Specialist Mathematics · Past QCAA questions All simple familiar questions
2024 · Paper 2 · Simple familiar

Question 12

[4 marks] Technology-active
Topic guidePremiumUnit 3 Topic 5.1 · Matrix algebra and systems of equations
Game plansPremium5.2 Choosing the route
The question

A system of linear equations is given by

$$\begin{aligned}x-2y-2z&=-6\\-3x-y+z&=2\\2x+3y-5z&=10\end{aligned}$$

a)

Express the system of equations as a matrix equation of the form $AX=B$, where $A$ is a $3\times3$ matrix and both $X$ and $B$ are $3\times1$ column vectors.

[1 mark] b)

Use matrix algebra to express $X$ in terms of $A$ and $B$.

[1 mark] c)

Use your result from Question 12b) to determine the solution of the system of equations.

[1 mark] d)

Verify your result from Question 12c) using one of the given linear equations.

[1 mark]
Watch the situation first

Three equations become one matrix equation

Watch the coefficients, unknowns and constants sort themselves into three matrices, then see the inverse do the solving.

A system of three equations in three unknowns Coefficients go into A, unknowns into X, constants into B Multiply both sides on the left by A−1 One point satisfies all three equations x − 2y − 2z= −6 −3x − y + z= 2 2x + 3y − 5z= 10 1−2−2 −3−11 23−5 A xyz X = −6210 B X = A−1B = −23−1

The calculator does the inverse and the multiplying. Your job is to set up the matrices in the right order.

Now for the mathematics

Test a point against all three equations

Choose values for $x$, $y$ and $z$. Multiplying by $A$ gives the left-hand side of each equation, so you can compare the result with $B$.

QCAA marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Prefer to watch? Watch me work through it on YouTube (opens in a new tab)

Step 1 · part a)1 mark

Put the coefficients of $x$, $y$ and $z$ into $A$, one row per equation, in the order the equations are given. The unknowns go into $X$ and the constants into $B$.

$$\begin{bmatrix}1&-2&-2\\-3&-1&1\\2&3&-5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-6\\2\\10\end{bmatrix}$$

Watch the signs in the second equation. The coefficient of $-y$ is $-1$ and the coefficient of $+z$ is $1$. Writing $A$ and $B$ out separately is also accepted.

Marker: correctly expresses the equations of the form $AX=B$

Step 2 · part b)1 mark

Multiply both sides on the left by $A^{-1}$. Since $A^{-1}A=I$ and $IX=X$,

$$A^{-1}AX=A^{-1}B$$

$X=A^{-1}B$

The order matters. $BA^{-1}$ is not defined, because $B$ is $3\times1$ and $A^{-1}$ is $3\times3$. Writing the matrices out in full, as $X=\begin{bmatrix}1&-2&-2\\-3&-1&1\\2&3&-5\end{bmatrix}^{-1}\begin{bmatrix}-6\\2\\10\end{bmatrix}$, is also accepted. So is the inverse with its entries rounded to two decimal places.

$$X=\begin{bmatrix}0.05&-0.38&-0.10\\-0.31&-0.02&0.12\\-0.17&-0.17&-0.17\end{bmatrix}\begin{bmatrix}-6\\2\\10\end{bmatrix}$$

Marker: correctly expresses $X$ in terms of $A$ and $B$

Step 3 · part c)1 mark

Enter $A$ and $B$ in your calculator and work out $A^{-1}B$. You don’t need to write down the inverse, but this is what the calculator finds.

$$X=\frac{1}{42}\begin{bmatrix}2&-16&-4\\-13&-1&5\\-7&-7&-7\end{bmatrix}\begin{bmatrix}-6\\2\\10\end{bmatrix}=\begin{bmatrix}-2\\3\\-1\end{bmatrix}$$

$x=-2,\ y=3,\ z=-1$

Here $\det A=42$. It is not zero, so $A^{-1}$ exists and the system has exactly one solution. If your calculator says the matrix is singular, check what you entered for $A$. The point $(-2,\ 3,\ -1)$ is also accepted.

Marker: determines solution to equations

Step 4 · part d)1 mark

Substitute your solution into one of the original equations. I would use the first one.

$$x-2y-2z=(-2)-2(3)-2(-1)=-6$$

This equals the right-hand side, so the result is verified.

Either of the other equations works too: $-3(-2)-3+(-1)=2$ or $2(-2)+3(3)-5(-1)=10$. Finish with a statement that the result is verified. If an earlier error means your check fails, say the result is not verified.

Marker: verifies result by substituting result from 12c) into one of the given linear equations

Putting it all together

Each phrase of the question gave you something

“a matrix equation of the form $AX=B$”
$A$ holds the coefficients, $X$ the unknowns and $B$ the constants, all in matching order.
“Use matrix algebra to express $X$”
Multiply both sides on the left by $A^{-1}$ and show that step.
“Use your result from Question 12b)”
Calculate $A^{-1}B$ on your calculator and write the matrix product you calculated.
“Verify … using one of the given linear equations”
Substitute into one original equation and finish with a statement that the result is verified.
What makes this simple familiar

This is the standard inverse matrix method, split into its four steps, and your calculator does the arithmetic. Each part asks for one line of working.

If I had to guess where you might lose marks, it would be part b). Matrices can’t be divided, so “divide both sides by $A$” gets nothing, and $X=BA^{-1}$ has the product in the wrong order. Write the line $A^{-1}AX=A^{-1}B$ so the marker can see where $A^{-1}$ went.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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