IB Questionbank

User interface language: English | Español

Date	May Example question	Marks available	8	Reference code	EXM.3.AHL.TZ0.5
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Find	Question number	5	Adapted from	N/A

Question

This question will diagonalize a matrix and apply this to the transformation of a curve.

Let the matrix $M = \left( {\begin{array}{*{20}{c}} {\tfrac{5}{2}}&{\tfrac{1}{2}} \\ {\tfrac{1}{2}}&{\tfrac{5}{2}} \end{array}} \right)$ .

Let $\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{{ - 1}}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = {{\mathbf{R}}^{ - 1}}$ .

Let ${\mathbf{R}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right)$ .

Let $\left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 3 }}}&0 \\ 0&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} u \\ v \end{array}} \right)$ .

Hence state the geometrical shape represented by

Find the eigenvalues for $M$ . For each eigenvalue find the set of associated eigenvectors.

[8]

Show that the matrix equation $\left( {x\,\,\,y} \right){\mathbf{M}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right)$ is equivalent to the Cartesian equation $\frac{5}{2}{x^2} + xy + \frac{5}{2}{y^2} = 6$ .

[2]

Show that $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right)$ are unit eigenvectors and that they correspond to different eigenvalues.

[2]

c.i.

Hence, show that $M\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ .

[1]

c.ii.

Find matrix R.

[2]

d.i.

Show that ${\mathbf{M}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}$ .

[1]

d.ii.

Verify that $\left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){{\mathbf{R}}^{ - 1}}$ .

[3]

e.i.

Hence, find the Cartesian equation satisfied by $X$ and $Y$ .

[2]

e.ii.

Find the Cartesian equation satisfied by $u$ and $v$ and state the geometric shape that this curve represents.

[2]

State geometrically what transformation the matrix ${\mathbf{R}}$ represents.

[2]

the curve in $X$ and $Y$ in part (e) (ii), giving a reason.

[2]

h.i.

the curve in $x$ and $y$ in part (b).

[1]

h.ii.

Write down the equations of two lines of symmetry for the curve in $x$ and $y$ in part (b).

[2]

Markscheme

$\left| {\begin{array}{*{20}{c}} {\frac{5}{2} - \lambda }&{\frac{1}{2}} \\ {\frac{1}{2}}&{\frac{5}{2} - \lambda } \end{array}} \right| = 0 \Rightarrow {\left( {\frac{5}{2} - \lambda } \right)^2} - {\left( {\frac{1}{2}} \right)^2} = 0 \Rightarrow \frac{5}{2} - \lambda = \pm \frac{1}{2} \Rightarrow \lambda = 2\,\,{\text{or}}\,\,{\text{3}}$ M1M1A1A1

$\lambda = 2$ $\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{\frac{1}{2}} \\ {\frac{1}{2}}&{\frac{1}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right) \Rightarrow q = - p$ eigenvalues are of the form $t\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ M1A1

$\lambda = 3$ $\left( {\begin{array}{*{20}{c}} { - \frac{1}{2}}&{\frac{1}{2}} \\ {\frac{1}{2}}&{ - \frac{1}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right) \Rightarrow q = p$ eigenvalues are of the form $t\left( {\begin{array}{*{20}{c}} 1 \\ {1} \end{array}} \right)$ M1A1

[8 marks]

$\left( {x\,\,\,y} \right)\left( {\begin{array}{*{20}{c}} {\frac{5}{2}}&{\frac{1}{2}} \\ {\frac{1}{2}}&{\frac{5}{2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right) \Rightarrow \left( {\begin{array}{*{20}{c}} {\frac{5}{2}x + \frac{1}{2}y}&{\frac{1}{2}x + \frac{5}{2}y} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right)$ M1A1

$\Rightarrow \left( {\frac{5}{2}{x^2} + \frac{1}{2}xy + \frac{1}{2}xy + \frac{5}{2}{y^2}} \right) = \left( 6 \right) \Rightarrow \frac{5}{2}{x^2} + xy + \frac{5}{2}{y^2} = 6.$ AG

[2 marks]

$\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ corresponding to $\lambda = 2$ , $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right) = \frac{1}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ corresponding to $\lambda = 3$ R1R1

[2 marks]

c.i.

