Math 304–504Linear AlgebraLecture 7:Elementary matrices.Determinants.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 invertibl e if and only if all diagonal entries arenonzero: di6= 0 for 1 ≤ i ≤ n.If D is i nvertible 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 2×2 matricesDefinition. The determinant of a 2×2 matrixA =a bc dis det A = ad − bc.Theorem A matrix A =a bc dis invertibl e 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 conv erts the identitymatrix into the inverse matrix A−1.Theorem 3 For any n× n matrices A and B,BA = I ⇐⇒ AB = I .Example. A =3 −2 01 0 1−2 3 0.A convenient way to com pute the inver s e matrixA−1is to merge the matri ces A and I into one 3×6matrix (A | I ), and apply elementary row oper ationsto 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 1 0 1 0−2 3 0 0 0 1(A | I ) =3 −2 01 0 01 0 10 1 0−2 3 0 0 0 1As soon as the left half of the 3×6 matrix isconverted to the identity matrix, we have got theinverse matrix A−1in the right half.→1 0 0350250 1 0250350 0 1−351 −25Thus3 −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 el ementary row operation can besimulated as left multiplication by a certai n matrix.Elementary ma tr icesE =1...O1r1O...1row #iTo obtain the matrix EA from A, multi ply the i throw by r . To obtain the matrix AE from A, m ultiplythe ith column by r.Elementary ma tr icesE =1......O0 · · · 1.........0 · · · r · · · 1............0 · · · 0 · · · 0 · · · 1row #irow #jTo obtai n the matrix EA from A, add r times the ithrow to the jth row. To o btai n the matrix AE fromA, add r times the jth column to the ith column.Elementary ma tr icesE =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 co lumn 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 s ame sequence of operations to theidentity m atrix, we obtain the matri xB = EkEk−1. . . E2E1I = EkEk−1. . . E2E1.Thus BA = I , which im plies that B = A−1.Problem Sol ve the matrix equation XA + B = X ,where A =4 −21 1, B =5 23 0.Since B is a 2×2 matrix, it fo llows that XA and Xare also 2×2 matrices.XA + B = X ⇐⇒ X − XA = B⇐⇒ X (I − A) = B ⇐⇒ X = B(I − A)−1provided that I −A is an invertible matrix.I −A =−3 2−1 0,I −A =−3 2−1 0,det(I −A) = (−3) · 0 − 2 · (−1) = 2,(I −A)−1=120 −21 −3,X = B(I −A)−1=5 23 0120 −21 −3=125 23 00 −21 −3=122 −160 −6=1 −80 −3.DeterminantsDeterminant is a s calar assigned to each square matrix.Notation. The determinant of a matrixA = (aij)1≤i,j≤nis denoted det A ora11a12. . . a1na21a22. . . a2n............an1an2. . . ann.Principal property: det A = 0 if and only if thematrix A is singular.Definition i n low dimensionsDefinition. det ( a) = a,a bc d= ad − bc,a11a12a13a21a22a23a31a32a33= a11a22a33+ a12a23a31+ a13a21a32−−a13a22a31− a12a21a33− a11a23a32.+ :* ∗ ∗∗* ∗∗ ∗ *,∗* ∗∗ ∗** ∗ ∗,∗ ∗** ∗ ∗∗ * ∗.− :∗ ∗*∗ * ∗* ∗ ∗,∗* ∗* ∗ ∗∗ ∗ *,* ∗ ∗∗ ∗ *∗ * ∗.Examples: 2×2 matrices1 00 1= 1,3 00 −4= − 12,−2 50 3= − 6,7 05 2= 14,0 −11 0= 1,0 04 1= 0,−1 3−1 3= 0,2 18 4= 0.a11a12a13a21a22a23a31a32a33= a11a22a33+ a12a23a31+ a13a21a32−−a13a22a31− a12a21a33− a11a23a32.a11a12a13a21a22a23a31a32a33→a11a12a13a11a12a21a22a23a21a22a31a32a33a31a32+1 2 3 ∗ ∗∗1 2 3 ∗∗ ∗1 2 3−∗ ∗ 1 2 3∗ 1 2 3 ∗1 2 3 ∗ ∗This rule works only for 3×3 matrices!Examples: 3×3 matrices3 −2 01 0 1−2 3 0= 3 · 0 · 0 + (−2) · 1 · (−2) + 0 · 1 · 3 −− 0 · 0 · (−2) − (−2) · 1 · 0 − 3 · 1 · 3 = 4 − 9 = −5,1 4 60 2 50 0 3= 1 · 2 · 3 + 4 · 5 · 0 + 6 · 0 · 0 −− 6 · 2 · 0 − 4 · 0 · 3 − 1 · 5 · 0 = 1 · 2 · 3 = 6.General definitionThe
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