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Matrices

Practice exam-style IB Math AI questions for Matrices, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

Let

A=(2−13041),B=(1230−21)A=\begin{pmatrix}2&-1&3\\0&4&1\end{pmatrix},\qquad B=\begin{pmatrix}1&2\\3&0\\-2&1\end{pmatrix}
A

the orders of ABAB and BABA.

[1]
B

ABAB.

[2]
C

BABA and why AA and BB do not commute.

[2]
Question 2
HL • Paper 1
Easy
Calculator Permitted
HL • Paper 1
Easy
Calculator Permitted

A shop sells three models of headphones. The numbers sold during two weeks are represented by

Q=(402530353228)Q=\begin{pmatrix}40&25&30\\35&32&28\end{pmatrix}

The selling prices and unit costs, in dollars, are

p=(12189),c=(7115)p=\begin{pmatrix}12\\18\\9\end{pmatrix},\qquad c=\begin{pmatrix}7\\11\\5\end{pmatrix}
A

the revenue in each week.

[2]
B

the profit in each week.

[2]
C

which week had the greater profit and by how much.

[1]
Question 3
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

A pair of numbers is encoded using y=Aq\mathbf{y}=A\mathbf{q}, where

A=(21−13)A=\begin{pmatrix}2&1\\-1&3\end{pmatrix}
A

A−1A^{-1}.

[2]
B

An encoded column is y=(1716)\mathbf{y}=\begin{pmatrix}17\\16\end{pmatrix}. Decode this column.

[2]
C

Encode the column (84)\begin{pmatrix}8\\4\end{pmatrix}.

[1]
Question 4
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The matrices AA, BB, CC and XX satisfy AX+B=CAX+B=C, where

A=(3121),B=(1−204),C=(9561)A=\begin{pmatrix}3&1\\2&1\end{pmatrix},\quad B=\begin{pmatrix}1&-2\\0&4\end{pmatrix},\quad C=\begin{pmatrix}9&5\\6&1\end{pmatrix}
A

A−1A^{-1}.

[2]
B

for XX.

[3]

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Question 5
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A manufacturer combines xx kg of material A, yy kg of material B and zz kg of material C. The mixture has a total mass of 100100 kg. The materials provide 22, 33 and 55 nutrient units per kg respectively, and the mixture must provide 380380 nutrient units. Their costs are $4, $6 and $3 per kg respectively, and the total cost is $410.

A

Write the information as a matrix equation of the form Ax=bA\mathbf{x}=\mathbf{b}.

[2]
B

Use an inverse matrix to xx, yy and zz.

[3]
C

the percentage by mass of material C in the mixture.

[1]
Question 6
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Consider the matrix

A=(k23k−1)A=\begin{pmatrix}k&2\\3&k-1\end{pmatrix}

where k∈Rk\in\mathbb{R}.

A

the values of kk for which AA is singular.

[2]
B

A−1A^{-1} when k=4k=4.

[2]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Matrices AA, BB and XX satisfy XA=BXA=B, where

A=(1235),B=(41−23)A=\begin{pmatrix}1&2\\3&5\end{pmatrix},\qquad B=\begin{pmatrix}4&1\\-2&3\end{pmatrix}
A

A−1A^{-1}.

[1]
B

for XX.

[2]
C

why using X=A−1BX=A^{-1}B is not valid.

[1]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

At a theatre, adult tickets cost $18, student tickets cost $12 and child tickets cost $7. A total of 250250 tickets were sold for $3470. The number of adult tickets sold was 4040 more than the number of student tickets sold.

A

Represent the information as a matrix equation, using aa, ss and cc for the numbers of adult, student and child tickets.

[2]
B

Use an inverse matrix to the number of each type of ticket sold.

[2]
C

the percentage of the tickets sold that were child tickets.

[1]

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Question 9
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let

M=(4213)M=\begin{pmatrix}4&2\\1&3\end{pmatrix}
A

the characteristic polynomial of MM.

