Determinants PDF

Introduction to Determinants M.A. Darwish Introduction to Determinants Determinants are considered as one of the important and easiest methods to solve many mathematical problems specially solving systems of three or more line...

Introduction to Determinants M.A. Darwish Introduction to Determinants Determinants are considered as one of the important and easiest methods to solve many mathematical problems specially solving systems of three or more linear equations. a11 a12 Definition 1: The symbol consists of the four numbers a11 , a12 , a21 , a21 a22 and a22 , arranged in two rows and two columns, is called a determinant of order two. The four numbers are called the elements of the determinant. Definition 2: The value of the determinant of order one is the number itself. Definition 3: The value of the determinant of order two is given by a11 a12  a11a22  a12 a21 a21 a22 3 2 Example 1:  (3)(5)  (2)(4)  15  8  7 4 5 Example 2: - 3 2 Example 3:  (3)(4)  (2)(6)  12  12  0 6 4 Notice that the value of the determinant is unique and may be positive as in Example 1 or negative as in Example 2 or zero as in Example 3. Example 4: Show that if we interchanged the rows and columns of the determinant of order two the value of the obtained determinant is the same as original one. a11 a12 Solution: Let the determinant be , so the determinant with the rows a21 a22 a11 a21 and columns interchanged is given by. Now, a12 a22 a11 a21 a11 a12  a11a22  a21a12  a11a22  a12 a21 . a12 a22 a21 a22 -1- Introduction to Determinants M.A. Darwish Example 5: Show that if the elements of one row/column are proportional to the elements of the other row/column, the value of the determinant is equal to zero. a b Solution: The determinant with proportional rows is given by. Now, ka kb a b  (a )(kb)  (b)(ka )  k (ab  ab)  0. ka kb c kc Also, the determinant with proportional columns is given by , and so d kd c kc  (c)(kd )  (kc)(d )  k (cd  cd )  0 d kd 2x  1 2x 1 Example 6: Find the values of x for which  0. 4x  2 x 1 Solution: We have - Cramer's Rule: Here, we solve two linear equations in two unknowns by using the determinants of second order. Consider the system of equations  a1 x  b1 y  c1  a2 x  b2 y  c2 (*) Now, it is possible to obtain c1b2  b1c2 a c c a x , y  1 2 1 2 ; a1b2  b1a2  0 a1b2  b1a2 a1b2  b1a2 Rewrite the values for x and y in terms of determinants of order two as follows: x y x ,y ;  0   a1 b1 where   is called the determinant of coefficients in which the a2 b2 elements are the coefficients of x and y arranged as given in equations (*), -2- Introduction to Determinants M.A. Darwish c1 b1 x  is the same as the determinant of coefficients  with the column of c2 b2 the coefficients for the variable x in  is replaced with the column of constants a1 c1 on the right-hand side of equations (*), and  y  is the same as the a2 c2 determinant of coefficients  with the column of the coefficients for the variable y in  is replaced with the column of constants on the right-hand side of equations (*). The method of solution of linear equations by determinants is called Cramer's rule. Notice that, in the case   0 , Cramer's rule can't be used to solve the system. 2 x  3 y  1 Example 7: Solve the system .  x y 3 Solution: We have 2 3     (2)(1)  (3)(1)  2  3  5  0 1 1 1 3 x     (1)(1)  (3)(3)  1  9  10 3 1 2 1 y     (2)(3)  (1)(1)  6  1  5 1 3 10 5 x  2, y   1 5 5 Thus, the solution of the system is (2,  1). 5 x  2 y  14  0 Example 8: Solve the system .  3y  2x  3  0 Solution:  5 x  2 y  14 Rewrite the system as .  2 x  3 y  3 We have -3- Introduction to Determinants M.A. Darwish 5 2     (5)(3)  (2)(2)  15  4  19  0 2 3 14 2 x     (14)(3)  (2)(3)  42  6  36 3 3 5 14 y     (5)(3)  (14)(2)  15  28  43 2 3 36 43 x ,y 19 19 Thus, the solution of the system is (36 19,  43 19). a11 a12 a13 Determinants of order three: The symbol a21 a22 a23 consists of nine a31 a32 a33 numbers arranged in three rows and three columns, is called a determinant of order three. The four numbers are called the elements of the determinant. The determinant of order three can be determined by expressing it in terms of second order determinants which is known as expansion of a determinant along a row/column. There are six ways of expanding a determinant of order three corresponding to each of three rows and three columns and each way gives the same value. a11 a12 a13 Consider the determinant | A | a21 a22 a23 , we expanding it along the first a31 a32 a33 row as follows: a22 a23 a21 a23 a21 a22 | A | (1)11 a11  (1)1 2 a12  (1)13 a13 a32 a33 a31 a33 a31 a32  a11 (a22 a33  a23a32 )  a12 (a21a33  a23 a31 )  a13 (a21a32  a22 a31 ) -4- Introduction to Determinants M.A. Darwish 2 3 1 Example 9: Evaluate 1 1 2. 0 1 1 Solution: 2 3 1 1 2 1 2 1 1 1 1 2  2 3  1 1 0 1 0 1 0 1 1  2(1  2)  3(1  0)  (1  0)  6  3 1  8 There is an alternative method commonly used for evaluating the determinant of order three. We copy the first and second columns of the determinant to form fourth and fifth columns. The determinant is then obtained by adding (or subtracting) the products of the six diagonals, as the following diagram. a11 a12 a13 | A | a21 a22 a23  (a11a22 a33  a12 a23a31  a13a21a32 )  (a31a22 a13  a32 a23a11  a33a21a12 ) a31 a32 a33 2 3 1 Example 10: Evaluate 1 1 2. 0 1 1 Solution: We write 2 2 1 2 3 1 1 2 1 1 0 1 1 0 1 Then, -5- Introduction to Determinants M.A. Darwish 2 3 1 1 1 2  [(2)(1)(1)  (3)(2)(0)  (1)(1)(1)] 0 1 1  [(0)(1)(1)  (1)( 2)(2)  (1)( 1)(3)]  (2  0  1)  (0  4  3)  1  (7)  8 Cramer's Rule: Cramer's rule for three linear equations in three unknowns by using the determinants of order three. Consider the system of equations  a1 x  b1 y  c1 z  d1  a2 x  b2 y  c2 z  d 2 (**) a x  b y  c z  d  3 3 3 3 Then x y  x ,y , z  z ;  0    a1 b1 c1 d1 b1 c1 a1 d1 c1 where   a2 b2 c2 ,  x  d2 b2 c2 ,  y  a2 d2 c2 and a3 b3 c3 d3 b3 c3 a3 d3 c3 a1 b1 d1  z  a2 b2 d2. a3 b3 d3  x  3y  z  2  Example 11: Solve the system 2 x  y  2 z  1. x  y  z  0  Solution: We have 1 3 1 1 2 2 2 2 1   2 1 2  3  1 1 1 1 1 1 1 1 1  (1  2)  3(2  2)  (2  1) -6- Introduction to Determinants M.A. Darwish  8  0 2 3 1 1 2 1 2 1 1  x  1 1 2  2 3   6  3 1  8 1 1 0 1 0 1 0 1 1 1 2 1 1 2 2 2 2 1  y  2 1 2  2   1  8  1   8 0 1 1 1 1 0 1 0 1 1 3 2 1 1 2 1 2 1  z  2 1 1  3 2  1 3  2  0 1 0 1 0 1 1 1 1 0 x 8 y 8 z 0 x   1 y  1 z  0  8  8  8 Thus, the solution of the system is (1,1, 0). -7-