 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
Question from Karlena, a student: This is a question from the math class that I want to get into in college so please help me out. I am supposed to write the augmented matrix of the system and use the matrix method to solve the system. I must show my work algebraically x+y+2z=30 |
Hi Karlena,
The augmented matrix of a linear system of equations contains the coefficients of the variables augmented by a column for the constants. For your system the augmented matrix is
1 |
1 |
2 |
30 |
2 |
3 |
2 |
53 |
1 |
2 |
3 |
47 |
(I have written the matrix as a table as it is much easier for me to manipulate a table than the standard matrix notation.)
The task now is to perform elementary row operations to transform the matrix to have zeros below the diagonal and if preferably with 1 as the diagonal value in each row. (The diagonal of interest starts at the upper left corner so currently the diagonal entries are 1, 3 nd 3.) There are many choices for row operation but here is what I did.
Multiply the first row by -2 and add it to the second row. I get
1 |
1 |
2 |
30 |
0 |
1 |
-2 |
-7 |
1 |
2 |
3 |
47 |
Multiply the first row by -1 and add it to the third row. I get
1 |
1 |
2 |
30 |
0 |
1 |
-2 |
-7 |
0 |
1 |
1 |
17 |
Multiply the second row by -1 and add it to the third row. I get
1 |
1 |
2 |
30 |
0 |
1 |
-2 |
-7 |
0 |
0 |
3 |
24 |
Divide the third row by 3. I get
1 |
1 |
2 |
30 |
0 |
1 |
-2 |
-7 |
0 |
0 |
1 |
8 |
At this point rewrite the system as equations rather than in matrix form.
The last equation tells me z = 8. Substitute z = 8 into the second equation and solve for y. Substitute this value for y and z = 8 into the first equation and solve for x.
Verify your answer.
Harley
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.