Inverse Matrices and Determinants

Overview • Review last lecture – Background for matrix multiplication – General rule and examples of matrix multiplication Matrix Multiplication, Inverse Matrices, and Determinants • Definition, computation and use of determinants • Use of determinants in computing matrix inverses • Example problems 2 Review last lecture • Vector has a magnitude and a direction – Can represent by components Review Matrix Basics x2 f f1  f2 • f∙dx = |f||dx|cos() x1 • f = f1i + f2 j+ f3k or f = [f1 f2 f3] • Unit vectors i = [1 0 0], j = [0 1 0], k = [0 0 1] in x, y, z directions • Dot product: f∙dx = f1dx1 + f2dx2 + f3dx3  a11 a12 a a22  21 a31 a32 A         a n1 a n 2 a13   a1m  a23   a2 m   a33   a3m            an 3   anm  • Array of numbers with n rows and m columns • Components are a(row)(column) • Size of matrix (n x m) is number of rows and columns • Square matrix: m = n 3 Review Diagonal Matrix a1 0 0 a 2  0 0 A      0 0 0 0 a3   0 0 0    0         an      Review Matrix Operations • The diagonal matrix A is a square matrix with nonzero components only on the principal diagonal • Components of A are aiij, where ij is the Kroenecker delta 4 • Can add or subtract matrices if they are the same size – C = A  B only valid if A, B, and C have the same size (rows and columns) – Components of C, cij = aij  bij • Multiplication by a scalar: C = xA – C and A have the same size (rows and columns) – Components of C = xA, cij = xaij 1 i  j 0 i  j  ij   5 6 Page 1 Review Null (0)/Unit (I) Matrices 0 0  0 0    0 0 0   0 0 0   0 0 0 0           0 0   0 1 0  0 I    0 Unit and Null Matrix sizes • • • • • • For any matrix, A, A + 0 = 0 + A = A; IA = AI = A and 0 A = A 0 = 0 • The unit (or identity) matrix is a square matrix; the null matrix need not be square 0 0   0 1 0   0 0 1   0           0 0   1 Given that A is an (n x m) matrix The I in IA = A is (n x n) The I in AI = A is (m x m) The 0 in 0 + A = A or A + 0 = A is (n x m) The 0 in A0 is (m x arbitrary size) – The product A0 is also a zero matrix whose size is (n x null matrix columns) • The 0 in 0A is (arbitrary size x n) – The product 0A is also a zero matrix whose size is (null matrix rows x m) 7 Review Matrix Transpose Review Row/Column Vectors • Transpose of A denoted as AT (or A’) • Reverse rows and columns; for B = AT • Matrices with only one row or only one column are called row or column vectors (or matrices) – bij = aji – If A is (n x m), B = AT is (m by n)  3 12  6 A  14  2 0  8 r  r11  r1  3 14  T A   12  2    6 0  r12 r2 r13   r1m  r3   rm   c11   c1  c   c   21   2  c  c31    c3            cn1  cn  • Single subscript is usual notation, but implied “1” for single row or column is important for two-subscript formulas 9 Review Multiplying Matrices • For matrix multiplication, C = AB – A has n rows and p columns – B has p rows and m columns – C has n rows and m columns 10 Matrix Multiplication Example • For matrix multiplication, C = AB p cij   aik bkj k 1 (i  1, n; j  1, m) • For C = AB, we get cij by adding products of terms in row i of A (left matrix) by terms in column j of B (right matrix) • cij = ai1b1j + ai2b2j + ai3b3j + ai4b4j + … • In general, AB ≠ BA • “Premultiply” by matrix on left and “postmultiply” by matrix on right – A has n rows and p columns – B has p rows and m columns – C has n rows and m columns p cij   aik bkj k 1 (i  1, n; j  1, m) • Example  3 4 B  1 2   6 1  3(3)  0(1)  6(6) 3( 4)  0( 2)  6(1)    27 6  AB     4(3)  2(1)  0(6) 4( 4)  2( 2)  0(1)  10 12  3 0  6 A  4  2 0  • Can we find BA? What is its size? 3 x 3 11 12 Page 2 Matrix Multiplication Exercise Matrix Multiplication Exercise II • Consider the following matrices • What is the size of C = AB? 1 2 3 A  0  1 4    1 1 0 0 2  2 0 B  1 3 1 0    2 1  3 1 1 2 3 A  0  1 4    1 1 0 • Can you find AB, BA, neither or both? • We can find AB, because A has three columns and B has three rows • We cannot find BA because B has four columns and A has three rows 0 2  2 0 B  1 3 1 0    2 1  3 1 • C = AB has three rows (like A) and four columns (like B) • What is c11? • c11 = (1)(0) + (2)(1) + (3)(2) = 8 13 Matrix Multiplication Exercise III • Find c11, c12, c13, and c14 in C = AB 1 2 3 A  0  1 4    1 1 0 • • • • 14 Matrix Multiplication Exercise IV • Find c21, c22, c23, and c24 in C = AB 0 2  2 0 B  1 3 1 0    2 1  3 1 1 2 3 A  0  1 4    1 1 0 c11 = (1)(0) + (2)(1) + (3)(2) = 8 c12 = (1)(2) + (2)(3) + (3)(1) = 11 c13 = (1)(-2) + (2)(1) + (3)(-3) = -9 c14 = (1)(0) + (2)(0) + (3)(1) = 3 • • • • 0 2  2 0 B  1 3 1 0    2 1  3 1 c21 = (0)(0) + (-1)(1) + (4)(2) = 7 c22 = (0)(2) + (-1)(3) + (4)(1) = 1 c23 = (0)(-2) + (-1)(1) + (4)(-3) = -13 c24 = (0)(0) + (-1)(0) + (4)(1) = 4 15 Matrix Multiplication Exercise V • Find c31, c32, c33, and c34 in C = AB 1 2 3 A  0  1 4    1 1 0 • • • • 16 Matrix Multiplication Exercise VI • Solution for matrix product C = AB 0 2  2 0 B  1 3 1 0    2 1  3 1 1 2 3 A  0  1 4    1 1 0 c31 = (1)(0) + (1)(1) + (0)(2) = 1 c32 = (1)(2) + (1)(3) + (0)(1) = 5 c33 = (1)(-2) + (1)(1) + (0)(-3) = -1 c34 = (1)(0) + (1)(0) + (0)(1) = 0 0 2  2 0 B  1 3 1 0    2 1  3 1 8 11  9 3 C  7 1  13 4   1 5  1 0 17 18 Page 3 Coordinate transformations Coordinate transformations II • Recall previous equations y1  a11 x1  a12 x2 z1  b11 y1  b12 y 2 y 2  a21 x1  a22 x2 z 2  b21 y1  b22 y2 • Define matrices so that y = Ax and z = By • Show that matrix definitions give transformation results y1  a11x1  a12 x2 z1  b11 y1  b12 y2 y2  a21x1  a22 x2 z2  b21 y1  b22 y2 a  x   a x  a x  y  a y   1   Ax   11 12   1    11 1 12 2   y2  a21 a22   x2  a21 x1  a22 x2  a  b  a b A   11 12  B   11 12  a21 a22  b21 b22  x  y  z  x   1 y   1 z   1  x2   y2   z2  b  y  b y  b y  b z  z   1   By   11 12   1    11 1 12 2   z2  b21 b22   y2  b21 y1  b22 y2  19 Coordinate Transformations III • From matrix equations y = Ax and z = By, we have z = BAx = Cx with C = BA a A   11 a21 a12  a22  b  b B   11 12  b21 b22  c  c C   11 12  c21 c22  c  b a  b a b11a12  b12 a22  c C   11 12    11 11 12 21  c21 c22  b21a11  b22 a21 b21a12  b22 a22  c  x  c x  c x  z  c z   1   Cx   11 12   1    11 1 12 2   z2  c21 c22   x2  c21 x1  c22 x2  20 Determinants • Looks like a matrix but isn’t a matrix • A square array of numbers with a rule for computing a single value for the array Example at  a11 a12 a13  a11 a12 a13 right shows Det a21 a22 a23   a21 a22 a23 calculation of a31 a32 a33  a31 a32 a33 Det(A), the determinant of a11a22 a33  a21a32 a13  a31a12 a23 3 x 3 matrix A   a a a  a a a  a a a 11 32 23 21 12 33 31 22 13 21 General Determinant 22 More Determinants • The value of an n by n determinant is found by taking a sum of n! terms • Each term in the sum has • Useful in obtaining algebraic expressions for matrix operations, but not useful for numerical computation – a product of n determinant elements, aij – an element from each row and an element from each column: a1a2a3a4… • Subscripts … are a permutation of 1234… • Terms are positive or negative if the subscript permutation is even or odd • Practical calculation formulas follow 23 a  a a a Det A  Det  11 12   11 12 a21 a22  a21 a22  a11a22  a21a12 • An n x n determinant has n2 Minors, Mij, obtained by deleting row i and column j • Cofactors, Cij = (-1)i+jMij used in general expressions for determinants 24 Page 4 General Rule for Determinants Determinant Behavior • Any size determinant can be evaluated by any of the following equations • A determinant is zero if any row or any column contains all zeros. • If one row or one column of a determinant is multiplied by a constant, k, the value of the determinant is multiplied by the same constant. n n n n i 1 j 1 i 1 j 1 Det A   (1)i  j aij M ij  (1)i  j aij M ij   aij Cij  aij Cij • Can pick any row or any column • Choose row or column with most zeros to simplify calculations • Can apply equation recursively; evaluate a 5 x 5 determinant as a sum of 4 x 4 determinants then get 4 x 4’s in terms of 3 x 3’s – Note the implication for matrices: if a matrix is multiplied by a constant, k, then each matrix element is multiplied by k. If A is an n x n matrix, Det(kA) = knDet(A). 