[2]
B

the eigenvalues of MM.

[1]
C

an eigenvector corresponding to each eigenvalue.

[2]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A two-region model is represented by

sn+1=Msn,M=(0.720.180.280.82)\mathbf{s}_{n+1}=M\mathbf{s}_n,\qquad M=\begin{pmatrix}0.72&0.18\\0.28&0.82\end{pmatrix}

The total population of the two regions is 50005000.

A

Given that s0=(30002000)\mathbf{s}_0=\begin{pmatrix}3000\\2000\end{pmatrix}, s1\mathbf{s}_1.

[2]
B

the invariant with total population 50005000.

[2]
C

the eigenvalue associated with the invariant and your answer.

[1]
Question 11
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The matrix

M=(k213)M=\begin{pmatrix}k&2\\1&3\end{pmatrix}

has eigenvalue 55.

A

the value of kk.

[2]
B

an eigenvector corresponding to the eigenvalue 55.

[1]
C

the other eigenvalue of MM.

[1]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The matrix

M=(3122)M=\begin{pmatrix}3&1\\2&2\end{pmatrix}

has eigenvalues 44 and 11. Corresponding eigenvectors are (11)\begin{pmatrix}1\\1\end{pmatrix} and (1−2)\begin{pmatrix}1\\-2\end{pmatrix} respectively.

A

matrices PP and DD such that M=PDP−1M=PDP^{-1}.

[2]
B

P−1P^{-1}.

[1]
C

M5M^5.

[3]

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Question 13
HL • Paper 2
Medium
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HL • Paper 2
Medium
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Each year, 85%85\% of the residents of town A remain in A and 15%15\% move to town B. Of the residents of town B, 80%80\% remain in B and 20%20\% move to town A. Initially, town A has 60006000 residents and town B has 40004000 residents. Assume the total population remains constant. Let sn=(AnBn)\mathbf{s}_n=\begin{pmatrix}A_n\\B_n\end{pmatrix}, where AnA_n and BnB_n are the populations of towns A and B, respectively, after nn years.

A

the transition matrix MM such that sn+1=Msn\mathbf{s}_{n+1}=M\mathbf{s}_n.

[1]
B

the population of each town after 88 years.

[2]
C

the long-term population of each town.

[3]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The populations, in hundreds, of two interacting species are modelled by

sn+1=Msn,M=(1.10.20.10.8)\mathbf{s}_{n+1}=M\mathbf{s}_n,\qquad M=\begin{pmatrix}1.1&0.2\\0.1&0.8\end{pmatrix}

Initially, s0=(400100)\mathbf{s}_0=\begin{pmatrix}400\\100\end{pmatrix}.

A

s1\mathbf{s}_1.

[1]
B

s6\mathbf{s}_6.

[2]
C

the dominant eigenvalue of MM.

[1]
D

the long-term ratio of the first species population to the second species population, and the dominant eigenvalue.

[2]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let

M=(2112)M=\begin{pmatrix}2&1\\1&2\end{pmatrix}

The eigenvalues of MM are 33 and 11.

A

corresponding eigenvectors and write M=PDP−1M=PDP^{-1}.

[2]
B

an expression for MnM^n, where nn is a positive integer.

[2]
C

the least positive integer nn for which the element in row 11, column 22 of MnM^n is greater than 10001000.

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A simplified population model is defined by

sn+1=Msn,M=(1110),s0=(10)\mathbf{s}_{n+1}=M\mathbf{s}_n,\qquad M=\begin{pmatrix}1&1\\1&0\end{pmatrix},\qquad \mathbf{s}_0=\begin{pmatrix}1\\0\end{pmatrix}
A

the eigenvalues of MM in exact form.

[2]
B

an eigenvector corresponding to each eigenvalue.

[2]
C

Using diagonalization, s10\mathbf{s}_{10}.

