MATH 304Linear AlgebraLecture 8:Inverse matrix (continued).Elementary matrices.Transpose of a matrix.Inverse matrixDefinition. Let A be an n×n matrix. The inverseof A is an n×n matrix, denoted A−1, such thatAA−1= A−1A = I .If A−1exists then the matrix A is called invertible.Otherwise A is called singular.Inverting diagonal matricesTheorem A diagonal matrix D = diag(d1, . . . , dn)is invertible if and only if all diagonal entries arenonzero: di6= 0 for 1 ≤ i ≤ n.If D is invertible then D−1= diag(d−11, . . . , d−1n).d10 . . . 00 d2. . . 0............0 0 . . . dn−1=d−110 . . . 00 d−12. . . 0............0 0 . . . d−1nInverting 2-by-2 matricesDefinition. The determinant of a 2×2 matrixA =a bc dis det A = ad − bc.Theorem A matrix A =a bc dis invertible ifand only if det A 6= 0.If det A 6= 0 thena bc d−1=1ad − bcd −b−c a.Fundamental results on inverse matricesTheorem 1 Given a square matrix A, the following areequivalent:(i) A is invertible;(ii) x = 0 is the only solution of the matrix equation Ax = 0;(iii) the row echelon form of A has no zero rows;(iv) the reduced row echelon form of A is the identity matrix.Theorem 2 Suppose that a sequence of elementary rowoperations converts a matrix A into the identity matrix.Then the same sequence of operations converts the identitymatrix into the inverse matrix A−1.Theorem 3 For any n×n matrices A and B,BA = I ⇐⇒ AB = I .Row echelon form of a square matrix:∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗∗ ∗ ∗∗ ∗∗∗ ∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗∗ ∗∗invertible case noninvertible caseExample. A =3 −2 01 0 1−2 3 0.To check whether A is invertible, we convert it torow echelon form.Interchange the 1st row with the 2nd row:1 0 13 −2 0−2 3 0Add −3 times the 1st row to the 2nd row:1 0 10 −2 −3−2 3 0Add 2 times the 1st row to the 3rd row:1 0 10 −2 −30 3 2Multiply the 2nd row by −1/2:1 0 10 1 1.50 3 2Add −3 times the 2nd row to the 3rd row:1 0 10 1 1.50 0 −2.5Multiply the 3rd row by −2/5:1 0 101 1.50 01We already know that the matrix A is invertible.Let’s proceed towards reduced row echelon form.Add −3/2 times the 3rd row to the 2nd row:1 0 10 1 00 0 1Add −1 times the 3rd row to the 1st row:1 0 00 1 00 0 1To obtain A−1, we need to apply the followingsequence of elementary row operations to theidentity matrix:• interchange the 1st row with the 2nd row,• add −3 times the 1st row to the 2nd row,• add 2 times the 1st row to the 3rd row,• multiply the 2nd row by −1/2,• add −3 times the 2nd row to the 3rd row,• multiply the 3rd row by −2/5,• add −3/2 times the 3rd row to the 2nd row,• add −1 times the 3rd row to the 1st row.A convenient way to compute the inverse matrixA−1is to merge the matrices A and I into one 3×6matrix (A | I ), and apply elementary row operationsto this new matrix.A =3 −2 01 0 1−2 3 0, I =1 0 00 1 00 0 1(A | I ) =3 −2 01 0 01 0 10 1 0−2 3 0 0 0 13 −2 01 0 01 0 10 1 0−2 3 00 0 1Interchange the 1st row with the 2nd row:1 0 10 1 03 −2 01 0 0−2 3 00 0 1Add −3 times the 1st row to the 2nd row:1 0 10 1 00 −2 −3 1 −3 0−2 3 00 0 1Add 2 times the 1st row to the 3rd row:1 0 10 1 00 −2 −31 −3 00 3 20 2 1Multiply the 2nd row by −1/2:1 0 10 1 00 1 1.5−0.5 1.5 00 3 20 2 1Add −3 times the 2nd row to the 3rd row:1 0 10 1 00 1 1.5 −0.5 1.5 00 0 −2.5 1.5 −2.5 1Multiply the 3rd row by −2/5:1 0 10 1 00 1 1.5−0.5 1.5 00 0 1−0.6 1 −0.4Add −3/2 times the 3rd row to the 2nd row:1 0 10 1 00 1 00.4 0 0.60 0 1−0.6 1 −0.4Add −1 times the 3rd row to the 1st row:1 0 00.6 0 0.40 1 0 0.4 0 0.60 0 1 −0.6 1 −0.4Thus3 −2 01 0 1−2 3 0−1=3502525035−351 −25.That is,3 −2 01 0 1−2 3 03502525035−351 −25=1 0 00 1 00 0 1,3502525035−351 −253 −2 01 0 1−2 3 0=1 0 00 1 00 0 1.Why does it work?1 0 00 2 00 0 1a1a2a3b1b2b3c1c2c3=a1a2a32b12b22b3c1c2c3,1 0 03 1 00 0 1a1a2a3b1b2b3c1c2c3=a1a2a3b1+3a1b2+3a2b3+3a3c1c2c3,1 0 00 0 10 1 0a1a2a3b1b2b3c1c2c3=a1a2a3c1c2c3b1b2b3.Proposition Any elementary row operation can besimulated as left multiplication by a certain matrix.Elementary matricesE =1...O1r1O...1row #iTo obtain the matrix EA from A, multiply the ithrow by r . To obtain the matrix AE from A, multiplythe ith column by r .Elementary matricesE =1......O0 · · · 1.........0 · · · r · · · 1............0 · · · 0 · · · 0 · · · 1row #irow #jTo obtain the matrix EA from A, add r times the ithrow to the jth row. To obtain the matrix AE fromA, add r times the jth column to the ith column.Elementary matricesE =1 O...0 · · · 1.........1 · · · 0...O 1row #irow #jTo obtain the matrix EA from A, interchange theith row with the jth row. To obtain AE from A,interchange the ith column with the jth column.Why does it work?Assume that a square matrix A can be converted tothe identity matrix by a sequence of elementary rowoperations. ThenEkEk−1. . . E2E1A = I ,where E1, E2, . . . , Ekare elementary matricescorresponding to those operations.Applying the same sequence of operations to theidentity matrix, we obtain the matrixB = EkEk−1. . . E2E1I = EkEk−1. . . E2E1.Thus BA = I , which implies that B = A−1.Transpose of a matrixDefinition. Given a matrix A, the transpose of A,denoted AT, is the matrix whose rows are columnsof A (and whose columns are rows of A). That is,if A = (aij) then AT= (bij), where bij= aji.Examples.1 2 34 5 6T=1 42 53 6,789T= (7, 8, 9),4 77 0T=4 77 0.Properties of transposes:•
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