well-defined.AssumeB and C are both inverses of A. Then:C = C In= C AB = InB = B• Do not wri teAB. Why?Be cause it is uncl ear whether it should me an A B−1or B−1A.• If AB = I, then BA = I (and so A−1= B). N ot easy to show at this stage.Example 4. The matr ix A =0 10 0is not invertible. W hy?Solution.0 10 0a bc d=c d0 011Example 5. If A =a bc d, thenA−1=1ad − bcd −b−c aprovided that ad − bc0.Let’s check that:1ad − bcd −b−c aa bc d=1ad − bcad − bc 00 −cb + ad= I2Note.•A 1 × 1 matrix [a] is i nvertiblea0.• A 2 × 2 matrixa bc dis invertiblead − bc0.We wi ll encounter the quantities on the right agai n when we discuss determinants.Armin [email protected] syst ems using matrix inverseTheorem 6 . Let A be invertible. Then the system Ax = b has the uniqu e solutionx = A−1b.Proof. Multiply both sides of Ax = b with A−1(from the left!). Example 7. Solve−7x1+3x2= 25x1−2x2= 1using matri x invers ion.Solution. In matrix form Ax = b, this sy stem is−7 35 −2x =21.Computing the inverse:−7 35 −2−1=1−1−2 −3−5 −7=2 35 7Recall thata bc d−1=1ad − bcd −b−c a.Hence, the s o lution is:x = A−1b =2 35 721=717Armin [email protected] for computing the inverseTo solv eAx = b, we do row reduction on [A b].To solv eAX = I, we do row redu ct ion on [A I].To c ompute A−1: Gauss–Jordan me thod• Form the augmented matrix [A I].• Compute the reduced echelon form. (i.e. G auss– Jordan elimination)• If A i s invertible, the result is of the formI A−1 .Example 8. Find the inverse of A =2 0 0−3 0 10 1 0, if it exist s.Solution. By row reduct ion:[A I] I A−1 2 0 0 1 0 0−3 0 1 0 1 00 1 0 0 0 1 1 0 0120 00 1 0 0 0 10 0 1321 0Hence, A−1=120 00 0 1321 0.Example 9. Let’s do the previous example step by step.[A I] I A−1 2 0 0 1 0 0−3 0 1 0 1 00 1 0 0 0 1>R2→R2+32R12 0 0 1 0 00 0 1321 00 1 0 0 0 1>R1→12R1R2↔R31 0 0120 00 1 0 0 0 10 0 1321 0Armin [email protected]. Here is another way to see why this algorithm works:• Each r ow reduction corresponds to multiplying w i t h an elementary matrix E:[A I] [E1A E1I] [E2E1A E2E1] • So at each step:[A I] [FA F]with F = ErE2E1• If we manage to reduce [A I]to [I F], this meansFA = I and henc e A−1= F .Some properti es of matrix inversesTheorem 10. Suppose A and B are in vertible. Then:• A−1is invertible and (A−1)−1= A.Why? BecauseA A−1= I• ATis invertible and (AT)−1= (A−1)T.Why? Because(A−1)TAT= (A A−1)T= IT= I (R ecall that (AB)T= BTAT.)• AB is i nvertible and (AB)−1= B−1A−1.Why? Because (B−1A−1)(A B) = B−1I B = B−1B = IArmin
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