[2]

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Question 17
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
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A bakery produces rolls, cakes and pies. The numbers produced on Monday and Tuesday are represented by

Q=(12080601507090)Q=\begin{pmatrix}120&80&60\\150&70&90\end{pmatrix}

The flour and sugar required per item, in kilograms, are represented by

R=(0.200.050.350.080.150.03)R=\begin{pmatrix}0.20&0.05\\0.35&0.08\\0.15&0.03\end{pmatrix}

where the first column represents flour and the second column represents sugar.

A
I.

the order of the matrix QRQR.

[1]
II.

QRQR and its elements.

[2]
B

Flour costs 1.40 dollars per kilogram and sugar costs 2.20 dollars per kilogram. the ingredient cost for each day.

[2]
C

On Wednesday, 400400 items are produced, using 105105 kg of flour and 2424 kg of sugar. Let xx, yy and zz be the numbers of rolls, cakes and pies produced. Use an inverse matrix to xx, yy and zz.

[3]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A coding system encodes a block q\mathbf q of three numbers using y=Aq\mathbf y=A\mathbf q, where

A=(120011201)A=\begin{pmatrix}1&2&0\\0&1&1\\2&0&1\end{pmatrix}
A
I.

det⁡(A)\det(A).

[1]
II.

why every encoded block can be decoded uniquely.

[2]
B
I.

A−1A^{-1}.

[2]
II.

Decode y=(131314)\mathbf y=\begin{pmatrix}13\\13\\14\end{pmatrix}.

[1]
C

A second coding matrix is

B=(120011131)B=\begin{pmatrix}1&2&0\\0&1&1\\1&3&1\end{pmatrix}

this matrix is unsuitable for unique decoding.

[2]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A two-component production column q\mathbf q is processed first by a packaging matrix PP and then by a transport matrix TT, where

P=(211302),T=(10.520.210.5),q=(4030)P=\begin{pmatrix}2&1\\1&3\\0&2\end{pmatrix}, \qquad T=\begin{pmatrix}1&0.5&2\\0.2&1&0.5\end{pmatrix}, \qquad \mathbf q=\begin{pmatrix}40\\30\end{pmatrix}
A
I.

PqP\mathbf q.

[2]
II.

T(Pq)T(P\mathbf q).

[2]
B

TPTP and that (TP)q=T(Pq)(TP)\mathbf q=T(P\mathbf q).

[2]
C

The final output is (295182)\begin{pmatrix}295\\182\end{pmatrix}. Use an inverse matrix to recover the original production column.

[2]
Question 20
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

Two digital filters act on a signal vector. Their matrices are

S=(1201),T=(2011)S=\begin{pmatrix}1&2\\0&1\end{pmatrix},\qquad T=\begin{pmatrix}2&0\\1&1\end{pmatrix}

and the initial signal is v=(34)\mathbf{v}=\begin{pmatrix}3\\4\end{pmatrix}.

A
I.

the output when filter SS is applied first and filter TT second.

[2]
II.

the output when the order of the filters is reversed.

[2]
B

STST and TSTS. why the order in which the filters are applied is significant, despite matrix multiplication being associative.

[4]

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Question 21
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Three sensors measure environmental quantities xx, yy and zz. Their readings satisfy

2x+y−z=7,x+3y+2z=11,3x−y+z=3.\begin{aligned} 2x+y-z&=7,\\ x+3y+2z&=11,\\ 3x-y+z&=3. \end{aligned}
A

(a)

I.

Write the system as a matrix equation Ax=bA\mathbf x=\mathbf b.

[2]
II.

Use an inverse matrix to the system.

[3]
B

A recalibration changes the right-hand-side vector by Δb=(01−1)\Delta\mathbf b=\begin{pmatrix}0\\1\\-1\end{pmatrix}.

I.

the corresponding change Δx\Delta\mathbf x in the solution.

[2]
II.

the recalibrated values of xx, yy and zz. If you did not obtain Δx\Delta\mathbf x, use (−0.20.40)\begin{pmatrix}-0.2\\0.4\\0\end{pmatrix}.

