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2024 Paper 3 Q5
D: 1500.0 B: 1500.0
trace determinant matrix differential equation 2x2 matrices commutator inverse matrix proof

In this question, \(\mathbf{M}\) and \(\mathbf{N}\) are non-singular \(2 \times 2\) matrices. The \emph{trace} of the matrix \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is defined as \(\operatorname{tr}(\mathbf{M}) = a + d\).

  1. Prove that, for any two matrices \(\mathbf{M}\) and \(\mathbf{N}\), \(\operatorname{tr}(\mathbf{MN}) = \operatorname{tr}(\mathbf{NM})\) and derive an expression for \(\operatorname{tr}(\mathbf{M}+\mathbf{N})\) in terms of \(\operatorname{tr}(\mathbf{M})\) and \(\operatorname{tr}(\mathbf{N})\).
The entries in matrix \(\mathbf{M}\) are functions of \(t\) and \(\dfrac{\mathrm{d}\mathbf{M}}{\mathrm{d}t}\) denotes the matrix whose entries are the derivatives of the corresponding entries in \(\mathbf{M}\).
  1. Show that \[\frac{1}{\det \mathbf{M}} \frac{\mathrm{d}}{\mathrm{d}t}(\det \mathbf{M}) = \operatorname{tr}\!\left(\mathbf{M}^{-1} \frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t}\right).\]
  2. In this part, matrix \(\mathbf{M}\) satisfies the differential equation \[\frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t} = \mathbf{MN} - \mathbf{NM},\] where the entries in matrix \(\mathbf{N}\) are also functions of \(t\). Show that \(\det \mathbf{M}\), \(\operatorname{tr}(\mathbf{M})\) and \(\operatorname{tr}(\mathbf{M}^2)\) are independent of \(t\). In the case \(\mathbf{N} = \begin{pmatrix} t & t \\ 0 & t \end{pmatrix}\), and given that \(\mathbf{M} = \begin{pmatrix} A & B \\ C & D \end{pmatrix}\) when \(t = 0\), find \(\mathbf{M}\) as a function of \(t\).
  3. In this part, matrix \(\mathbf{M}\) satisfies the differential equation \[\frac{\mathrm{d}\mathbf{M}}{\mathrm{d}t} = \mathbf{MN},\] where the entries in matrix \(\mathbf{N}\) are again functions of \(t\). The trace of \(\mathbf{M}\) is non-zero and independent of \(t\). Is it necessarily true that \(\operatorname{tr}(\mathbf{N}) = 0\)?

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1990 Paper 3 Q6
D: 1700.0 B: 1500.0
linear transformations eigenvectors eigenvalues matrix conics ellipse tangent-normal area inverse matrix

The transformation \(T\) from \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} X \\ Y \end{pmatrix}\) is given by \[ \begin{pmatrix}X\\ Y \end{pmatrix}=\frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}. \] Show that \(T\) leaves the vector \(\begin{pmatrix} 1\\ 2 \end{pmatrix}\) unchanged in direction but multiplied by a scalar, and that \(\begin{pmatrix} 2\\ -1 \end{pmatrix}\) is similarly transformed. The circle \(C\) whose equation is \(x^{2}+y^{2}=1\) transforms under \(T\) to a curve \(E\). Show that \(E\) has equation \[ 8X^{2}+12XY+17Y^{2}=80, \] and state the area of the region bounded by \(E\). Show also that the greatest value of \(X\) on \(E\) is \(2\sqrt{17/5}.\) Find the equation of the tangent to \(E\) at the point which corresponds to the point \(\frac{1}{5}(3,4)\) on \(C\).

Solution: \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}1\\ 2 \end{pmatrix} \\ &= \frac25\begin{pmatrix}9 - 4\\ -2+12 \end{pmatrix} \\ &= \begin{pmatrix}2\\ 4 \end{pmatrix} \\ &= 2 \begin{pmatrix}1\\ 2 \end{pmatrix} \end{align*} \begin{align*} T\begin{pmatrix}1\\ 2 \end{pmatrix} &= \frac{2}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}2\\ -1 \end{pmatrix} \\ &= \frac25\begin{pmatrix}18+2\\ -4-6 \end{pmatrix} \\ &= \begin{pmatrix}8\\ -4 \end{pmatrix} \\ &= 4 \begin{pmatrix}2\\ -1 \end{pmatrix} \end{align*} Consider $T^{-1} = \frac{5}{2} \frac{1}{50}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\(, so \)T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} = \begin{pmatrix}x\\ y \end{pmatrix}$ and so: \begin{align*} x^2 + y^2 & = \begin{pmatrix}x& y \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix} \\ &= \begin{pmatrix}X& Y \end{pmatrix} (T^{-1})^T T^{-1} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}X\\ Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix}\begin{pmatrix}6 & 2\\ 2 & 9 \end{pmatrix} \begin{pmatrix}6X+2Y\\ 2X+9Y \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}6(6X+2Y)+2(2X+9Y)\\ 2(6X+2Y)+9(2X+9Y) \end{pmatrix} \\ &= \frac{1}{400}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}40X+30Y\\ 30X +85Y \end{pmatrix} \\ &= \frac{1}{80}\begin{pmatrix}X& Y \end{pmatrix} \begin{pmatrix}8X+6Y\\ 6X +17Y \end{pmatrix} \\ &= \frac{1}{80} \l 8X^2 + 12XY + 17Y^2\r \end{align*} Therefore \(8X^2 + 12XY + 17Y^2 = 80\). The area will be \(\det T \cdot \pi = \frac{4}{25} \cdot 50 \cdot \pi = 8 \pi\). Differentiating we obtain \(2 \cdot 8 \cdot X \cdot \frac{dX}{dY} + 2 \cdot 6 \cdot X + 2 \cdot 6 \cdot Y \cdot \frac{dX}{dY} + 2 \cdot 17 Y \Rightarrow \frac{dX}{dY} = -\frac{6X + 17Y}{8X+6Y}\), at a maximum (or minimum, \(6X = -17Y\)). Therefore \begin{align*} \Rightarrow && 8X^2 + 12 \cdot \frac{6}{17}X^2 + 17 ( -\frac{6}{17} X)^2 &= 80 \\ \Rightarrow && \frac{100}{17}X^2 &= 80 \\ \Rightarrow &&X^2 &= \frac{17 \cdot 4}{5} \\ \Rightarrow && |X| = 2 \sqrt {\frac{17}{5}} \end{align*} The point \(\frac15 (3,4)\) maps to \begin{align*} \frac{2}{5}\frac{1}{5}\begin{pmatrix}9 & -2\\ -2 & 6 \end{pmatrix}\begin{pmatrix}3\\ 4 \end{pmatrix} &= \frac{2}{25} \begin{pmatrix}19\\ 18 \end{pmatrix} \end{align*} So the point is \((\frac{38}{25}, \frac{36}{25})\), with gradient \(\frac{dY}{dX} = -\frac{8X+6Y}{6X + 17Y}\) which is \(-\frac{8 \cdot 19+6 \cdot 18}{6\cdot 19 + 17 \cdot 18} = -\frac{13}{21}\) therefore the equation is \(21Y+13X = 50\)

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