Modelling with Matrices
Translating a real-world scenario into a system of linear equations, writing it as $M\mathbf{x}=\mathbf{b}$, and solving with the inverse matrix.
Solving a system of equations with a matrix
Many real situations — mixing ingredients in fixed ratios, dividing money between accounts, tracking how a population is split into groups — can be described by several linear equations in several unknowns that all hold at once. Collecting the unknowns into a column vector $\mathbf{x}$, the coefficients into a matrix $M$, and the right-hand sides into a column vector $\mathbf{b}$, the whole system becomes a single matrix equation $M\mathbf{x}=\mathbf{b}$. Provided $\det M\neq0$, multiplying both sides on the left by $M^{-1}$ solves it in one step: $\mathbf{x}=M^{-1}\mathbf{b}$.
The system is $M\mathbf{x}=\mathbf{b}$ where $\mathbf{b}=\begin{pmatrix}2200\\2500\\4200\end{pmatrix}$, so $\mathbf{x}=M^{-1}\mathbf{b}$.
So $A=500$, $B=600$, $C=300$ litres. Since all three values are positive, the blend is physically realisable — a negative value here would signal the requested order is impossible with this recipe.
Modelling percentage change across several categories
A very common exam context asks you to find three unknown initial quantities $x$, $y$, $z$ (amounts of money, populations, items) from three pieces of information: a total they sum to, $x+y+z=T$; a relationship between two of them, such as "$x$ is $k$ more than $y$", giving $x-y=k$; and the overall change once each quantity grows or shrinks by a given percentage, $r_1x+r_2y+r_3z=C$, where each rate $r_i$ is written as a decimal (positive for an increase, negative for a decrease) and $C$ is the total change. Writing these three equations as $M\mathbf{x}=\mathbf{b}$ and solving with $\mathbf{x}=M^{-1}\mathbf{b}$ (usually with a calculator that inverts a $3\times3$ matrix) recovers the original quantities.
Let $P$, $Q$, $R$ be the amounts (in £) originally invested. Total invested: $P+Q+R=4500$. $P$ is £300 more than $Q$: $P-Q=300$. Change in value, using each rate as a decimal: $0.02P+0.04Q-0.01R=102$.
Solving this (with a calculator that inverts the $3\times3$ matrix, or by substituting $P=Q+300$ into the other two equations and eliminating) gives $P=2025$, $Q=1725$, $R=750$.
Check: $2025+1725+750=4500$ ✓, and the change in value is $0.02(2025)+0.04(1725)-0.01(750)=40.5+69-7.5=102$ ✓.
Let $x$, $y$, $z$ be the initial numbers of red, roe and fallow deer. $x+y+z=1800$, $x-y=150$, and $0.03x-0.02y+0.05z=51$.
Substituting $x=y+150$ into the other two equations and eliminating $z$ gives $x=550$, $y=400$, $z=850$.
Check: $550+400+850=1800$ ✓, and the change is $0.03(550)-0.02(400)+0.05(850)=16.5-8+42.5=51$ ✓.
Whenever a matrix model produces the values of physical quantities (money, populations, litres), check the sign and size of the answer make sense in context. A negative amount of money, a negative population, or a value that doesn't satisfy the original total usually signals an arithmetic slip — but if the arithmetic checks out, a negative or non-integer answer can also expose a genuine limitation of the model itself, which is worth stating explicitly if asked. It's also good practice, as in the examples above, to substitute your answer back into all three original equations as a check.
Exam-style questions
Question 1 6 marks
A pet-food company blends three grains, $x$, $y$, $z$ kg, to meet nutritional targets, giving the system $3x+2y+3z=800$, $2x+y+2z=700$, $4x+3y+3z=700$. Given $N=\begin{pmatrix}3&2&3\\2&1&2\\4&3&3\end{pmatrix}$ has inverse $N^{-1}=\begin{pmatrix}-3&3&1\\2&-3&0\\2&-1&-1\end{pmatrix}$, find $x$, $y$, $z$.
