Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 29

MATH152 April 2011

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Question A 29
For questions A29 and A30 below, decide if the statements are true or false. In each case give brief justification of your conclusion. For all $n\times {n}$ matrices A and B such that det(A)=det(B)=0, det(A+B)=0 also.

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
Recall it is much easier to prove something is false, since all you require is one counter example. Can you think of an example such that this is not true? You can choose any value for $n,$ whichever makes finding a counter example the easiest for you.

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. As mentioned in the hint, if we can find a counter example to the statement for any $n$, then the statement is false since, if true, it must hold for ALL $n$. Consider $2\times {2}$ matrices with all entries zero except for the alternating diagonals. Specifically, we mean let ${\begin{aligned}A={\begin{bmatrix}a&0\\0&0\end{bmatrix}},\qquad B={\begin{bmatrix}0&0\\0&b\end{bmatrix}}.\end{aligned}}$ We choose these matrices because we know that the determinant of a diagonal matrix is just the product of its diagonal elements. Of course, since these are $2\times {2}$ matrices we could easily compute any determinant, but these matrices also satisfy a condition of the statement which is that det(A)=det(B)=0. We can then compute det(A+B), ${\textrm {det}}(A+B)={\begin{vmatrix}a&0\\0&b\end{vmatrix}}=ab$ which is not zero for arbitrary a and b with $a\neq {0}$, $b\neq {0}$. Therefore we have found a whole class of $2\times {2}$ matrices that satisfy det(A)=det(B)=0 but which do not satisfy det(A+B)=0. Since the statement claims to be true for ALL $n\times {n}$ matrices we can use our counter example to verify the statement is false. Note that only one counterexample would be enough, so your solution is just as good, if you choose numbers for $a$ and $b$, e.g. $a=b=1.$