$M\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right) = 2\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right)\,\,{\text{and}}\,\,M\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right) = 3\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right) \Rightarrow M\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ A1AG

[1 mark]

c.ii.

Determinant is 1. ${\mathbf{R}} = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{{ - 1}}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)$ M1A1

[2 marks]

d.i.

${\mathbf{M}}{{\mathbf{R}}^{ - 1}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ so post multiplying by ${\mathbf{R}}$ gives ${\mathbf{M}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}$ M1AG

[1 mark]

d.ii.

$\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{{ - 1}}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}x - \tfrac{1}{{\sqrt 2 }}y} \\ {\tfrac{1}{{\sqrt 2 }}x + \tfrac{1}{{\sqrt 2 }}y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}x - \tfrac{1}{{\sqrt 2 }}y}&{\tfrac{1}{{\sqrt 2 }}x + \tfrac{1}{{\sqrt 2 }}y} \end{array}} \right)$ M1A1

and $\left( {\begin{array}{*{20}{c}} x&y \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\tfrac{{ - 1}}{{\sqrt 2 }}}&{\tfrac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\tfrac{1}{{\sqrt 2 }}x - \tfrac{1}{{\sqrt 2 }}y}&{\tfrac{1}{{\sqrt 2 }}x + \tfrac{1}{{\sqrt 2 }}y} \end{array}} \right)$ completing the proof A1AG

[3 marks]

e.i.

$\left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){\mathbf{M}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right) \Rightarrow \left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right) \Rightarrow \left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right) = \left( 6 \right)$

$\Rightarrow \left( {2{X^2} + 3{Y^2}} \right) = \left( 6 \right) \Rightarrow \frac{{{X^2}}}{3} + \frac{{{Y^2}}}{2} = 1$ M1A1

[2 marks]

e.ii.

$\frac{X}{{\sqrt 3 }} = u{\text{,}}\,\,\frac{Y}{{\sqrt 2 }} = v \Rightarrow {u^2} + {v^2} = 1$ , a circle (centre at the origin radius of 1) A1A1

[2 marks]

A rotation about the origin through an angle of 45° anticlockwise. A1A1

[2 marks]

an ellipse, since the matrix represents a vertical and a horizontal stretch R1A1

[2 marks]

h.i.

an ellipse A1

[1 mark]

h.ii.

$y = x$ , $y = - x$ A1A1

[2 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

h.i.

[N/A]

h.ii.

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Eigenvalues and eigenvectors

Show 60 related questions

22M.2.AHL.TZ1.4a.iii:
Hence state the long-term velocity of the particle.
22M.2.AHL.TZ2.5b:
Find the eigenvalues of $M$ .
22M.2.AHL.TZ2.5c:
Find the eigenvectors of $M$ .
22M.1.AHL.TZ1.11a:
Find an eigenvector corresponding to the eigenvalue of $1$ . Give your answer in the form $(\begin{matrix} a \\ b \end{matrix})$ , where $a, b \in ℤ$ .
EXM.3.AHL.TZ0.5g:
State geometrically what transformation the matrix ${\mathbf{R}}$ represents.
EXM.2.AHL.TZ0.23a:
Find the values of $\lambda$ for which the matrix (A − $\lambda$ I) is singular.
22M.2.AHL.TZ1.4a.ii:
Find the eigenvalues for the matrix $(\begin{matrix} 0 & 1 \\ - 6 & - 5 \end{matrix})$ .
22M.2.AHL.TZ2.7b:
Find the eigenvalues of the system of coupled first order equations given in part (a).
EXN.2.AHL.TZ0.5a.i:
the eigenvalues.
21N.2.AHL.TZ0.6a:
Given that $y = \dot{x}$ , show that $\dot{y} = - 1.25 x - 3 y$ .
EXN.2.AHL.TZ0.5d:
Write down a condition on the size of the initial population of $Y$ if it is to avoid its population reducing to zero.
EXN.2.AHL.TZ0.5e.i:
Find the value of $t$ at which $x = 0$ .
21N.2.AHL.TZ0.6c.i:
Find the eigenvalues of matrix $A$ .
22M.2.AHL.TZ1.4a.i:
Use the substitution $y = \frac{d x}{d t}$ to show that this equation can be written as