[2]
Question 22
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The matrices AA, BB, CC and XX satisfy AXB=CAXB=C, where

A=(1201),B=(2011),C=(15573)A=\begin{pmatrix}1&2\\0&1\end{pmatrix},\quad B=\begin{pmatrix}2&0\\1&1\end{pmatrix},\quad C=\begin{pmatrix}15&5\\7&3\end{pmatrix}
A
I.

A−1A^{-1} and B−1B^{-1}.

[2]
II.

X=A−1CB−1X=A^{-1}CB^{-1}.

[2]
B

XX.

[2]
C

A student claims that X=B−1CA−1X=B^{-1}CA^{-1}. why this rearrangement is invalid.

[2]
Question 23
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Customers subscribe to either a basic service or a premium service. Each month, 80%80\% of basic subscribers remain on basic and 75%75\% of premium subscribers remain on premium. Initially there are 540540 basic and 360360 premium subscribers. The total number of subscribers remains constant.

A
I.

the transition matrix MM and initial s0\mathbf s_0 such that sn+1=Msn\mathbf s_{n+1}=M\mathbf s_n.

[2]
II.

the after one month.

[1]
B
I.

the eigenvalues of MM and an eigenvector corresponding to each.

[3]
II.

an expression for the number BnB_n of basic subscribers after nn months.

[2]
C

the long-term numbers of basic and premium subscribers.

[2]
Question 24
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A directed network has adjacency matrix

A=(0110001110010100)A=\begin{pmatrix} 0&1&1&0\\ 0&0&1&1\\ 1&0&0&1\\ 0&1&0&0 \end{pmatrix}

The element in row ii, column jj is 11 when there is a direct route from location ii to location jj, and is 00 otherwise.

1

2

3

4

1

0

1

1

0

2

0

0

1

1

3

1

0

0

1

4

0

1

0

0

A
I.

A2A^2.

[2]
II.

the element in row 11, column 44 of A2A^2.

[2]
B
I.

the total number of routes of exactly three stages starting at location 22.

[2]
II.

why A3A^3 and A2AA^2A must be equal.

[1]
C

why an entry of 22 in a power of AA does not mean that there are two direct routes between the corresponding locations.

[2]

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Question 25
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Four unknown calibration constants are represented by

x=(wxyz)\mathbf x=\begin{pmatrix}w\\x\\y\\z\end{pmatrix}

They satisfy Ax=bA\mathbf x=\mathbf b, where

A=(1111210304123021),eqquadb=(17283123)A=\begin{pmatrix} 1&1&1&1\\ 2&1&0&3\\ 0&4&1&2\\ 3&0&2&1 \end{pmatrix}, eqquad \mathbf b=\begin{pmatrix}17\\28\\31\\23\end{pmatrix}
A
I.

det⁡(A)\det(A) and why the system has a unique solution.

[2]
II.

Use an inverse matrix to the calibration constants.

[2]
B

After a change in conditions, the constant column becomes

b′=(19353128)\mathbf b'=\begin{pmatrix}19\\35\\31\\28\end{pmatrix}
I.

the new calibration column x′\mathbf x'.

[2]
II.

Δx=x′−x\Delta\mathbf x=\mathbf x'-\mathbf x.

[1]
C

that AΔx=b′−bA\Delta\mathbf x=\mathbf b'-\mathbf b and what this demonstrates.

[2]
Question 26
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A workshop produces standard, touring and cargo bicycles. Let xx, yy and zz denote the daily numbers produced. The production constraints are represented by

x+y+z=60,2x+3y+5z=180,4x+2y+6z=200.\begin{aligned} x+y+z&=60,\\ 2x+3y+5z&=180,\\ 4x+2y+6z&=200. \end{aligned}
A
I.

Write the constraints in the form Ax=bA\mathbf{x}=\mathbf{b}.

[2]
II.

Use an inverse matrix to the daily production of each type of bicycle.

