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

This preview shows page 142 - 148 out of 151 pages.

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
Image of page 142
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
Image of page 143
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
Image of page 144
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
Image of page 145
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
Image of page 146
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.
Image of page 147
Image of page 148

You've reached the end of your free preview.

Want to read all 151 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes