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Lect1-07-web

# Lect1-07-web - MATH 304 Fall 2011 Linear Algebra Homework...

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Unformatted text preview: MATH 304, Fall 2011 Linear Algebra Homework assignment #3 (due Thursday, September 22) All problems are from Leon’s book (8th edition). Section 2.1: 3b, 3c, 3d, 3e, 3f, 3g Section 2.2: 3f, 4, 7c, 7d MATH 304 Linear Algebra Lecture 7: Evaluation of determinants. The Vandermonde determinant. Cramer’s rule. Determinants Determinant is a scalar assigned to each square matrix. Notation. The determinant of a matrix A = (aij )1≤i ,j ≤n is denoted det A or a11 a12 a21 a22 . . . . . . an1 an2 . . . a1n . . . a2n ... . . . . . . . ann Principal property: det A = 0 if and only if a system of linear equations with the coeﬃcient matrix A has a unique solution. Equivalently, det A = 0 if and only if the matrix A is invertible. Explicit deﬁnition in low dimensions Deﬁnition. det (a) = a, ab = ad − bc , cd a11 a12 a13 a21 a22 a23 = a11a22 a33 + a12a23a31 + a13a21a32 − a31 a32 a33 −a13a22a31 − a12 a21a33 − a11a23 a32. +: −: * ∗ ∗ ∗ ∗ * ∗ * ∗ ∗ * ∗ ∗ ∗ ∗ , ∗ * * ∗ * ∗ , * ∗ ∗ * ∗ ∗ * ∗ ∗ ∗ * , ∗ ∗ ∗ , * ∗ * ∗ ∗ ∗ * * ∗ ∗ ∗ ∗ * * ∗ . ∗ ∗ * . ∗ Properties of determinants Determinants 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 scalar to another row, the determinant remains the same; • if we interchange two rows of a matrix, the determinant changes its sign. Properties of determinants Tests for singularity: • if a matrix A has a zero row then det A = 0; • if a matrix A has two identical rows then det A = 0; • if a matrix has two proportional rows then det A = 0. Properties of determinants Special matrices: • det I = 1; • the determinant of a diagonal matrix is equal to the product of its diagonal entries; • the determinant of an upper triangular matrix is equal to the product of its diagonal entries. Properties of determinants Determinant 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 a scalar, the determinant is multiplied by the same scalar; • adding a scalar multiple of one column to another does not change the determinant; • interchanging two columns of a matrix changes the sign of its determinant; • if a matrix A has a zero column or two proportional columns then det A = 0. Row and column expansions Given an n×n matrix A = (aij ), let Mij denote the (n − 1)×(n − 1) submatrix obtained by deleting the i th row and the j th column of A. Theorem For any 1 ≤ k , m ≤ n we have that n (−1)k +j akj det Mkj , det A = j =1 (expansion by kth row ) n (−1)i +maim det Mim . det A = i =1 (expansion by mth column) Signs for row/column expansions + − + − . . . − + − + . . . + − + − . . . − + − + . . . ··· · · · · · · · · · ... 123 Example. A = 4 5 6. 789 Expansion by the 1st row: ∗2∗ ∗∗ 3 1 ∗∗ ∗ 5 6 4 ∗ 6 4 5 ∗ ∗ 89 7∗9 78 ∗ det A = 1 56 46 45 −2 +3 89 79 78 = (5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0. 123 Example. A = 4 5 6. 