Advanced Math/Discrete Mathematics
Expert: Ahmed Salami - 10/19/2008
QuestionChapter 6- Systems of Linear Equations and Matrices
Write the augmented matrix of the system and use the matrix method to solve the system. If the system has an infinite number of solutions, express them in terms of the parameter z.
3x - 2y - 8z = 1
9x - 6y -24z =-2
x - y + z = 1
AnswerHi Jeffrey,
The set of equations
3x - 2y - 8z = 1
9x - 6y -24z = -2
x - y + z = 1
is unfortunately unsolvable because of the first two equations.
Multiply the first by 3 and you get
9x - 6y - 24z = 3
Compare with the second equation
9x - 6y - 24z = -2
and you can quickly see that there exists no solutions for the two
equations. I'm, however of the opinion that this is a sort of typo
from you and would be expecting a correction.
In case we want to proceed with finding a solution, the set of
equations is written in matrix form as
| 3 -2 -8 | |x| | 1 |
| 9 -6 -24| |y| = |-2 |
| 1 -1 1 | |z| | 1 |
The augmented form of this is written
| 3 -2 -8 ! 1 |
| 9 -6 -24 !-2 |
| 1 -1 1 ! 1 |
We need to make the equation easily solvable by using the method of
matrix row transformations.
Let Ri represents row i, then doing the first transformation
R1 = R1 - 2R3
the matrix becomes
| 1 0 -10 !-1 |
| 9 -6 -24 !-2 |
| 1 -1 1 ! 1 |
doing R2 = R2 - 9R3, we have
| 1 0 -10 !-1 |
| 0 3 -33 !-11|
| 1 -1 1 ! 1 |
doing R3 = R3 - R1, we have
| 1 0 -10 !-1 |
| 0 3 -33 !-11|
| 0 -1 11 ! 2 |
doing R3 = 3R3 + R2, we have
| 1 0 -10 !-1 |
| 0 3 -33 !-11|
| 0 0 0 !-5 |
Transforming back to the usual matrix form
| 1 0 -10 | |x| |-1 |
| 0 3 -33 | |y| = |-11|
| 0 0 0 | |z| |-5 |
The set of equations become
x + 0y - 10z = -1
0x + 3y - 33z = -11
0x + 0y + 0z = -5
The last equation is clearly impossible for it claims that 0 = -5
With a set of consistent equations, the method proves to be
excellent. You can get more to read here;
http://www.okc.cc.ok.us/maustin/Matrix_Solutions/Matrix%20Solution%20of%20Linear...
www.math.tamu.edu/~fnarc/psfiles/rowred2005.pdf
You can always get back to me.
Regards.