Science:Math Exam Resources/Courses/MATH307/December 2008/Question 03

MATH307 December 2008

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Question 03
For the parameter α, consider the linear system: ${\begin{aligned}x-y+z&=2,\\-2x+y+\alpha z&=-3,\\x+\alpha y-z&=1.\end{aligned}}$ Determine when the system has a unique solution, no solution, or infinitely many solutions. (Don’t determine the actual solutions!)

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
What is the augmented matrix that corresponds to this problem?

Hint 2
Row-reduce the augmented matrix, then consider special choices for α separately.

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. Let A be the augmented matrix, that is, $A={\begin{bmatrix}1&-1&1&2\\-2&1&\alpha &-3\\1&\alpha &-1&1\end{bmatrix}}$ To begin the row-reduction, we add two multiples of the first line to the second line and also subtract the first line from the third to obtain ${\begin{bmatrix}1&-1&1&2\\-2&1&\alpha &-3\\1&\alpha &-1&1\end{bmatrix}}\sim {\begin{bmatrix}1&-1&1&2\\0&-1&\alpha +2&1\\0&\alpha +1&-2&-1\end{bmatrix}}\sim {\begin{bmatrix}1&-1&1&2\\0&1&-(\alpha +2)&-1\\0&\alpha +1&-2&-1\end{bmatrix}}$ Then, subtract a factor of (α+2) times the second row from the last row and simplify ${\begin{aligned}{\begin{bmatrix}1&-1&1&2\\0&1&-(\alpha +2)&-1\\0&\alpha +1&-2&-1\end{bmatrix}}&\sim {\begin{bmatrix}1&-1&1&2\\0&1&-(\alpha +2)&-1\\0&0&(-(\alpha +2))(-(\alpha +1))-2&-1+\alpha +1\end{bmatrix}}\\&\sim {\begin{bmatrix}1&-1&1&2\\0&1&-(\alpha +2)&-1\\0&0&\alpha (\alpha +3)&\alpha \end{bmatrix}}\end{aligned}}$ We can now investigate what happens for different values of α. Notice that the first line doesn't depend on α and so there is nothing to conclude for the first line. For the second line, α could make one of the terms vanish but there will still be a solution to that equation if this happens. Therefore, the real power of α comes from the third line. We can distinguish three cases for α: When α = 0 the matrix will be ${\begin{bmatrix}1&-1&1&2\\0&1&-2&-1\\0&0&0&0\end{bmatrix}}$, and there will be infinitely many solutions. When α = -3 the matrix will be ${\begin{bmatrix}1&-1&1&2\\0&1&1&-1\\0&0&0&-3\end{bmatrix}}$, and there will be no solution. When α is neither 0 nor -3, there will be a unique solution.

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