$(\begin{matrix} \frac{d x}{d t} \\ \frac{d y}{d t} \end{matrix}) = (\begin{matrix} 0 & 1 \\ - 6 & - 5 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})$ .
21N.2.AHL.TZ0.6c.ii:
Find the eigenvectors of matrix $A$ .
EXM.3.AHL.TZ0.5i:
Write down the equations of two lines of symmetry for the curve in $x$ and $y$ in part (b).
EXM.2.AHL.TZ0.23b.iii:
Use the result from part (b) (ii) to explain why A is non-singular.
EXM.3.AHL.TZ0.5h.i:
the curve in $X$ and $Y$ in part (e) (ii), giving a reason.
EXM.3.AHL.TZ0.5d.i:
Find matrix R.
EXM.1.AHL.TZ0.42:
Given the matrix A = $\left( {\begin{array}{*{20}{c}} 3&2 \\ { - 1}&0 \end{array}} \right)$ find the values of the real number $k$ for which ${\text{det}}\left( {A - kI} \right) = 0$ where $I = \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)$
21M.2.AHL.TZ1.5a:
It is sunny today. Find the probability that it will be sunny in three days’ time.
21M.2.AHL.TZ1.5b:
Find the eigenvalues and eigenvectors of $T$ .
21M.2.AHL.TZ1.5c.ii:
Write down the matrix $D$ .
EXM.3.AHL.TZ0.5h.ii:
the curve in $x$ and $y$ in part (b).
EXM.3.AHL.TZ0.5f:
Find the Cartesian equation satisfied by $u$ and $v$ and state the geometric shape that this curve represents.
19M.1.AHL.TZ0.F_3a.i:
Show that the eigenvalues of A are real if ${\left( {a - d} \right)^2} + 4bc \geqslant 0$ .
19M.1.AHL.TZ0.F_3b.i:
Determine the eigenvalues of B.
EXM.3.AHL.TZ0.5b:
Show that the matrix equation $\left( {x\,\,\,y} \right){\mathbf{M}}\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( 6 \right)$ is equivalent to the Cartesian equation $\frac{5}{2}{x^2} + xy + \frac{5}{2}{y^2} = 6$ .
EXM.3.AHL.TZ0.5c.i:
Show that $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \end{array}} \right)$ are unit eigenvectors and that they correspond to different eigenvalues.
EXM.2.AHL.TZ0.23c:
Use the values from part (b) (i) to express A⁴ in the form $p$ A+ $q$ I where $p$ , $q \in \mathbb{Z}$ .
19M.1.AHL.TZ0.F_3b.ii:
Determine the corresponding eigenvectors.
20N.1.AHL.TZ0.F_4b.ii:
Assuming that $A$ is non-singular, use the result in part (b)(i) to show that

$A^{- 1} = \frac{1}{β} (α I - A)$ .
EXM.2.AHL.TZ0.23b.i:
Find the value of $m$ and of $n$ .
EXM.1.AHL.TZ0.47:
Given that A = $\left( {\begin{array}{*{20}{c}} 3&{ - 2} \\ { - 3}&4 \end{array}} \right)$ and I = $\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)$ , find the values of $\lambda$ for which (A – $\lambda$ I) is a singular matrix.
EXM.3.AHL.TZ0.5d.ii:
Show that ${\mathbf{M}} = {{\mathbf{R}}^{ - 1}}\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right){\mathbf{R}}$ .
19M.1.AHL.TZ0.F_3a.ii:
Deduce that the eigenvalues are real if A is symmetric.
20N.1.AHL.TZ0.F_4b.i:
Verify that

