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

MATH152 April 2011

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[hide] 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.

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[show] Hint
Do we know anything about cross-products that produce the zero vector?

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[show] 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).