Science:Math Exam Resources/Courses/MATH221/December 2011/Question 01 (b)

MATH221 December 2011

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Question 01 (b)
Consider the system of equations in the variables x1, x2, x3: ${\begin{aligned}&x_{1}+2x_{2}+3x_{3}&=1\\&x_{1}+x_{2}+2x_{3}&=2\\&2x_{1}+3x_{2}+(5+t)x_{3}&=2\end{aligned}}$ For those t for which the system is consistent, give the solution set in parametric form.

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
You can determine the solution of a system from its row-reduced form. Parametric just means you keep carrying the variable t, so your solution is a general formula that gives the solution of the system in terms of t.

Checking a solution serves two purposes: helping you if, after having used the hint, 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. In the first part of the question, we calculated the row reduced form of the augmented matrix of the system and obtained: $\left[{\begin{array}{cccc}1&2&3&1\\0&1&1&-1\\0&0&t&-1\end{array}}\right]$ For which we saw that unless t = 0, there is always a unique solution (since the matrix has full rank in that case). The third row of the matrix translates into $\displaystyle tx_{3}=-1$ Which is equivalent to $x_{3}=-{\frac {1}{t}}$ (recall that we suppose here that t is NOT equal to 0.) The second row gives $\displaystyle x_{2}+x_{3}=-1$ and so $\displaystyle x_{2}-{\frac {1}{t}}=-1$ which gives $\displaystyle x_{2}=-1+{\frac {1}{t}}={\frac {1-t}{t}}$ And finally, the first row gives $\displaystyle x_{1}+2x_{2}+3x_{3}=1$ Substituting for x2 and x3 we obtain $\displaystyle x_{1}+{\frac {2(1-t)}{t}}-{\frac {3}{t}}=1$ And hence $\displaystyle x_{1}={\frac {t-2(1-t)+3}{t}}={\frac {3t+1}{t}}$ And we obtain that the system has a unique solution (when t is not zero) which depends on t and the solution is $\left[{\begin{array}{c}x_{1}\\\\x_{2}\\\\x_{3}\end{array}}\right]=\left[{\begin{array}{c}{\frac {3t+1}{t}}\\\\{\frac {1-t}{t}}\\\\-{\frac {1}{t}}\end{array}}\right]$