Matrix arithmetic, determinants and inverses: VCE General Mathematics Unit 4 Matrices

Element $c_{21}$ of a product is row 2 of the left matrix multiplied by column 1 of the right matrix, element by element, then added.

Row 2 of $A$ is $\begin{bmatrix} 1 & 6 \end{bmatrix}$ and column 1 of $B$ is $\begin{bmatrix} 3 \\ 5 \end{bmatrix}$, so $c_{21} = 1 \times 3 + 6 \times 5 = 33$: option C. $84\%$ of students answered correctly. Option A is $c_{11}$ (row 1 of $A$), and option D pairs the entries in the wrong order.

Source: VCAA 2025 General Mathematics Examination 1, Question 26, and the 2025 examination report.

An inverse exists only when the determinant is not zero.

$\det(E) = mn - (-9)(4) = mn + 36$.

The inverse fails to exist when $mn + 36 = 0$, that is when $mn = -36$. Test each option: A gives $36$, B gives $36$, C gives $3 \times (-12) = -36$, D gives $36$. Only option C makes the determinant $-36 + 36 = 0$. $76\%$ answered correctly; the most common wrong answer was D ($13\%$), from losing track of the double negative in $-(-9)(4)$.

Source: VCAA 2025 General Mathematics Examination 1, Question 27, and the 2025 examination report.

Write every order first: $A$ is $2 \times 2$, $B$ is $3 \times 1$, $C$ is $3 \times 2$, $D$ is $1 \times 3$.

• A: $A + B$ needs equal orders ($2 \times 2$ and $3 \times 1$), so it is undefined.
• B: $C^T$ is $2 \times 3$. $(2 \times 3)(3 \times 1)$ gives $2 \times 1$, then $(2 \times 1)(1 \times 3)$ gives $2 \times 3$. Defined.
• C: $BD$ is $(3 \times 1)(1 \times 3) = 3 \times 3$, which cannot be added to the $3 \times 2$ matrix $C$.
• D: $DA$ is $(1 \times 3)(2 \times 2)$; the inner numbers $3$ and $2$ differ, so it is undefined.

The answer is B ($70\%$ correct).

Source: VCAA 2025 General Mathematics Examination 1, Question 31, and the 2025 examination report.

• (1) is false: a square matrix with determinant $0$, such as $\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$, has no inverse.
• (2) is true: for the permutation matrix $P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, $P^{-1} = P^T = P$.
• (3) is true: this is the condition for an inverse.
• (4) is true: $I^{-1} = I$, because $I \times I = I$.

Three statements are true: option D. Only $39\%$ answered correctly; $42\%$ chose C, usually by rejecting statement 2 or statement 4.

Source: VCAA 2023 General Mathematics Examination 1, Question 30, and the 2023 examination report.

(a) The child price is in row 3, column 1: $n_{31}$. (1 mark; $48\%$. Writing a capital $N_{31}$ or $n_{13}$ lost the mark.)

(b) $\begin{bmatrix} 0 & 2 & 2 \end{bmatrix} \times N = 2 \times 15 + 2 \times 8 = 46$, the cost of two adult and two child tickets bought separately. Subtract the family ticket, $\begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \times N = 36$: $46 - 36 = 10$. The missing row is $\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$. (1 mark; $44\%$.)

(c) Each price must be multiplied by its own quantity and kept in its own row, so $K$ is the diagonal matrix

Source: VCAA 2023 General Mathematics Examination 2, Question 8, and the 2023 examination report.

(a) Divide every element by 20, which is scalar multiplication by $\frac{1}{20}$:

(b) $R \times \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \end{bmatrix}^T$ adds the five revenues to give the total, $4 712 700. Increasing a total by $25\%$ means multiplying by $1.25$, so $t = 1.25$ (target $5 890 875). (1 mark; $47\%$. Many students wrote $0.25$, which would give only the increase, not the new total.)

Source: VCAA 2023 General Mathematics Examination 2, Question 9 (parts a and b), and the 2023 examination report.

(a) $3 \times 5$ (three rows, five columns). (1 mark; $96\%$.)

(b) A row of ones on the left adds down each column: $W = \begin{bmatrix} 31 & 34 & 32 & 31 & 24 \end{bmatrix}$ is the total number of children enrolled at the centre on each day of the week. (1 mark; $57\%$. Many answers described row totals, which would need $E$ multiplied on the right by a column of ones.)