$\begin{pmatrix}x\\y\\z\end{pmatrix}=N^{-1}\begin{pmatrix}800\\700\\700\end{pmatrix}=\begin{pmatrix}-3(800)+3(700)+1(700)\\2(800)-3(700)+0(700)\\2(800)-1(700)-1(700)\end{pmatrix}$ (M1, A1)
$=\begin{pmatrix}-2400+2100+700\\1600-2100+0\\1600-700-700\end{pmatrix}=\begin{pmatrix}400\\-500\\200\end{pmatrix}$ (M1, A1)
So $x=400$, $y=-500$, $z=200$. Since $y$ is negative, this blend is not physically realisable as stated — a modelling limitation, not an arithmetic one. (A1, B1)
Question 2 7 marks
Amelia invested £5000 across three savings accounts, $D$, $E$ and $F$. She invested £250 more in $D$ than in $E$. After one year, account $D$ had increased in value by $3\%$, account $E$ had increased in value by $1.5\%$, and account $F$ had decreased in value by $2\%$. The total value of Amelia's savings had increased by £57. Form and solve a matrix equation to find how much money was invested by Amelia in each account.
$D+E+F=5000$, $D-E=250$, $0.03D+0.015E-0.02F=57$ (M1, A1, A1)
$\begin{pmatrix}1&1&1\\1&-1&0\\0.03&0.015&-0.02\end{pmatrix}\begin{pmatrix}D\\E\\F\end{pmatrix}=\begin{pmatrix}5000\\250\\57\end{pmatrix}$ (M1)
Substituting $D=E+250$ and eliminating gives $D=1950$, $E=1700$, $F=1350$. (M1, A1)
Check: $1950+1700+1350=5000$ and $0.03(1950)+0.015(1700)-0.02(1350)=58.5+25.5-27=57$ ✓. (A1)
Question 3 7 marks
A butterfly conservation trust is monitoring three species in a meadow: common blues, small whites and orange tips. Initially there are $2400$ butterflies in total, and there are $200$ more common blues than small whites. After one breeding season, the number of common blues had increased by $5\%$, the number of small whites had decreased by $3\%$, and the number of orange tips had increased by $2\%$. Overall, the population had increased by $41$. Form and solve a matrix equation to find the number of each species in the initial population.
Let $b$, $w$, $o$ be the initial numbers of common blues, small whites and orange tips. $b+w+o=2400$, $b-w=200$, $0.05b-0.03w+0.02o=41$ (M1, A1, A1)
$\begin{pmatrix}1&1&1\\1&-1&0\\0.05&-0.03&0.02\end{pmatrix}\begin{pmatrix}b\\w\\o\end{pmatrix}=\begin{pmatrix}2400\\200\\41\end{pmatrix}$ (M1)
Substituting $b=w+200$ and eliminating gives $b=850$, $w=650$, $o=900$. (M1, A1)
Check: $850+650+900=2400$ and $0.05(850)-0.03(650)+0.02(900)=42.5-19.5+18=41$ ✓. (A1)
Question 4 8 marks
A charity trustee monitors donations received through three fundraising channels: online, events and mail. In one year, total donations across the three channels were £18 000, and the amount raised online was £1200 more than the amount raised through events. The following year, online donations increased by $4\%$, events donations increased by $8\%$, and mail donations decreased by $6\%$. The total raised across all three channels increased by £372.
(a) Form and solve a matrix equation to find the amount raised by each channel in the first year. 7 marks
(b) State one assumption made by this model. 1 mark
(a) Let $O$, $E$, $M$ be the amounts (£) raised online, at events, and by mail. $O+E+M=18000$, $O-E=1200$, $0.04O+0.08E-0.06M=372$ (M1, A1, A1)
$\begin{pmatrix}1&1&1\\1&-1&0\\0.04&0.08&-0.06\end{pmatrix}\begin{pmatrix}O\\E\\M\end{pmatrix}=\begin{pmatrix}18000\\1200\\372\end{pmatrix}$ (M1)
Substituting $O=E+1200$ and eliminating gives $O=6750$, $E=5550$, $M=5700$. (M1, A1)
Check: $6750+5550+5700=18000$ and $0.04(6750)+0.08(5550)-0.06(5700)=270+444-342=372$ ✓. (A1)
(b) Any valid limitation, e.g. the model assumes each channel's percentage rate of change stays constant across the whole channel, when in reality different individual donations within a channel could change by different amounts. (B1)