789 Expansion by the 2nd 1∗ ∗2∗ 4 ∗ 6 ∗ 5 7∗9 7∗ det A = −2 column: 3 1∗3 ∗ 4 ∗ 6 9 ∗8∗ 46 13 13 +5 −8 79 79 46 = −2(4 · 9 − 6 · 7) + 5(1 · 9 − 3 · 7) − 8(1 · 6 − 3 · 4) = 0. 123 Example. A = 4 5 6. 789 Subtract the 1st row from the 2nd row and from the 3rd row: 123 123 123 4 5 6 = 3 3 3 = 3 3 3 =0 789 789 666 since the last matrix has two proportional rows. 12 3 Another example. B = 4 5 6 . 7 8 13 Let’s do some row reduction. Add −4 times the 1st row to the 2nd row: 12 3 123 4 5 6 = 0 −3 −6 7 8 13 7 8 13 Add −7 times the 1st row to the 3rd row: 123 123 0 −3 −6 = 0 −3 −6 0 −6 −8 7 8 13 123 123 0 −3 −6 = 0 −3 −6 7 8 13 0 −6 −8 Expand the determinant by the 1st column: 123 −3 −6 0 −3 −6 = 1 −6 −8 0 −6 −8 Thus det B = = (−3)(−2) 12 −3 −6 = (−3) −6 −8 −6 −8 12 = (−3)(−2)(−2) = −12 34 2 −2 0 3 −5 3 2 1 Example. C = 1 −1 0 −3, det C =? 2 0 0 −1 Expand the determinant by the 3rd column: 2 −2 0 3 2 −2 3 −5 3 2 1 = −2 1 −1 −3 1 −1 0 −3 2 0 −1 2 0 0 −1 Add −2 times the 2nd row to the 1st row: 009 2 −2 3 det C = −2 1 −1 −3 = −2 1 −1 −3 2 0 −1 2 0 −1 009 2 −2 3 det C = −2 1 −1 −3 = −2 1 −1 −3 2 0 −1 2 0 −1 Expand the determinant by the 1st row: 009 1 −1 det C = −2 1 −1 −3 = −2 · 9 20 2 0 −1 Thus det C = −18 1 −1 = −18 · 2 = −36 20 Problem. For what values of a will the following system have a unique solution? x + 2y + z = 1 −x + 4y + 2z = 2 2x − 2y + az = 3 The system has a unique solution if and only if the coeﬃcient matrix is invertible. 1 21 A = −1 4 2, det A =? 2 −2 a 1 21 A = −1 4 2, 2 −2 a Add −2 times the 12 −1 4 2 −2 det A =? 3rd column to the 2nd column: 1 0 1 1 0 2 2 = −1 2 −2 − 2a a a Expand the determinant by the 2nd column: 1 0 1 11 0 2 = −(−2 − 2a) det A = −1 −1 2 2 −2 − 2a a Hence det A = −(−2 − 2a) · 3 = 6(1 + a). Thus A is invertible if and only if a = −1. More properties of determinants Determinants and matrix multiplication: • if A and B are n×n matrices then det(AB ) = det A · det B ; • if A and B are n×n matrices then det(AB ) = det(BA); • if A is an invertible matrix then det(A−1) = (det A)−1. Determinants and scalar multiplication: • if A is an n×n matrix and r ∈ R then det(rA) = r n det A. Examples −1 2 1 1 00 X = 0 2 −2, Y = −1 3 0. 0 0 −3 2 −2 1 det X = (−1) · 2 · (−3) = 6, det Y = det Y T = 3, det(XY ) = 6 · 3 = 18, det(YX ) = 3 · 6 = 18, det(Y −1) = 1/3, det(XY −1) = 6/3 = 2, det(XYX −1) = det Y = 3, det(X −1 Y −1 XY ) = 1, det(2X ) = 23 det X = 23 · 6 = 48, det(−3X T XY −4) = (−3)3 · 6 · 6 · 3−4 = −12. The Vandermonde determinant Deﬁnition. The Vandermonde determinant is the determinant of the following matrix n 2 1 x1 x1 · · · x1 −1 n 2 1 x2 x2 · · · x2 −1 n 2 V = 1 x3 x3 · · · x3 −1 , . ... . . . .. . . . . . . 2 n 1 xn xn · · · xn −1 where x1, x2, . . . , xn ∈ R. Equivalently, V = (aij )1≤i ,j ≤n, where aij = xij −1. Examples. • 1 x1 1 x2 • 2 1 x1 x1 2 1 x2 x2 2 1 x3 x3 = x2 − x1 . 1 x1 0 2 = 1 x2 x2 − x1 x2 2 1 x3 x3 − x1 x3 1 0 0 2 = 1 x2 − x1 x2 − x1 x2 2 1 x3 − x1 x3 − x1 x3 = (x2 − x1 ) = 1 x2 2 x3 − x1 x3 − x1 x3 = (x2 − x1 )(x3 − x1 )(x3 − x2 ). 