p1 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Determinants and Inverses Consider the weight suspended by wires problem: One (poor) way is to find inverse or ‘reciprocal’ of A, A -1. It is defined that A -1 *A = A*A -1 = I the identity matrix If we know A -1, then by pre-multiplying the equation: A -1 * A * T = A -1 * Ythus T = A -1 * Y How to find the inverse - if one exists (just as 1/0 not exists) ? We shall consider square matrices and one method of finding the inverse - using determinants and cofactor matrices. Note: determinants are not just used to find inverses. In fact, determinants are more useful than inverses.

p2 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Determinants determinant of A = det(A) = |A| |a 11 | = a 11

p3 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Cofactors for finding Determinants Det of matrix without ith row and jth column is defined as M ij. e.g for 3*3 matrix The cofactor of element i,j, C ij, is (-1) i+j M ij (-1) i+j = 1 or -1 Then the determinant of any n*n matrix is defined as a 11 * C 11 + a 12 * C a 1n * C 1n Determinant of a 2*2 matrix is: a 11 * C 11 + a 12 * C 12 = a 11 * (-1) 1+1 a 22 + a 12 * (-1) 1+2 a 21 = a 11 * a 22 - a 12 * a 21 Determinant of a 3*3 matrix: a 11 * C 11 + a 12 * C 12 + a 13 * C 13

p4 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Adjoint Matrix of a square matrix A, Adj A If C ij is the cofactor of a ij, then Adj A, = [C ji ] = [C ij ] T. then the matrix of cofactors of A is: i.e. the transpose of the above

p5 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Matrix Inverse (poor method)

p6 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Example The cofactors are:

p7 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Properties (A -1 ) -1 = A (A T ) -1 = (A -1 ) T (A * B) -1 = B -1 * A -1 If A T = A -1 then matrix A is an orthogonal matrix.

p8 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Applying Inverses to Example Systems Suspended Mass i.e. T 1 = 300N and T 2 = 360N CheckT 1 * T 2 *0.8 = = 0 T 1 * T 2 *0.6 = = 300  Good!

p9 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Electronic Circuit |A| = 18 * (-10-15) - 10 * (0 --15) + 0 * (0-1) = -600

p10 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Thus, i 1 = 0.5A, Check: 18 * i * i 2 = = 12 i 2 = 0.3A -10 * i * i 3 = = 0 and i 3 = 0.2A-i 1 + i 2 + i 3 = = 0 Exercise: Use matrix inversion to solve

p11 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Stochastic Matrix: what was situation in 1990? By post-multiplying both sides by inverse of transition matrix

p12 RJM 07/08/02EG1C2 Engineering Maths: Matrix Algebra 3 Note, 1/0.3 is a scalar, and a k b = k a b