[2]
B

A revised plan leaves the total number of bicycles unchanged but increases the right-hand sides of the second and third constraints by 66 and 44 respectively. the revised production plan and the change.

[4]
Question 27
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Three-number message blocks are encoded using y=Aq\mathbf{y}=A\mathbf{q}, where

A=(110011101)A=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}
A
I.

det⁡(A)\det(A) and whether the coding can be reversed uniquely.

[2]
II.

Decode the received block y=(151922)\mathbf{y}=\begin{pmatrix}15\\19\\22\end{pmatrix}.

[2]
B

A second coding system uses

B=(110011121)B=\begin{pmatrix}1&1&0\\0&1&1\\1&2&1\end{pmatrix}

this coding system can lose information by finding two distinct plaintext blocks that produce the same encoded block.

[4]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A two-sensor calibration model is A(k)x=bA(k)\mathbf{x}=\mathbf{b}, where

A(k)=(k14k−2)A(k)=\begin{pmatrix}k&1\\4&k-2\end{pmatrix}
A
I.

the values of kk for which the calibration matrix is singular.

[2]
II.

For k=3k=3 and b=(1115)\mathbf{b}=\begin{pmatrix}11\\15\end{pmatrix}, x\mathbf{x}.

[2]
B

why calibration becomes unreliable when kk is very close to either value found in part (a)(i), even if A(k)−1A(k)^{-1} technically exists.

[4]

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Question 29
HL • Paper 3
Hard
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HL • Paper 3
Hard
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An economy has agricultural and manufacturing sectors. The internal consumption matrix is

C=(0.20.10.30.2)C=\begin{pmatrix}0.2&0.1\\0.3&0.2\end{pmatrix}

and external demand is d=(430360)\mathbf{d}=\begin{pmatrix}430\\360\end{pmatrix}. Total output x\mathbf{x} satisfies x=Cx+d\mathbf{x}=C\mathbf{x}+\mathbf{d}.

A
I.

Rearrange the model into the form Ax=dA\mathbf{x}=\mathbf{d} and AA.

[2]
II.

Use an inverse matrix to the total output required from each sector.

[2]
B

the internal demand CxC\mathbf{x} and why total output must exceed external demand in both sectors.

[4]
Question 30
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A directed transport network has three stations. Its adjacency matrix is

A=(011101010)A=\begin{pmatrix}0&1&1\\1&0&1\\0&1&0\end{pmatrix}

where the element in row ii, column jj is the number of direct routes from station ii to station jj.

A
I.

A2A^2.

[2]
II.

the element in row 22, column 22 of A2A^2.

[2]
B

A traveller may use at most two stages, and a journey of zero stages is allowed only when the starting and finishing stations are the same. a matrix whose entries count all such journeys. the number of permitted journeys from station 11 to station 33.

[4]
Question 31
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A design point is transformed using

R=(0−110),S=(2001)R=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad S=\begin{pmatrix}2&0\\0&1\end{pmatrix}

Here RR is a rotation and SS is a horizontal stretch.

A
I.

the image of (13)\begin{pmatrix}1\\3\end{pmatrix} when RR is applied first and SS second.

[2]
II.

the image when the stretch is applied first and the rotation second.

[2]
B

A region has area 1212 square units. its area after either composite transformation and why the two composites have the same area scale factor but different effects on points.

[4]
Question 32
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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The amplitudes of two coupled vibration modes after nn cycles are represented by sn=Mns0\mathbf{s}_n=M^n\mathbf{s}_0, where

M=(5222),s0=(50)M=\begin{pmatrix}5&2\\2&2\end{pmatrix},\qquad \mathbf{s}_0=\begin{pmatrix}5\\0\end{pmatrix}
A
I.

the eigenvalues of MM.

[2]
II.

a corresponding eigenvector for each eigenvalue.

[2]
B

Express sn\mathbf{s}_n in terms of nn. the limiting ratio of its first component to its second component.

