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I have been asked the following question in a tutorial:

Let A be a 3x3 matrix which is invertible. Show that you can always perform a rotation of 3-space to make the last row of A be [0 0 A33]

I haven't the faintest idea how to do this and have been deeply confused during lectures. I have tried speaking to my lecturer in person when I've previously had problems, but I never understand him. (He's new and has difficulty pitching at first-year level.)

Could anyone here show me how to do this and explain, as simply and clearly as possible, how they derived the answer? (I need to be able to apply this in a test this Thursday.) You guys are my last resource.

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    I guess it means that there is a rotation matrix $S$ (i.e. an orthogonal matrix with determinant $+1$) such that $SAS^{-1}$ has the property that its last row is of the form $\begin{pmatrix} 0 & 0 & A_{33}\end{pmatrix}$.2011-04-17
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    You can premultiply any 3-by-3 matrix $\mathbf A$ with a matrix of the form $$\mathbf Q=\bigl(\begin{smallmatrix}1&0&0\\0&c&s\\0&-s&c\end{smallmatrix}\bigr)$$ where $c^2+s^2=1$; the idea is to choose $c$ and $s$ appropriately so that a particular entry in the last row of $\mathbf Q\mathbf A$ is zero. Remember also that the product of two rotation matrices is also a rotation matrix.2011-04-17
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    Is A33 the same as in the original matrix or is it stating that it just needs to be non zero?2011-04-17

3 Answers 3