2 x 7 y z 4 x 4 y z 9 x 3 y 3 We can express as AX B as follows 2 7 1 1 4 1 1 3

# 2 x 7 y z 4 x 4 y z 9 x 3 y 3 we can express as ax b

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2 x + 7 y + z = 4 x + 4 y - z = 9 x + 3 y = 3 We can express as AX = B as follows 2 7 1 1 4 - 1 1 3 0 x y z = 4 9 3 . We have already seen that A - 1 = - 3 2 - 3 2 11 2 1 2 1 2 - 3 2 1 2 - 1 2 - 1 2 . 139 Using inverse of matrix method to solve X = A - 1 B , i.e. x y z = - 3 2 - 3 2 11 2 1 2 1 2 - 3 2 1 2 - 1 2 - 1 2 4 9 3 = ( - 3 2 )(4) + ( - 3 2 )(9) + ( 11 2 )(3) ( 1 2 )(4) + ( 1 2 ) + ( - 3 2 )(3) ( 1 2 )(4) + ( - 1 2 )(9) + ( - 1 2 )(3) = - 6 - 27 2 + 33 2 2 + 9 2 - 9 2 2 - 9 2 - 3 2 = - 3 2 - 4 . Hence x = - 3, y = 2 and z = - 4. Cramer’s Rule This method uses determinant to solve the system of linear equations. Consider the system of linear equation AX = B with unique solution. Then det A exist, say | A | = D . Let D x = | A x | where A x is a matrix obtained by replacing the column of coefficients of x by B . Similarly D y = | A y | where A y is a matrix obtained by replacing the column of coefficients of y by B ; D y = | A y | in similar fashion, etc. Then we can obtain unique solution x = D x D ; y = D y D ; z = D z D ; etc. where D 6 = 0 . 140 Example 5.5.9. 2 x + 7 y + z = 4 x + 4 y - z = 9 x + 3 y = 3 2 7 1 1 4 - 1 1 3 0 x y z = 4 9 3 . Then D = | A | = 2 7 1 1 4 - 1 1 3 0 = 2 7 1 3 11 0 1 3 0 = 1 × 3 11 1 3 = 3 × 3 - 11 × 1 = 9 - 11 = - 2 . D x = | A x | = 4 7 1 9 4 - 1 3 3 0 = 4 7 1 13 11 0 3 3 0 = 1 × 13 11 3 3 = 13 × 3 - 11 × 3 = 39 - 33 = 6 . 141 D y = | A y | = 2 4 1 1 9 - 1 1 3 0 = 2 4 1 3 13 0 1 3 0 = 1 × 3 13 1 3 = 3 × 3 - 13 × 1 = 9 - 13 = - 4 . D z = | A z | = 2 7 4 1 4 9 1 3 3 = 0 1 - 2 1 4 9 0 - 1 - 6 = - 1 × 1 - 2 - 1 - 6 = - [1 × ( - 6) - ( - 1) × ( - 2)] = - ( - 6 - 2) = 8 . So x = D x D = 6 - 2 = - 3 , y = D y D = - 4 - 2 = 2 , z = D z D = 8 - 2 = - 4 . 5.6 3 × 3 systems with parameters Consider the system of linear equations: 142 Example 5.6.1. x + 3 y + (2 - 3 k ) z = 3 2 x + 7 y + (4 - 7 k ) z = 7 2 x + ky + ( k + 2) z = 2 k - 1 Determine the values of k R such that (a) The system has no solution (b) The system has infinitely many solutions (c) The system has unique (only one) solution (consistent). Solution: We perform Gauss-Jordan elimination of the augmented matrix to obtain: 1 3 2 - 3 k 2 7 4 - 7 k 2 k k + 2 3 7 2 k - 1 1 3 2 - 3 k 0 1 - k 0 k - 6 7 k - 2 3 1 2 k - 7 R 2 - 2 R 1 R 3 - 2 R 1 1 0 2 0 1 - k 0 0 k 2 + k - 2 0 1 k - 1 R 1 - 3 R 2 R 3 - ( k - 6) R 2 1 0 2 0 1 - k 0 0 ( k + 2)( k - 1) 0 1 k - 1 Then (a) The system has no solution if ( k + 2)( k - 1) = 0 and k - 1 6 = 0, so k = - 2 and k 6 = 1, since for k = - 2, 143 we obtain 1 0 2 0 1 2 0 0 0 0 1 - 3 which is clearly impossible. (b) The system has infinitely many solutions if ( k + 2)( k - 1) = 0 and k - 1 = 0, thus k = 1. (c) The system has unique solution if ( k + 2)( k - 1) 6 = 0 thus k 6 = - 2 and k 6 = 1. Tutorials 5.6.2. 1. Use Gauss-Jordan Elimination to find values of x , y and z where possible and describe whether each type of solution is either unique, no solution or infinitely many solutions. (a) x + 3 y + 2 z = 4 - 2 x - 7 y + z = 9 3 x + 10 y + 9 z = 19 (b) x + 2 y + z = 2 2 x + 3 y + 3 z = 3 - 3 x - 4 y - 5 z = - 5 (c) x + 2 y + 3 z = 3 2 x + 3 y + 8 z = 4 5 x + 8 y + 19 z = 11 (d) From ( a ), ( b ) and ( c ) above, where you found a unique solution, use the inverse of matrix method to confirm the solution.  #### You've reached the end of your free preview.

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