2 x2 − x1 x2 − x1 x2 2 x3 − x1 x3 − x1 x3 = (x2 − x1 )(x3 − x1 ) 1 x2 1 x3 Theorem 1 1 1 . . . 1 x1 x2 x3 . . . xn 2 x1 2 x2 2 x3 . . . 2 xn n · · · x1 −1 n · · · x2 −1 n (xj − xi ). · · · x3 −1 = ... . . 1≤i <j ≤n . n · · · xn −1 Corollary The Vandermonde determinant is not equal to 0 if and only if the numbers x1 , x2, . . . , xn are distinct. Let x1, x2, . . . , xn be distinct real numbers. Theorem For any b1 , b2, . . . , bn ∈ R there exists a unique polynomial p (x ) = a0 +a1 x + · · · +an−1 x n−1 of degree less than n such that p (xi ) = bi , 1 ≤ i ≤ n. n −1 2 a0 + a1 x1 + a2 x1 + · · · + an−1x1 = b1 n 2 a0 + a1 x2 + a2 x2 + · · · + an−1x2 −1 = b2 ············ 2 n a0 + a1 xn + a2 xn + · · · + an−1xn −1 = bn a0 , a1, . . . , an−1 are unknowns. The coeﬃcient matrix is the Vandermonde matrix. A general system of n linear equations in n variables: a11x1 + a12x2 + · · · + a1n xn = b1 a21x1 + a22x2 + · · · + a2n xn = b2 ⇐⇒ Ax = b, ········· an1 x1 + an2 x2 + · · · + ann xn = bn where a11 a12 a a 21 22 A = . . . . . . an1 an2 . . . a1n x1 b1 b x . . . a2n 2 2 . . . . , x = . , b = . . . . . . . bn xn . . . ann Cramer’s rule a11x1 + a12x2 + · · · + a1n xn = b1 a21x1 + a22x2 + · · · + a2n xn = b2 ⇐⇒ Ax = b ········· an1 x1 + an2 x2 + · · · + ann xn = bn Theorem Assume that the matrix A is invertible. Then the only solution of the system is given by det Ai , i = 1, 2, . . . , n , xi = det A where the matrix Ai is obtained by substituting the vector b for the i th column of A. Example. x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 13z = 1 Augmented matrix of the system: 12 3 0 (A | b) = 4 5 6 0 7 8 13 1 As obtained earlier in this lecture, det A = −12. Since det A = 0, there exists a unique solution of the system. Example. x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 13z = 1 12 3 0 4 5 6 0 7 8 13 1 By Cramer’s rule, x= 02 3 05 6 1 8 13 12 3 45 6 7 8 13 23 56 −3 1 = = = −12 −12 4 Example. x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 13z = 1 12 3 0 4 5 6 0 7 8 13 1 By Cramer’s rule, y= 10 3 40 6 7 1 13 12 3 45 6 7 8 13 − = 13 46 6 1 = =− −12 −12 2 Example. x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 13z = 1 12 3 0 4 5 6 0 7 8 13 1 By Cramer’s rule, z= 120 450 781 12 3 45 6 7 8 13 12 45 −3 1 = = = −12 −12 4 System of linear equations: x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 13z = 1 Solution: (x , y , z ) = 11 1 ,− , . 4 24 Determinants and the inverse matrix Given an n×n matrix A = (aij ), let Mij denote the (n − 1)×(n − 1) submatrix obtained by deleting the i th row and the j th column of A. The cofactor matrix of A is an n×n matrix A = (αij ) deﬁned by αij = (−1)i +j det Mij . Theorem AT A = AAT = (det A)I . Corollary If det A = 0 then the matrix A is invertible and A−1 = (det A)−1 AT . AAT = (det A)I means that n k +j akj j =1 (−1) n k +j amj j =1 (−1) det Mkj = det A det Mkj = 0 for all k , for m = k . ...
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