[4]

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Question 33
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Consider the parameter-dependent system

A(k)(xy)=(34),A(k)=(k12k−1)A(k)\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}, \qquad A(k)=\begin{pmatrix}k&1\\2&k-1\end{pmatrix}

where k∈Rk\in\mathbb R.

A
I.

the values of kk for which A(k)A(k) is singular.

[2]
II.

A(k)−1A(k)^{-1} for all other values of kk.

[2]
B
I.

why the system has no solution when k=2k=2.

[1]
II.

the system when k=2.01k=2.01, giving the coordinates to three significant figures.

[2]
C

Discuss why solutions close to k=2k=2 may be highly sensitive to small measurement errors.

[2]
Question 34
HL • Paper 2
Hard
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HL • Paper 2
Hard
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Let

M=(5222)M=\begin{pmatrix}5&2\\2&2\end{pmatrix}
A
I.

the eigenvalues of MM.

[2]
II.

an eigenvector corresponding to each eigenvalue.

[2]
B
I.

Write matrices PP and DD such that M=PDP−1M=PDP^{-1}.

[2]
II.

an expression for MnM^n, where n∈Z+n\in\mathbb Z^+.

[2]
C

the least positive integer nn for which the element in row 11, column 22 of MnM^n exceeds 10510^5.

[2]
Question 35
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The populations of two interacting organisms are modelled by

sn+1=Msn,M=(1.20.30.20.7),s0=(500200)\mathbf s_{n+1}=M\mathbf s_n, \qquad M=\begin{pmatrix}1.2&0.3\\0.2&0.7\end{pmatrix}, \qquad \mathbf s_0=\begin{pmatrix}500\\200\end{pmatrix}

Population values are measured in thousands.

A
I.

the eigenvalues of MM.

[2]
II.

a corresponding eigenvector for each eigenvalue.

[2]
B
I.

Express s0\mathbf s_0 as a linear combination of the eigenvectors.

[2]
II.

an expression for sn\mathbf s_n. If you did not obtain the coefficients, use a=12007a=\frac{1200}{7} and b=−1007b=-\frac{100}{7}.

[3]
C

the long-term ratio of the first population to the second population and the dominant eigenvalue.

[2]
Question 36
HL • Paper 2
Hard
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HL • Paper 2
Hard
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For k∈Rk\in\mathbb R, let

Mk=(k21k+1)M_k=\begin{pmatrix}k&2\\1&k+1\end{pmatrix}
A
I.

the eigenvalues of MkM_k are k+2k+2 and k−1k-1.

[2]
II.

an eigenvector corresponding to each eigenvalue.

[2]
B
I.

Write matrices PP and DkD_k such that Mk=PDkP−1M_k=PD_kP^{-1}.

[1]
II.

an expression for MknM_k^n.

[2]
C

For k=0.4k=0.4, the long-term ratio of the two components of Mkn(41)M_k^n\begin{pmatrix}4\\1\end{pmatrix}.

[2]

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Question 37
HL • Paper 2
Hard
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HL • Paper 2
Hard
Calculator Permitted

A simplified ranking algorithm updates the scores of two webpages using

rn+1=Mrn,M=(0.60.90.40.1),r0=(0.50.5)\mathbf r_{n+1}=M\mathbf r_n, \qquad M=\begin{pmatrix}0.6&0.9\\0.4&0.1\end{pmatrix}, \qquad \mathbf r_0=\begin{pmatrix}0.5\\0.5\end{pmatrix}
A
AI.

the eigenvalues of MM.

[2]
AII.

an eigenvector corresponding to each eigenvalue.

[2]
B
BI.

r0=113(94)−526(1−1)\mathbf r_0=\frac1{13}\begin{pmatrix}9\\4\end{pmatrix}-\frac5{26}\begin{pmatrix}1\\-1\end{pmatrix}
[2]
BII.

an expression for rn\mathbf r_n.

[2]
C

the least value of nn for which each score differs from its limiting value by less than 0.0010.001.

