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TAMU MATH 304 - Lect1-08web

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MATH 304Linear AlgebraLecture 8:Elementary matrices.Transpose of a matrix.Determinants.General 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 tha t a sequence of elementary rowoperations converts a matrix A into the identity matrix.Then the same sequence o f 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 caseWhy does it work?1 0 00 2 00 0 1a1a2a3b1b2b3c1c2c3=a1a2a32b12b22b3c1c2c3,1 0 03 1 00 0 1a1a2a3b1b2b3c1c2c3=a1a2a3b1+3a1b2+3a2b3+3a3c1c2c3,1 0 00 0 10 1 0a1a2a3b1b2b3c1c2c3=a1a2a3c1c2c3b1b2b3.Proposition Any elementary row operation can besimulated as left multipli cati on by a certain matrix.Elementary matricesE =1...O1r1O...1row #iTo obtain the matrix EA from A, multiply the i throw by r . To obtain the matrix AE from A,multiply the ith colum n by r .Elementary ma tr icesE =1......O0 · · · 1.........0 · · · r · · · 1............0 · · · 0 · · · 0 · · · 1row #irow #jTo obtain the matrix EA from A, add r times theith row to the jth row. To obtain the matrix AEfrom A, add r times the jth col umn to the i thcolumn.Elementary ma tr icesE =1 O...0 · · · 1.........1 · · · 0...O 1row #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 matricessimulating those operations.Applying the s ame sequence of operations to theidentity m atrix, 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 i s ,if A = (aij) then AT= (bij), where bij= aji.Examples.1 2 34 5 6T=1 42 53 6,789T= (7, 8, 9) ,4 77 0T=4 77 0.Properties of transposes:• (AT)T= A• (A + B)T= AT+ BT• (rA)T= rAT• (AB)T= BTAT• (A1A2. . . Ak)T= ATk. . . AT2AT1• (A−1)T= (AT)−1Definition. A square matrix A is said to besymmetric if AT= A.For example, any diagonal matrix is symmetric.Proposition For any square matrix A the matricesB = AATand C = A + ATare sym metric.Proof:BT= (AAT)T= (AT)TAT= AAT= B,CT= (A + AT)T= AT+ (AT)T= AT+ A = C .DeterminantsDeterminant is a scalar assigned to each square matrix.Notation. The determinant of a matrixA = (aij)1≤i,j≤nis denoted det A ora11a12. . . a1na21a22. . . a2n............an1an2. . . ann.Principal property: det A 6= 0 if and only if asystem of linear equations with the coefficientmatrix A has a unique solution. Equivalently,det A 6= 0 if and only if the matrix A is inverti ble.Definition i n low dimensionsDefinition. det (a) = a ,a bc d= ad − bc,a11a12a13a21a22a23a31a32a33= a11a22a33+ a12a23a31+ a13a21a32−−a13a22a31− a12a21a33− a11a23a32.+ :* ∗ ∗∗* ∗∗ ∗*,∗* ∗∗ ∗** ∗ ∗,∗ ∗** ∗ ∗∗* ∗.− :∗ ∗*∗ * ∗* ∗ ∗,∗* ∗* ∗ ∗∗ ∗ *,* ∗ ∗∗ ∗ *∗ * ∗.Examples: 2×2 matr ices1 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.Examples: 3×3 matr ices3 −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 =


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TAMU MATH 304 - Lect1-08web

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