Science:Math Exam Resources/Courses/MATH152/April 2012/Question 02 (b)

MATH152 April 2012

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Question 02 (b)
Consider the following system of linear equations ${\begin{aligned}2x-y+2z&=a\\x+3y-6z&=b\\3x+2y-4z&=c,\end{aligned}}$ where a,b, and c $\in \mathbb {R}$. For what values of a, b, and c is the system homogeneous? Solve the resulting homogeneous system by calculating the reduced row echelon form of the augmented matrix.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
A homogeneous problem is one in which the source vector is zero.

Hint 2
The source vector is the vector $\mathbf {b}$ in $A\mathbf {x} =\mathbf {b}$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 obtain the homogeneous problem by setting a=b=c=0. We can then solve the system by Gaussian elimination. First we switch the second and first row so that we have a 1 as the first entry. Doing this we get, $\left[{\begin{array}{ccc|c}1&3&-6&0\\2&-1&2&0\\3&2&-4&0\end{array}}\right]$ Next we want zeros in the rest of the entries of the first column. To do this we subtract 2 multiples of row 1 from row 2 and 3 multiples of row 1 from row 3 $\left[{\begin{array}{ccc|c}1&3&-6&0\\0&-7&14&0\\0&-7&14&0\end{array}}\right]$ Now divide the second row by -7 to get a 1 in the second column of the second row $\left[{\begin{array}{ccc|c}1&3&-6&0\\0&1&-2&0\\0&-7&14&0\end{array}}\right]$ Next subtract 3 multiples of row 2 from row 1 and add 7 multiples of row 2 to row 3 $\left[{\begin{array}{ccc|c}1&0&0&0\\0&1&-2&0\\0&0&0&0\end{array}}\right]$ The system is now in reduced row echelon form and we can now solve the system. From the first two rows we get ${\begin{array}{cccc}x&&&=0\\&y&-2z&=0\end{array}}$ The third row is a little more interesting. No matter what we multiply the third row with, we will get the zero vector which implies that we have a free variable. This free variable can't be x since we know that x=0 and so it will be either y or z. Let's choose it to be the z variable and set $\displaystyle {}z=t$ to be a parameter. Using the relation from row 2 then we get $\displaystyle {}y=2z=2t$ We can therefore write that the solution to the homogeneous problem is $\mathbf {x} ={\begin{bmatrix}x\\y\\z\end{bmatrix}}=t{\begin{bmatrix}0\\2\\1\end{bmatrix}}$ for any number t.

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