[2]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A damped displacement model is given by

sn+1=Msn,eqquadM=(01−0.180.9),eqquads0=(107)\mathbf s_{n+1}=M\mathbf s_n, eqquad M=\begin{pmatrix}0&1\\-0.18&0.9\end{pmatrix}, eqquad \mathbf s_0=\begin{pmatrix}10\\7\end{pmatrix}

The first component of sn\mathbf s_n is the displacement, in millimetres, at step nn.

A
I.

the eigenvalues of MM.

[2]
II.

an eigenvector corresponding to each eigenvalue.

[2]
B
I.

Express s0\mathbf s_0 as a linear combination of the eigenvectors.

[2]
II.

an expression for the displacement dnd_n.

[2]
C

the first step at which the magnitude of the displacement is less than 0.10.1 mm, and why the displacement approaches zero.

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A reversible numerical filter is represented by

F=(120−1)F=\begin{pmatrix}1&2\\0&-1\end{pmatrix}

A column v\mathbf v is updated repeatedly according to vn+1=Fvn\mathbf v_{n+1}=F\mathbf v_n.

A
I.

F2=IF^2=I.

[2]
II.

F−1F^{-1}.

[1]
B
I.

the eigenvalues of FF and a corresponding eigenvector for each.

[2]
II.

Diagonalize FF and obtain a formula for FnF^n.

[2]
C

Given v0=(43)\mathbf v_0=\begin{pmatrix}4\\3\end{pmatrix}, v2025\mathbf v_{2025} and the periodic behaviour of the sequence.

[2]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Matrices AA, BB and XX satisfy AXB=CAXB=C, where

A=(1102),B=(2011),C=(113142)A=\begin{pmatrix}1&1\\0&2\end{pmatrix},\quad B=\begin{pmatrix}2&0\\1&1\end{pmatrix},\quad C=\begin{pmatrix}11&3\\14&2\end{pmatrix}
A
I.

A−1A^{-1} and B−1B^{-1}.

[2]
II.

for XX.

[2]
B

A student claims that X=B−1A−1CX=B^{-1}A^{-1}C. the error and the correct solution by substitution.

[4]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The numbers of books borrowed from two sections of a library are modelled by

sn+1=Msn,M=(0.80.30.20.7),s0=(800200)\mathbf{s}_{n+1}=M\mathbf{s}_n,\qquad M=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix},\qquad \mathbf{s}_0=\begin{pmatrix}800\\200\end{pmatrix}
A
I.

11 and 0.50.5 are eigenvalues of MM, and a corresponding eigenvector for each.

[2]
II.

sn=(600400)+200(0.5)n(1−1)\mathbf{s}_n=\begin{pmatrix}600\\400\end{pmatrix}+200(0.5)^n\begin{pmatrix}1\\-1\end{pmatrix}
[2]
B

the least value of nn for which each component is within one book of its long-term value. the long-term .

[4]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The numbers of active users of two competing online platforms are modelled by

sn+1=Msn,M=(1.20.40.10.9),s0=(9010)\mathbf{s}_{n+1}=M\mathbf{s}_n,\qquad M=\begin{pmatrix}1.2&0.4\\0.1&0.9\end{pmatrix},\qquad \mathbf{s}_0=\begin{pmatrix}90\\10\end{pmatrix}

The components are measured in thousands.

A
I.

the eigenvalues of MM.

[2]
II.

an expression for sn\mathbf{s}_n.

[2]
B

the first value of nn for which the model predicts more than one million active users in total. one reason why predictions far beyond this point should be treated cautiously.

[4]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A conic is represented by

(xy)M(xy)=12,M=(5−2−22)\begin{pmatrix}x&y\end{pmatrix} M \begin{pmatrix}x\\y\end{pmatrix}=12, \qquad M=\begin{pmatrix}5&-2\\-2&2\end{pmatrix}
A
I.

the eigenvalues of MM.

[2]
II.

unit eigenvectors and an orthogonal matrix PP that diagonalizes MM.

