Matrices and linear transformations – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(\mathbf{AB}=\begin{pmatrix}2+2&0-5\\3-8&0+20\end{pmatrix}\) [1] \(=\) \(\begin{pmatrix}4&-5\\-5&20\end{pmatrix}\) [1]
   2. \(\det\mathbf A=8+3=11\), \(\det\mathbf B=5-0=5\) [1]; \(\det(\mathbf{AB})=80-25=55=11\times5\) ✓ [1]

   Method: each entry of \(\mathbf{AB}\) is a row of \(\mathbf A\) times a column of \(\mathbf B\); for example the top-left entry is \(2\times1+(-1)\times(-2)=4\). For a \(2\times2\) matrix the determinant is \(ad-bc\), so \(\det\mathbf A=2\times4-(-1)\times3=11\).

   The product rule for determinants reflects area scale factors: applying \(\mathbf B\) then \(\mathbf A\) multiplies areas by 5 and then by 11. Note that \(\mathbf{BA}\ne\mathbf{AB}\) in general, but both have determinant 55.

2. Answer to Question 2 (4 marks)

   1. \(k(k\cdot1-1\cdot1)-1(2\cdot1-1\cdot0)+0\) [1] \(=k^2-k-2\) ✓ [1]
   2. \((k-2)(k+1)=0\): \(k=2\) or \(k=-1\) [1]
   3. \(\det=9-3-2=4\); volume \(=8\times4=\) 32 cubic units [1]

   Method: expand along the first row, using \(\det\begin{pmatrix}k&1\\1&1\end{pmatrix}=k-1\) and \(\det\begin{pmatrix}2&1\\0&1\end{pmatrix}=2\), with the sign pattern \(+\,-\,+\). A matrix is singular when its determinant is zero, so \(\mathbf N\) has no inverse for these values of \(k\).

   In (c), the determinant is the volume scale factor: the cube has volume \(2^3=8\), and multiplying by \(|\det\mathbf N|=4\) gives 32. Check (b): with \(k=2\), \(4-2-2=0\) ✓.

3. Answer to Question 3 (4 marks)

   1. \(T_1\): \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) [1]; \(T_2\): \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\) [1]
   2. \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\) \(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\) [1]: reflection in the line \(y=-x\) [1]

   Method: the columns of a transformation matrix are the images of \(\binom10\) and \(\binom01\). A clockwise rotation of \(90^\circ\) sends \(\binom10\) to \(\binom0{-1}\) and \(\binom01\) to \(\binom10\). For \(T_1\) followed by \(T_2\), the matrix is \(\mathbf T_2\mathbf T_1\): the first transformation is written on the right.

   Check: \((1,0)\mapsto(0,-1)\mapsto(0,-1)\), and \((0,-1)\) is the reflection of \((1,0)\) in \(y=-x\) ✓. The determinant is \(-1\), as expected for a reflection.

4. Answer to Question 4 (4 marks)

   1. \(\mathbf M\binom t{-t}=\binom{2t-t}{2t-3t}=\binom t{-t}\) ✓ [1]
   2. \((x,mx)\mapsto(2x+mx,\;2x+3mx)\); \(2+3m=m(2+m)\) [1], so \(m^2-m-2=0\), \(m=-1\) or \(m=2\): \(y=2x\) [1]
   3. \((1,2)\mapsto(4,8)\), so points are multiplied by 4: invariant line, not a line of invariant points [1]

   Method: a line \(y=mx\) is invariant if the image of every point on it lies on the same line, which gives an equation for \(m\). A line of invariant points is stronger: each point must map to itself. In (b), the root \(m=-1\) recovers the line from (a), a useful check. The eigenvalue for \(y=2x\) is 4, so points move away from the origin along that line.

5. Answer to Question 5 (4 marks)

   1. \(P'(8,4)\), \(Q'(2,6)\) [1]
   2. Field: \(\tfrac12\times4\times2=4\) square units (\(40\,000\) m²) [1]; \(\det\mathbf M=6-1=5\), so image area 20 square units [1]
   3. \(\mathbf M^{-1}=\dfrac15\begin{pmatrix}3&-1\\-1&2\end{pmatrix}\), \(\dfrac15\binom{15-5}{-5+10}\): \((2,1)\), i.e. 200 m east and 100 m north of \(O\) [1]

   Method: the determinant of \(\mathbf M\) is the area scale factor, so the image area is \(5\times4\). In real units the image represents \(200\,000\) m², five times the true field. In (c), solve \(\mathbf M\mathbf x=\binom55\) by premultiplying by \(\mathbf M^{-1}\). Check: \(\mathbf M\binom21=\binom{4+1}{2+3}=\binom55\) ✓.

Total : 20 marks