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FP2: Matrices help

http://mei.org.uk/files/papers/fp210ju_irue.pdf

Does anyone on question 3 part (ii) for the last 2 marks so find x,y & z how you do it and why you can do what the mark scheme does for this part?
We know that M(331)=2(331)M\begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix}

So: (331)=12M(331)=M(3/23/21/2)\begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} = \frac{1}{2} M \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} = M \begin{pmatrix} 3/2 \\ -3/2 \\ 1/2 \end{pmatrix}

Hence:

M(xyz)=(331)M(xyz)=M(3/23/21/2)M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ 1 \end{pmatrix} \Rightarrow M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = M \begin{pmatrix} 3/2 \\ -3/2 \\ 1/2 \end{pmatrix} \Rightarrow \cdots

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