|
From B1(b) we have that the augmented matrix is
![{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc|c}1&-2&0&0\\8&0&8&6\\16&16&16&14\end{array}}\right]\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3888f32f491cc175f2b0690749ebb5afbe3fa488)
which we now need to solve by Gaussian elimination. This means putting pivots along the diagonals of the matrix with value 1. Notice the pivot on the first row is the first column which already has a 1 and so we can leave that unchanged. We want all of the entries below this 1 (and in the same column) to vanish. To do this we subtract 8 multiples of row 1 from row 2 (8 because that's the number in the first column of the second row). For a similar reason, we subtract 16 multiples of row 1 from row 3. Therefore the new resulting augmented matrix is
![{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc|c}1&-2&0&0\\0&16&8&6\\0&48&16&14\end{array}}\right].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6967b4ca01b7914691e7efdc34594695282c5e8a)
We move to the second row, where the pivot is in the second column. We want this value to be 1 and so we multiply the second row by 1/16 to get
![{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc|c}1&-2&0&0\\0&1&1/2&3/8\\0&48&16&14\end{array}}\right].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1e9e7055f355807a50e4d413f97f812a021e5a7a)
We now must make the entries below this pivot (and in the same column) vanish. To do this we subtract 48 multiples of row 2 from row 3
![{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc|c}1&-2&0&0\\0&1&1/2&3/8\\0&0&-8&-4\end{array}}\right].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b4091fce24205f9b1b19aa4d88b3ded6f135d70f)
Now, if wanted we could remove the entries above the pivot as well. This is known as Gauss-Jordan elimination. By doing this we'd end up with zeros everywhere except for the augmented portion and possibly for the pivots. This is a very easy way to see the final solution since there are no more operations to perform. We will continue with just Gaussian elimination, which will result in an upper triangular matrix with zeros below each pivot. To find the solution from there will require a little more work, but we are compensating that by doing fewer row operations on the matrix.
Finally then, by continuing Gaussian elimination, we want to make the pivot in the third row (the number in the third column) have a value 1. We therefore multiply the third row by -1/8 to get
![{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc|c}1&-2&0&0\\0&1&1/2&3/8\\0&0&1&1/2\end{array}}\right].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/532a6d7bf32b43297613a7b7aba57261908827e8)
We must now answer the problem which is to find how many rooms each person can paint in an hour ( ). From the last row we have that
and so Keiko paints 1/2 rooms per hour (or 1 room every 2 hours). From the second row we get (using Keiko's now known speed),
and so Mark paints 1/8 rooms per hour (or 1 room every 8 hours). We can use both of these speeds now with the first row to get,
and so Hartosh paints 1/4 rooms per hour (or 1 room every 4 hours). We have found the number of rooms that Hartosh, Mark, and Keiko can paint in an hour to be 1/4, 1/8, and 1/2 respectively.
| 
