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

MATH152 April 2011

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Question B 04 (d)
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 $\mathbb {R} ^{3}$. Find any nonzero vector x such that T(x)=0.

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
Do we know anything about cross-products that produce the zero vector?

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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 part(b), we have that $A={\begin{bmatrix}0&-1&-2\\1&0&1\\2&-1&0\end{bmatrix}}$ and in part(c) we concluded that det(A)=0. Therefore A has a null space and there are many solutions such that Ax=0. We could perform some sort of Gaussian elimination and determine the entire null-space but the question asks for any vector and so we are permitted to just guess. In fact we saw in part(c) that any vector parallel to a had a zero cross product. Therefore $T(\mathbf {a} )=\mathbf {a} \times \mathbf {a} =\mathbf {0}$. Note that, since T(a) = Aa we can also verify this using the matrix representation: ${\begin{bmatrix}0&-1&-2\\1&0&1\\2&-1&0\end{bmatrix}}{\begin{bmatrix}1\\2\\-1\end{bmatrix}}={\begin{bmatrix}0\\0\\0\end{bmatrix}}.$ Therefore a vector such that T(x)=0 is x=a (or any vector of the form ka, due to the linearity of T).