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

MATH152 April 2011

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[hide] 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).

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[show] 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?

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[show] Solution

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 $\mathbf {a} \times (k\mathbf {a} )=k(\mathbf {a} \times \mathbf {a} )=\mathbf {0} .$