Game plan 5 · Unit 3

Matrices

Which side to multiply on, which route to solve by, and what the determinant does and doesn’t tell you.

Reach for it when

You see a matrix, $\det A$, $A^{-1}$, a system of linear equations, an augmented matrix, “Leslie”, “dominance”, or a population or ranking context.

Part 1

Matrix equations

Game plan 5.1

Solving a matrix equation

  1. 1Gather every term containing $X$ on one side.
  2. 2Factorise, keeping $X$ on whichever side it already sits. If $X$ is on the left in every term, it factors out to the left. If it is on the right, it factors out to the right.
  3. 3If $X$ is on the right of the bracket, multiply both sides on the left by the inverse. If $X$ is on the left, multiply on the right. Matrix multiplication doesn’t commute, so the side is the whole answer.
  4. 4Calculate the inverse by hand for a $2\times2$ matrix on Paper 1, and with technology beyond that. Then multiply out in the correct order.
  5. 5Check the size of your answer before you write it down.

In the Topic Guides

Part 2

Solving systems: choose the route

Game plan 5.2

Choosing the route

  1. 1Tidy the system first, whichever method you use. Put the variables on the left in the same order every time, the constant on the right, and write a zero for any variable that doesn’t appear. An equation given as $3x+3z=y$ must become $3x-y+3z=0$ before you can use it.
  2. 2Choose the route from the wording. “Express as $AX=B$”, “use matrix algebra”, or a plain technology-active solve means the inverse route. “Use a Gaussian technique”, an augmented matrix in the question, a stated row operation, or a question about which case it is means the Gaussian route.
  3. 3Inverse route: write three separate objects. $A$ is the coefficients, $X$ is the column of variables and $B$ is the column of constants, with no partition line and no fourth column. Then $X=A^{-1}B$.
  4. 4Gaussian route: write the augmented matrix with the constants in a final column behind a ruled partition, and reduce towards row echelon form. If the question specifies a row operation, use exactly that one, and apply it across the whole row, including past the partition.
  5. 5Read the bottom row, identify the case (Game plan 5.3), and finish from there.

In the Topic Guides

Part 3

The three cases

Row-reduce, then read the last row. It tells you which of three pictures you have.

Game plan 5.3

Identifying the case

  1. 1If the last row gives a value, there is a unique solution. The three planes meet at a point.
  2. 2If the last row reads $0=\text{non-zero}$, there is no solution. Either two planes are parallel and distinct, or the three form a triangular prism: each pair meets in a line, but the three lines never coincide.
  3. 3If the last row reads $0=0$, there are infinitely many solutions. One variable is free. Set it to a parameter $\lambda$, back-substitute for the others, and the planes share a common line, or coincide.
  4. 4$\det A=0$ only tells you the solution is not unique. It can’t tell “none” from “infinitely many”, so eliminate to find out.

Try it

Read the last row, see the picture

The last row gives a value, so there is one solution. The three planes meet at a single point.

The last row reads $0=0$. One variable is free, and the three planes share a common line.

The last row reads $0=5$. Each pair of planes meets in a line, but the three lines are parallel and never meet.

The last row reads $0=5$. Two of the planes are parallel and distinct, so there is no common point.

The last three cases all have $\det A=0$. Only elimination tells them apart.

When there are infinitely many, write the solution as a line

  1. Set the free variable to a parameter: $z=\lambda$.
  2. Back-substitute to write $x$ and $y$ in terms of $\lambda$.
  3. Collect the constants and the $\lambda$-coefficients into $\boldsymbol r=\boldsymbol a+\lambda\boldsymbol d$, the same object as in Game plan 3.2.

The solution set is a line, so the QCAA can ask for it in vector form. This is where matrices link to vectors. It has been examined exactly this way: find the value that gives infinitely many solutions, then state them as a vector equation of a line.

