Determinants of Order 2 and 3 PDF

10. Determinants  Determinant of order 2  a11 a12  Consider a 2  2 matrix: A=  a21 a22   ▪Determinant of A, denoted | A | , is a number and can be evaluated by...

10. Determinants  Determinant of order 2  a11 a12  Consider a 2  2 matrix: A=  a21 a22   ▪Determinant of A, denoted | A | , is a number and can be evaluated by a11 a12 | A |= = a11 a22 − a12 a21 a21 a22 ▪easy to remember (for order 2 only).. a11 a12 | A |= = + a11 a22 − a12 a21 a21 a22 1 2 Example: Evaluate the determinant: 3 4 1 2 = 1  4 − 2  3 = −2 3 4 The following properties are true for determinants of any order. 1. If every element of a row (column) is zero, e.g., 1 2 = 1 0 − 2  0 = 0 , then |A| = 0. 0 0 2. |AT| = |A| determinant of a matrix = that of its transpose 3. |AB| = |A||B| Example: Show that the determinant of any orthogonal matrix is either +1 or –1. Solution For any orthogonal matrix, A AT = I. Since |AAT| = |A||AT | = 1 and |AT| = |A|, so |A|2 = 1 or |A| = 1.   a11 a12  For any 2x2 matrix A =  a   21 a22  −1 1  a22 − a12  Its inverse can be written as A =  A  − a21 a11   −1 0 Example: Find the inverse of A =  1 2  The determinant of A is -2 −1  2 0   −1 0  −1 = = 2   1 / 2 A Hence, the inverse of A is  −1 −1 1 / 2  How to find an inverse for a 3x3 matrix?  Determinants of order 3 1 2 3 4 6 Consider an example: A =  5   7 8 9  Its determinant can be obtained by: + − + − + − 1 2 3 + − + 4 5 1 2 1 2 A = 4 5 6 =3 −6 +9 7 8 7 8 4 5 7 8 9 = 3 ( −3) − 6 ( −6) + 9 ( −3) = 0 You are encouraged to find the determinant by using other rows or columns  Inverse of a 33 matrix Each element a ij of a square matrix has a minor which is the value of the determinant obtained from the matrix after eliminating the ith row and jth column to which the element is common. The cofactor of element a ij is then given as the minor of a ij multiplied by i+ j (-1 ) Example 1 2 3 If A= 0 4 5  1  0 6  + − + − + − The cofactor for each element of matrix A: + − + 4 5 A11 = = 24 0 5 0 4 0 6 A12 =- =5 A13 = = -4 1 6 1 0 2 3 1 3 1 2 A21 = = -12 A22 = =3 A23 =- =2 0 6 1 6 1 0 2 3 1 3 1 2 A31 = = -2 A32 =- = -5 A33 = =4 4 5 0 5 0 4  Adjoint of a square matrix Let square matrix C be constructed from the square matrix A where the elements of C are the respective cofactors of the elements of A so that if: A =  aij  and Aij is the cofactor of aij then C = ( Aij )   Then the transpose of C is called the adjoint of A, denoted by adjA (or Adj (A)). 1 2 3 So, the matrix C of A= 0 4 5  is then given by:  24 5 -4  1  0 6  C =  -12 3 2   24 5 -4 T 24 -12 -2   -2 -5 4  adj A =  -12 3 2  =  5 3 -5   -2 -5 4   -4 2 4   1 2 3 Inverse matrix of A = 0 4 5  is given by: 1  0 6  T  24 5 -4  24 -12 -2 1 1  1  A-1 = adj (A) =  -12 3 2  =  5 3 -5   A A 22  -2  -5 4  -4  2 4  12 11 - 6 11 -1 11   =  5 22 3 22 -5 22   - 2 11 1 11 2 11  Example: An example in which cancellation is not valid Show that AC = BC but A  B Solution: Let C is noninvertible, 1 3 2 4  1 − 2 A=  , B= , C= (i.e., i.e., C-1 dos not exists) 0 1 2 3 − 1 2  1 3  1 − 2 − 2 4 AC =  = 0 1  − 1  2  − 1 2  2 4  1 − 2 − 2 4 BC =  = 2 3  − 1  2   −1 2  So A C = BC But A  B