Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 01 (c)

MATH152 April 2011

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Question B 01 (c)

Three identical fields of size one hectare are planted with different amounts of three kinds of wheat (types 1, 2 and 3). The first field is divided in three equal parts with one type of wheat planted on each part. It yields 12 tons of total harvest. The second field is divided in two equal parts with type 2 planted on one part and type 3 planted on the other. It yields 10 tons. The third field is divided in two equal parts with type 1 planted on one part and type 2 on the other. It yields 16 tons. Solve the system above using Gaussian elimination on the augmented matrix. What is the yield in tons per hectare for each type of wheat?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Gaussian elimination is achieved by putting the matrix into reduced row-echelon form.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We take our augmented matrix from part (b) and we put it into reduced row-echelon form. Recall that this procedure ensures that every entry above and below a pivot is a zero. We could also just do a row-echelon form which ensure that only numbers below pivots are zero. However, for the purpose of example we will highlight the full reduced row-echelon form step by step. We start with the augmented matrix $\left[{\begin{array}{ccc|c}{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}&12\\&&&\\0&{\frac {1}{2}}&{\frac {1}{2}}&10\\&&&\\{\frac {1}{2}}&{\frac {1}{2}}&0&16\end{array}}\right]$ We see that the first row can contain a pivot in its first column. We set that pivot to 1 by multiplying the first row by 3 (recall that elementary row operations do not change the solution to the linear system). $\left[{\begin{array}{ccc|c}{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}&12\\&&&\\0&{\frac {1}{2}}&{\frac {1}{2}}&10\\&&&\\{\frac {1}{2}}&{\frac {1}{2}}&0&16\end{array}}\right]{\stackrel {\rightarrow }{3R1}}\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&{\frac {1}{2}}&{\frac {1}{2}}&10\\&&&\\{\frac {1}{2}}&{\frac {1}{2}}&0&16\end{array}}\right]$ We let $R_{i}$ indicate an operation on row i. We now have to clear out the rest of the entries in column 1 to be zero. We see that column one on row two is already zero and so we can move on to row three. To remove the entry in column 1 of row 3, we can subtract off 1/2 of row 1. Therefore $\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&{\frac {1}{2}}&{\frac {1}{2}}&10\\&&&\\{\frac {1}{2}}&{\frac {1}{2}}&0&16\end{array}}\right]{\stackrel {\rightarrow }{R3-1/2R1}}\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&{\frac {1}{2}}&{\frac {1}{2}}&10\\&&&\\0&0&-1/2&-2\end{array}}\right]$ We now move on to the second column and we seek to place a pivot in the second row. We want the pivot to have the value of 1 and so we multiply the whole row by 2. Therefore $\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&{\frac {1}{2}}&{\frac {1}{2}}&10\\&&&\\0&0&-1/2&-2\end{array}}\right]{\stackrel {\rightarrow }{2R2}}\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&1&1&20\\&&&\\0&0&-1/2&-2\end{array}}\right]$ To make the entries above and below this new pivot zero we must subtract row 2 from row 1 and we leave row three alone since it is already zero. $\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&1&1&20\\&&&\\0&0&-1/2&-2\end{array}}\right]{\stackrel {\rightarrow }{R1-R2}}\left[{\begin{array}{ccc|c}1&0&0&16\\&&&\\0&1&1&20\\&&&\\0&0&-1/2&-2\end{array}}\right]$ Finally for the last pivot in the third row and third column we have to multiply row three by -2. We will then have to subtract row three from row two so that the third column entry in the second row vanishes. We do not need to perform any operations on the first row since there is already a zero in the third column. $\left[{\begin{array}{ccc|c}1&0&0&16\\&&&\\0&1&1&20\\&&&\\0&0&-1/2&-2\end{array}}\right]{\stackrel {\rightarrow }{-2R3}}\left[{\begin{array}{ccc|c}1&0&0&16\\&&&\\0&1&1&20\\&&&\\0&0&1&4\end{array}}\right]{\stackrel {\rightarrow }{R2-R3}}\left[{\begin{array}{ccc|c}1&0&0&16\\&&&\\0&1&0&16\\&&&\\0&0&1&4\end{array}}\right]$ Having all entries zero except for the pivots we have our matrix in reduced row-echelon form. The advantage to putting the matrix in this form is that we can immediately determine the vector x. We have that, ${\begin{aligned}x_{1}&=16\\x_{2}&=16\\x_{3}&=4\end{aligned}}$ Therefore we conclude that the yield of the wheat is 16 tons/hectare for type 1, 16 tons/hectare for type 2 and 4 tons/hectare for type 3. Notice that if we were so inclined we could put our solution in for x in the original problem an we do indeed recover the right output vector b. If instead you opted to just put the matrix in row-echelon form, you would get $\left[{\begin{array}{ccc|c}1&1&1&36\\&&&\\0&1&1&20\\&&&\\0&0&1&4\end{array}}\right]$ which of course leads to the same conclusion but some arithmetic is involved. This type of situation is often the trade off in deciding which process to choose. Doing a row-echelon form is often quicker but because it only guarantees entries below pivots are zero, then arithmetic will almost always be required to get the solution. Conversely, the reduced row-echelon form takes longer to obtain but then the unknowns are clearly presented.