In the Topic Guides

Part 4

Leslie matrices

Game plan 5.4

Working with a Leslie matrix

  1. 1Build $L$ with the fecundity rates across the first row, the survival rates on the sub-diagonal, and zeros everywhere else.
  2. 2Project with $N_r=L^{r-1}N_1$ on your calculator, where $N_1$ is the initial population vector.
  3. 3For long-run behaviour, compare the total population in consecutive years to find the growth rate. Say whether it is growing, declining or stable, in context.

The shape of a Leslie matrix

$$L=\begin{bmatrix}F_1&F_2&F_3&F_4\\S_1&0&0&0\\0&S_2&0&0\\0&0&S_3&0\end{bmatrix}\qquad N_r=L^{r-1}N_1$$
  • Fecundity rates $F_i$, the births per female, go across the first row.
  • Survival rates $S_i$, the proportion surviving into the next age class, go on the sub-diagonal.
  • Everything else is zero. The matrix is always square, and its size is the number of age classes.
  • $N_1$ is the initial population vector, so $r$ counts from year 1.

Leslie matrices track female populations only, grouped into age classes. The sub-diagonal moves each age group up one class each cycle, so a survival rate never sits on the main diagonal.

In the Topic Guides

Part 5

Dominance matrices

Game plan 5.5

Ranking with a dominance matrix

  1. 1Build the dominance matrix $M$: a $1$ where the row player beat the column player, and a $0$ elsewhere.
  2. 2First-order dominance is the row sums of $M$. Second-order influence is the row sums of $M^2$.
  3. 3Rank by the row sums of $M+M^2$. State the ranking in context, and note any ties.

A dominance matrix, ranked

$$M=\begin{array}{c}\\A\\B\\C\\D\end{array}\!\!\begin{array}{c}\begin{array}{cccc}A&B&C&D\end{array}\\\begin{bmatrix}0&1&0&1\\0&0&1&0\\1&0&0&1\\0&1&0&0\end{bmatrix}\end{array}$$

A $1$ means the row player beat the column player.

$$\begin{array}{lcccc}&A&B&C&D\\\hline\text{row sums of }M&2&1&2&1\\\text{row sums of }M^2&2&2&3&1\\\text{row sums of }M+M^2&4&3&5&2\end{array}$$

$A$ and $C$ both won two games, so first-order dominance can’t separate them. The second-order sums break the tie, because $C$ beat $A$. The ranking is $C\gt A\gt B\gt D$. If a tie survives, keep going with $M^3$.

In the Topic Guides

Formulas

Formula panel

Each formula is labelled with how to treat it in the exam.

Matrix laws: commutative addition, identities, inverses and the distributive laws
On the sheet
The $2\times2$ determinant and inverse
On the sheet
$A$ is singular $\iff\det A=0$ $\iff$ no unique solution
Memorise
The three row operations: swap two rows, scale a row by a non-zero constant, add a multiple of another row
Memorise
Building a Leslie matrix (fecundity on the first row, survival on the sub-diagonal) and $N_r=L^{r-1}N_1$
Memorise
Dominance ranking by the row sums of $M+M^2$
Memorise
What each of the three cases looks like
Memorise

Watch for

$\det A=0$ doesn’t give you the answer. You still have to eliminate to separate “no solution” from “infinitely many”.

QCAA language

What the question says, and what it means

Question typeWhat you’ll seeWhat it means
Matrix equation$AX=B$, $XA=B$, $AX+BX=C$Game plan 5.1. Multiply on the correct side
SystemEquations in three unknowns, an augmented matrixGame plan 5.2. Use Gaussian elimination if there may not be a single solution
Cases“For what value of $k$ does the system have no solution”Game plan 5.3. Row-reduce and read the last row
LeslieAge groups, fecundity, survivalGame plan 5.4
DominanceTeams, wins, rankingGame plan 5.5

Back to the External Game plan or the Topic Guides.

Wording quoted from past QCAA Specialist Mathematics external assessments is © State of Queensland (QCAA), licensed under CC BY 4.0. Tangent Tuition is not affiliated with the QCAA.

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