FM Nov 2024 p12 q04
4146
☆
The matrices A , B and C are given by
\(A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix}, \ B = \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} \text{ and } C = \begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix}.\)
(a) Show that \(CAB = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}.\) [3]
(b) Find the equations of the invariant lines, through the origin, of the transformation in the \(x-y\) plane represented by \(CAB.\) [5]
Let \(M = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}.\)
(c) Give full details of the transformation represented by \(M.\) [2]
(d) Find the matrix \(N\) such that \(NM = CAB.\) [3]
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Solution
(a) To find \(CAB\), first calculate \(AB\):
\(AB = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix}\)
Then calculate \(CAB\):
\(CAB = \begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}\)
(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\):
\(-9x + 3mx = m(3x - 7y)\)
\(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9\)
\(m = \pm \frac{3}{7}\)
Thus, \(y = \frac{3}{7}x\) and \(y = -\frac{3}{7}x\).
(c) The matrix \(M\) represents a stretch parallel to the x-axis with scale factor 3.
(d) Find \(N\) such that \(NM = CAB\):
\(M^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\)
\(N = CAB \cdot M^{-1} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -7 \\ -3 & 3 \end{pmatrix}\)
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