[2]
B

Using coordinates (uv)=PT(xy)\begin{pmatrix}u\\v\end{pmatrix}=P^T\begin{pmatrix}x\\y\end{pmatrix}, the equation of the conic in uu and vv. its shape, semi-axis lengths and area.

[4]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A parameter-dependent update matrix is

M(a)=(a0.20.2a)M(a)=\begin{pmatrix}a&0.2\\0.2&a\end{pmatrix}

A is updated by sn+1=M(a)sn\mathbf{s}_{n+1}=M(a)\mathbf{s}_n.

A
I.

the eigenvalues and corresponding eigenvector directions of M(a)M(a).

[2]
II.

Write a diagonalization of M(a)M(a).

[2]
B

the set of values of aa for which every initial approaches the zero vector as n→∞n\to\infty. your answer using the eigenvalues.

[4]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
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Let

M(p)=(p11p)M(p)=\begin{pmatrix}p&1\\1&p\end{pmatrix}

where p∈Rp\in\mathbb{R}.

A
I.

the eigenvalues and corresponding eigenvectors of M(p)M(p).

[2]
II.

a formula for M(p)nM(p)^n, where nn is a positive integer.

[2]
B

For p=0.6p=0.6, an initial is s0=(73)\mathbf{s}_0=\begin{pmatrix}7\\3\end{pmatrix}. The evolves according to sn+1=M(0.6)sn\mathbf{s}_{n+1}=M(0.6)\mathbf{s}_n. Define dn=(sn)1−(sn)2d_n=(\mathbf{s}_n)_1-(\mathbf{s}_n)_2. an expression for sn\mathbf{s}_n and the alternating behaviour of dnd_n.

[4]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The values, in thousands of dollars, of two linked investment funds are modelled by

sn+1=Msn,M=(1.050.020.011.04),s0=(320140)\mathbf{s}_{n+1}=M\mathbf{s}_n,\qquad M=\begin{pmatrix}1.05&0.02\\0.01&1.04\end{pmatrix},\qquad \mathbf{s}_0=\begin{pmatrix}320\\140\end{pmatrix}
A
I.

the eigenvalues and a corresponding eigenvector for each.

[2]
II.

an expression for sn\mathbf{s}_n.

[2]
B

the first value of nn for which the combined modelled value exceeds one million dollars. the long-term ratio of the value of the first fund to the second.

[4]
Question 47
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let

M=(3212)M=\begin{pmatrix}3&2\\1&2\end{pmatrix}
A
I.

the characteristic polynomial and eigenvalues of MM.

[2]
II.

directly that M2−5M+4I=0M^2-5M+4I=0.

[2]
B
I.

a recurrence relating MnM^n, Mn−1M^{n-1} and Mn−2M^{n-2} for n≥2n\ge2.

[1]
II.

Using diagonalization, an explicit expression for MnM^n.

[3]
C

that the expression from part (b)(ii) satisfies the recurrence from part (b)(i).

[2]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A two-page ranking model uses the transition matrix

M(d)=(0.50.5+0.5d0.50.5−0.5d)M(d)=\begin{pmatrix}0.5&0.5+0.5d\\0.5&0.5-0.5d\end{pmatrix}

where 0<d<10<d<1. A ranking vector has components summing to 11 and is updated by rn+1=M(d)rn\mathbf{r}_{n+1}=M(d)\mathbf{r}_n.

A
I.

11 is an eigenvalue and the invariant ranking vector r∗\mathbf{r}_* in terms of dd.

[2]
II.

the second eigenvalue and a corresponding eigenvector.

[2]
B

For d=0.8d=0.8 and r0=(10)\mathbf{r}_0=\begin{pmatrix}1\\0\end{pmatrix}, the least value of nn for which both components of rn\mathbf{r}_n are within 0.0010.001 of their invariant values. why the rankings alternate around equilibrium.

[4]

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Financial Mathematics

Number Skills