25 26 Determinant Behavior II Determinant Behavior III • If one row (or one column) of a determinant is replaced by a linear combination of that row (or column) with another row (or column), the value of the determinant is not changed. • If two rows (or two columns) of a determinant are linearly dependent the value of the determinant is zero. • The determinant of the product of two matrices, A and B is the product of the determinants of the individual matrices: Det(AB) = Det(A) Det(B). • The determinant of transposed matrix is the same as the determinant of the original matrix: Det(AT) = Det(A). 27 Example of General Rule Example of General Rule II • Get determinant of a 3 x 3 matrix by expansion along last row a11 a12 a21 a22 a13 a23  a31C31  a32C32  a33C33 a31 a33 a32 C33  ( 1)33 M 33  C32  (1) 3 2 M 32 a11 a12 a21 a22 a a13   11 a21 a23 C22  (1) 2 2 M 22  C31  (1) 31 M 31  • Get determinant of a 3 x 3 matrix a31M 31  a32 M 32  a33 M 33  a a13 a a a a a31 12  a32 11 13  a33 11 12  a22 a23 a21 a23 a21 a22  a31 ( a12a23  a22a13 )  a32 (a11a23 a11 a13 a31 a33 a12 a13 a22 28 a23 29  a21a13 )  a33 (a11a22  a12a21 )  a31a12a23  a31a22a13  a32a11a23  a32a21a13  a33a11a22  a33a12a21 30 Page 5 Inverse of a Matrix Formula for Inverse of a Matrix • For a square matrix, A, an inverse matrix, A-1 may exist such that AA-1 = A-1A = I • For the algebraic equation ax = b, x = a-1b • For the matrix equation Ax = b, x = A-1b • Just as x = a-1b is not valid if a = 0, x = A-1b is not valid if A-1 does not exist – A-1 does not exist if DetA = 0 • The inverse is an important concept in analysis of linear systems, but is not used in computational work • Find the components of B = A-1, bij, from determinant of A and its cofactors C ji M ji  (1)i  j If B  A 1 , bij  Det ( A) Det ( A) • Use to get algebraic equations for components of inverse matrix • Matrix computations, if necessary, obtain components by alternative numerical algorithms – more next two weeks 31 Inverse of 2 x 2 Matrix 1 If B  A , b11  ( 1)11 b21  (1) 2 1 bij  C ji Det ( A) M 11 a22  Det ( A ) Det ( A) M 12 a21  Det ( A ) Det ( A) a  a A   11 12  a a  21 22  A 1   (1) i j Inverse of 3 x 3 Matrix B  A1, M ji bij  (1)i j Det(A) Apply M ji Det ( A) M 21 a12  Det ( A ) Det ( A) M 22 a11  b22  (1) 2 2 Det ( A ) Det ( A) b12  (1)1 2 1 a11a22  a21a12 32  a22  a12   a   21 a11   a11 a12 a a  21 22 a31 a32 1 a13  a23    a33  ( a22 a33  a32 a23 ) ( a32 a13  a33a12 ) ( a12 a23  a22 a13 )  (a a  a a ) (a a  a a ) (a a  a a )  11 33 31 13 21 13 11 23   31 23 33 21 ( a21a32  a31a22 ) ( a31a12  a11a32 ) ( a11a22  a21a12 )  a11a22 a33  a21a32 a13  a31a12 a23    (-1)(2+3)M32   a11a32 a23  a21a12 a33  a31a22 a13  33 Example Problem • Find A-1 for A at right 1 2 • Have the following -1 A formula for B = A 1 (1)i  j M ji C ji  bij   0 Det(A) Det(A) 34 Example Problem Det A 0 1 0 2 2 0 0 3 • General determinant formula: aijCij • Take sum over third row to simplify calculation of Det A 1 0  0  1 • Det A = (-1)3+1a31M31 + (-1)3+2a32M32+ (-1)3+3a33M33 + (-1)3+4a34M34 1 0 2 1   2 1 0 0 0 2 1  A 1 0 0 0  Det A  (1)31 (1) 1 0 0  0  0  0   2 3 1 0 2 3 1  0 2 1  (0)(0)(1)  (1)(3)(1)  (2)(2)(0) Det A  1 0 0 1  (2)(0)(1)  (1)(2)(1)  (0)(3)(0) 2 3 1 35 36 Page 6 Example Problem III • Apply: bij = (-1)i+jMji/Det A • Det A = 1 so bij = (-1)i+jMji 1 2 – Examples b12, b43 shown below A   1 0 2 1  0 1 2 b12  (1) M 21   0 0 0  0 2 3 1 b43  (1) 4 3 Example Problem Solution 0 2 1 1 0 0  0 0 0  2 3 1 Remove Remove for M21 for M34 1 0 2  3  8  0 M 34   2 1 0   11  0  0  0 0 2 3 37 1 2 A 1  0 • • • • 0 2 1 1 0 0 0 0 0  2 3 1 0 1 0 0 0 1 2 0   A 1    1  2 5 1    4  11  2 3 We can show that AA-1 = A-1A = I (AA-1)11 = 1·0 + 0·0 + 2·(-1) + 1·3 = 1 (A-1A)43 = 3·2 + 4·0 + (-11)·0 + (-2)·2 = 0 Only 14 left to check 38 Excel MINVERSE Function Original Matrix MINVERSE Result Select area for inverse and type MINVERSE formula in formula bar Press Control+ Shift+Enter MINVERSE Result (formula view) 39 Page 7