Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 04 (c)

MATH152 April 2011

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Question B 04 (c)
Let a=[1,2,-1]. Consider the transformation $T:\mathbb {R} ^{3}\to \mathbb {R} ^{3}$ defined by $T(\mathbf {x} )=\mathbf {x} \times \mathbf {a}$ for all x in $\mathbb {R} ^{3}$, where $\times$ is the cross-product in $\mathbb {R} ^{3}$. Compute det(A).

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
How can we compute a determinant of a matrix? Can we guess what the determinant will be based on the properties of cross-products?

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. Recall from part(b) that we computed the transformation matrix A as $A={\begin{bmatrix}0&-1&-2\\1&0&1\\2&-1&0\end{bmatrix}}$. To compute the determinant we can use co-factor expansion or any special tricks one may have for computing $3\times {3}$ determinants. ${\textrm {det}}(A)={\begin{vmatrix}0&-1&-2\\1&0&1\\2&-1&0\end{vmatrix}}=-1{\begin{vmatrix}-1&-2\\-1&0\end{vmatrix}}+2{\begin{vmatrix}-1&-2\\0&1\end{vmatrix}}=0$. Therefore we conclude that det(A)=0. Is there anyway we could have expected this before hand? Firstly, notice that the next part of the question part(d) asks for any nonzero vector that has a zero cross product. This implies that A has a null space and thus that det(A) is zero. However, this reason is somewhat cheating since it involves analyzing the wording of the problem. A more independent reasoning would be that if the determinant weren't zero, we would have an empty null space (or a unique solution to Ax=0). However, we know from the properties of cross-products that any vector in a cross product with itself is zero. Therefore we know that $\mathbf {a} \times \mathbf {a} =\mathbf {0}$, and hence T(a) = Aa = 0, i.e. a is in the null space of A. Hence det(A) must be zero. In fact, any vector parallel to $\mathbf {a}$ will also have a zero cross product since $\mathbf {a} \times (k\mathbf {a} )=k(\mathbf {a} \times \mathbf {a} )=\mathbf {0} .$