$A^{2} - α A + β I =$ 0.
21M.2.AHL.TZ2.7a:
Find the eigenvalues and corresponding eigenvectors of the matrix $(\begin{matrix} - 4 & 0 \\ 3 & - 2 \end{matrix})$ .
EXM.3.AHL.TZ0.5e.ii:
Hence, find the Cartesian equation satisfied by $X$ and $Y$ .
EXM.3.AHL.TZ0.5e.i:
Verify that $\left( {\begin{array}{*{20}{c}} X&Y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x&y \end{array}} \right){{\mathbf{R}}^{ - 1}}$ .
21M.2.AHL.TZ1.5d:
Hence find the long-term percentage of sunny days in Vokram.
EXM.2.AHL.TZ0.23b.ii:
Hence show that I = $\frac{1}{5}$ A (6I – A).
EXM.3.AHL.TZ0.5c.ii:
Hence, show that $M\left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \\ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}} \end{array}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 2&0 \\ 0&3 \end{array}} \right)$ .
20N.1.AHL.TZ0.F_4a:
By considering the determinant of a relevant matrix, show that the eigenvalues, $λ$ , of $A$ satisfy the equation

$λ^{2} - α λ + β = 0$ ,

where $α$ and $β$ are functions of $a, b, c, d$ to be determined.
21M.2.AHL.TZ2.4a:
Write down a transition matrix $T$ indicating the annual population movement between clinics.
21M.2.AHL.TZ2.7e:
Sketch a phase portrait for the general solution to the system of coupled differential equations for $- 6 \leq x \leq 6$ , $- 6 \leq y \leq 6$ .
21M.2.AHL.TZ2.7b:
Hence, write down the general solution of the system.
21M.2.AHL.TZ2.7c:
Determine, with justification, whether the equilibrium point $(0, 0)$ is stable or unstable.
21M.2.AHL.TZ2.4b:
Find a prediction for the ratio of the number of patients Doctor Black will have, compared to Doctor Green, after two years.
21M.2.AHL.TZ1.5c.i:
Write down the matrix $P$ .
21M.2.AHL.TZ2.4c:
Find a matrix $P$ , with integer elements, such that $T = P D P^{- 1}$ , where $D$ is a diagonal matrix.
21M.2.AHL.TZ2.4d:
Hence, show that the long-term transition matrix $T^{\infty}$ is given by $T^{\infty} = (\begin{matrix} \frac{10}{17} & \frac{10}{17} \\ \frac{7}{17} & \frac{7}{17} \end{matrix})$ .
21M.2.AHL.TZ2.4e:
Hence, or otherwise, determine the expected ratio of the number of patients Doctor Black would have compared to Doctor Green in the long term.
21M.2.AHL.TZ2.7d:
(i) at $(4, 0)$ .

(ii) at $(- 4, 0)$ .
EXN.2.AHL.TZ0.5e.ii:
Find the population of $Y$ at this value of $t$ . Give your answer to the nearest $10$ herbivores.
EXN.2.AHL.TZ0.5a.ii:
the eigenvectors.
EXN.2.AHL.TZ0.5b:
Hence write down the general solution of the system of equations.
EXN.2.AHL.TZ0.5c:
Sketch the phase portrait for this system, for $x, y > 0$ .

On your sketch show
- the equation of the line defined by the eigenvector in the first quadrant
- at least two trajectories either side of this line using arrows on those trajectories to represent the change in populations as t increases
21N.2.AHL.TZ0.6b:
Write down the matrix $A$ .
21N.2.AHL.TZ0.6d:
Given that when $t = 0$ the shock absorber is displaced $8 cm$ and its velocity is zero, find an expression for $x$ in terms of $t$ .

Hide related questions

Topic 1—Number and algebra

Question

Markscheme

Examiners report

Syllabus sections

View options