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TAMU MATH 304 - Lect1-07web

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MATH 304Linear AlgebraLecture 7:Evaluation of determinants.The Vandermonde determinant.Cramer’s rule.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 invertible.Explicit definition in low dimensionsDefinition. det (a) = a,a bc d= ad − bc,a11a12a13a21a22a23a31a32a33= a11a22a33+ a12a23a31+ a13a21a32−−a13a22a31− a12a21a33− a11a23a32.+ :* ∗ ∗∗ * ∗∗ ∗ *,∗* ∗∗ ∗ ** ∗ ∗,∗ ∗** ∗ ∗∗ * ∗.− :∗ ∗*∗ * ∗* ∗ ∗,∗* ∗* ∗ ∗∗ ∗ *,* ∗ ∗∗ ∗ *∗ * ∗.Properties of determinantsDeterminants and elementary row operations:• if a row of a matrix is multiplied by a scalar r,the determinant is also multiplied by r;• if we add a row of a matrix multiplied by a scalarto another row, the determinant remains the same;• if we interchange two rows of a matrix, thedeterminant changes its sign.Properties of determinantsTests for singularity:• if a matrix A has a zero row then det A = 0;• if a matrix A has two iden tical rows thendet A = 0;• if a matrix has two proportional rows thendet A = 0.Properties of determinantsSpecial matrices:• det I = 1;• the determinant of a diagonal matrix is equal tothe product of its diagonal entries;• the determinant of an upper triangular matrix isequal to the product of its diagonal entries.Properties of determinantsDeterminant of the transpose:• If A is a square matrix then det AT= det A.Columns vs. rows:• if one column of a matrix is multiplied by ascalar, the determinant is multiplied by the samescalar;• adding a scalar multiple of one column toanother does not change the determinant;• interchanging two columns of a matrix changesthe sign of its determinant;• if a matrix A has a zero column or twoproportional columns then det A = 0.Row and column expansionsGiven an n×n matrix A = (aij), let Mijdenote the(n − 1)×(n − 1) submatrix obtained by deleting theith row and the jth column of A.Theorem For any 1 ≤ k, m ≤ n we have thatdet A =nXj=1(−1)k+jakjdet Mkj,(expansion by kth row)det A =nXi=1(−1)i+maimdet Mim.(expansion by mth column)Signs for row/column expansions+ − + − · · ·− + − + · · ·+ − + − · · ·− + − + · · ·...............Example. A =1 2 34 5 67 8 9.Expansion by the 1st row:1 ∗ ∗∗ 5 6∗ 8 9∗2 ∗4 ∗ 67 ∗ 9∗ ∗34 5 ∗7 8 ∗det A = 15 68 9− 24 67 9+ 34 57 8= (5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0.Example. A =1 2 34 5 67 8 9.Expansion by the 2nd column:∗2 ∗4 ∗ 67 ∗ 91 ∗ 3∗ 5 ∗7 ∗ 91 ∗ 34 ∗ 6∗ 8 ∗det A = −24 67 9+ 51 37 9− 81 34 6= −2(4 · 9 − 6 · 7) + 5(1 · 9 − 3 · 7) − 8(1 · 6 − 3 · 4) = 0.Example. A =1 2 34 5 67 8 9.Subtract the 1st row from the 2nd row and fromthe 3rd row:1 2 34 5 67 8 9=1 2 33 3 37 8 9=1 2 33 3 36 6 6= 0since the last matrix has two proportional rows.Another example. B =1 2 34 5 67 8 13.Let’s do some row reduction.Add −4 times the 1st row to the 2nd row:1 2 34 5 67 8 13=1 2 30 −3 −67 8 13Add −7 times the 1st row to the 3rd row:1 2 30 −3 −67 8 13=1 2 30 −3 −60 −6 −8Expand the determinant by the 1st column:1 2 30 −3 −60 −6 −8= 1−3 −6−6 −8Thusdet B =−3 −6−6 −8= (−3)1 2−6 −8= (−3)(−2)1 23 4= (−3)(−2)(−2) = −12.Example. C =2 −2 0 3−5 3 2 11 −1 0 −32 0 0 −1, det C =?Expand the determinant by the 3rd column:2 −2 0 3−5 3 2 11 −1 0 −32 0 0 −1= −22 −2 31 −1 −32 0 −1Add −2 times the 2nd row to the 1st row:det C = −22 −2 31 −1 −32 0 −1= −20 0 91 −1 −32 0 −1Expand the determinant by the 1st row:det C = −20 0 91 −1 −32 0 −1= −2 · 91 −12 0Thusdet C = −181 −12 0= −18 · 2 = −36.Problem. For what values of a will the followingsystem have a unique solution?x + 2y + z = 1−x + 4y + 2z = 22x − 2y + az = 3The system has a unique solution if and only if t hecoefficient matrix is invertible.A =1 2 1−1 4 22 −2 a, det A =?Add −2 times the 3rd column to the 2nd column:1 2 1−1 4 22 −2 a=1 0 1−1 0 22 −2 − 2a aExpand the determinant by the 2nd column:det A =1 0 1−1 0 22 −2 − 2a a= −(−2 − 2a)1 1−1 2Hence det A = −(−2 − 2a) · 3 = 6(1 + a).Thus A is invertible if and only if a 6= −1.More properties of determinantsDeterminants and matrix multiplication:• if A and B are n×n matrices thendet(AB) = det A · det B;• if A and B are n×n matrices thendet(AB) = det(BA);• if A is an invertible matrix thendet(A−1) = (det A)−1.Determinants and scalar multiplication:• if A is an n×n matrix and r ∈ R thendet(rA) = rndet A.ExamplesX =−1 2 10 2 −20 0 −3, Y =1 0 0−1 3 02 −2 1.det X = (−1) · 2 · (−3) = 6, det Y = det YT= 3,det(XY ) = 6 · 3 = 18, det(YX ) = 3 · 6 = 18,det(Y−1) = 1/3, det(XY−1) = 6/3


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TAMU MATH 304 - Lect1-07web

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