(c) A diagonal matrix on the left multiplies each row of $C$ by its own factor: $1.25$, $1.5$ and $0.9$.

Source: VCAA 2025 General Mathematics Examination 2, Question 11, and the 2025 examination report.

Write the orders side by side and compare the inner numbers.

• $AB$: $(3 \times 2)(2 \times 4)$, inner $2 = 2$: defined, order $3 \times 4$.
• $BA$: $(2 \times 4)(3 \times 2)$, inner $4 \ne 3$: not defined.
• $BC$: $(2 \times 4)(4 \times 3)$, inner $4 = 4$: defined, order $2 \times 3$.
• $CB$: $(4 \times 3)(2 \times 4)$, inner $3 \ne 2$: not defined.
• $ABC$: $AB$ is $3 \times 4$, then $(3 \times 4)(4 \times 3)$ is defined, order $3 \times 3$.

(2 marks for all five; 1 mark for three or four.) Notice that $AB$ exists but $BA$ does not: order matters.

Multiply every element by the scalar, then subtract matching elements.

(1 mark for the scalar multiples, 1 mark for the answer.) The sign slip to watch is $0 - (-3) = 3$.

(a) Row by column:

(b)

(c) $AB \ne BA$: matrix multiplication is not commutative, even when both products exist and have the same order. Always multiply in the order the question gives. (1 mark)

(a) $\det(M) = k \times k - 8 \times 2 = k^2 - 16$. No inverse when $k^2 - 16 = 0$, so $k = 4$ or $k = -4$. (1 mark) A CAS solve of $\det(M) = 0$ gives both values; giving only $k = 4$ loses the mark.

(b) With $k = 5$, $\det(M) = 25 - 16 = 9$. Swap the leading diagonal, negate the other diagonal, divide by the determinant:

Check: $\begin{bmatrix} 5 & 8 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 5 & -8 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 25 - 16 & -40 + 40 \\ 10 - 10 & -16 + 25 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}$, and $\frac{1}{9}$ of this is $I$. (1 mark)

(a) $(2 \times 3)(3 \times 1)$ gives $2 \times 1$:

(b) $\begin{bmatrix} 1 & 1 \end{bmatrix} S = \begin{bmatrix} 21 & 22 & 21 \end{bmatrix}$: the total number of road, mountain and kids bikes sold across both stores. (1 mark)

(c) $D = \begin{bmatrix} 1.1 & 0 & 0 \\ 0 & 1.05 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, so $DP = \begin{bmatrix} 1320 \\ 997.5 \\ 280 \end{bmatrix}$ and

(a) $\begin{bmatrix} 6 & 4 \\ 3 & 8 \end{bmatrix} \begin{bmatrix} r \\ t \end{bmatrix} = \begin{bmatrix} 51 \\ 48 \end{bmatrix}$. (1 mark)

(b) $\det(A) = 6 \times 8 - 4 \times 3 = 36$, so $A^{-1} = \frac{1}{36}\begin{bmatrix} 8 & -4 \\ -3 & 6 \end{bmatrix}$ and

(c) The coefficient matrix would be $\begin{bmatrix} 6 & 4 \\ 3 & 2 \end{bmatrix}$, with determinant $6 \times 2 - 4 \times 3 = 0$. It has no inverse, so there is no unique solution: bouquet 3 is exactly half of bouquet 1 and gives no new information. (1 mark)

(a) $M^2 = MM = \begin{bmatrix} 0(0) + 1(1) & 0(1) + 1(1) \\ 1(0) + 1(1) & 1(1) + 1(1) \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$. (1 mark)

(b) $M^5 = \begin{bmatrix} 3 & 5 \\ 5 & 8 \end{bmatrix}$. (1 mark)

(c) Squaring each element of $M$ gives $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$ (unchanged, since $0^2 = 0$ and $1^2 = 1$), but $M^2 = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$. A matrix power means repeated matrix multiplication, not element-by-element powers.

(d) $G^2 = GG$ is $(2 \times 3)(2 \times 3)$; the inner numbers $3$ and $2$ differ, so the product is not defined. Only square matrices can be raised to a power. (1 mark for (c) and (d) together.)