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QLDSpecialist MathematicsSyllabus dot point

Topic 2: Vectors and matrices

Use matrices beyond order two, including the determinant and inverse of a three by three matrix, to represent and solve systems of linear equations, and interpret unique, infinite and no-solution cases geometrically

A focused answer to the QCE Specialist Mathematics Unit 3 dot point on systems of linear equations. Covers representing systems in matrix form, the determinant and inverse of a three by three matrix, solving by the matrix inverse, and the geometric meaning of unique, infinitely many and no solutions, with a verified worked example and the determinant-zero trap.

Generated by Claude Opus 4.76 min answer

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What this dot point is asking

QCAA wants you to handle matrices larger than 2Γ—22\times 2, especially to represent a system of three linear equations in three unknowns as Ax=bA\mathbf{x} = \mathbf{b}, solve it using the inverse x=Aβˆ’1b\mathbf{x} = A^{-1}\mathbf{b}, and interpret what happens when the determinant is zero. The geometric reading, three planes meeting in a point, in a line or not at all, ties the algebra to the vectors topic.

The answer

Matrix form of a system

A system such as

a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3 \begin{aligned} a_1 x + b_1 y + c_1 z &= d_1\\ a_2 x + b_2 y + c_2 z &= d_2\\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}

is written compactly as Ax=bA\mathbf{x} = \mathbf{b}, where AA is the 3Γ—33\times 3 matrix of coefficients, x=(x,y,z)T\mathbf{x} = (x, y, z)^{\mathsf T} and b=(d1,d2,d3)T\mathbf{b} = (d_1, d_2, d_3)^{\mathsf T}.

Determinant of a three by three matrix

Expanding along the first row,

det⁑A=a1∣b2c2b3c3βˆ£βˆ’b1∣a2c2a3c3∣+c1∣a2b2a3b3∣.\det A = a_1\begin{vmatrix} b_2 & c_2\\ b_3 & c_3\end{vmatrix} - b_1\begin{vmatrix} a_2 & c_2\\ a_3 & c_3\end{vmatrix} + c_1\begin{vmatrix} a_2 & b_2\\ a_3 & b_3\end{vmatrix}.

The alternating signs (+,βˆ’,++, -, +) are essential. The determinant decides invertibility: AA has an inverse if and only if det⁑Aβ‰ 0\det A \neq 0.

Inverse and solving

When det⁑Aβ‰ 0\det A \neq 0 the system has a unique solution

x=Aβˆ’1b.\mathbf{x} = A^{-1}\mathbf{b}.

In practice the inverse of a 3Γ—33\times 3 matrix is found from the adjugate divided by the determinant, or computed with technology, which QCAA permits in the relevant assessment. The point is to recognise the structure and read off the unique solution.

Geometric interpretation

Each equation ax+by+cz=dax + by + cz = d is a plane. Solving the system locates the common points of three planes.

  • Unique solution (det⁑Aβ‰ 0\det A \neq 0): the three planes meet at a single point.
  • No solution (det⁑A=0\det A = 0, system inconsistent): the planes have no common point, for example two are parallel, or they form a triangular prism arrangement.
  • Infinitely many solutions (det⁑A=0\det A = 0, system consistent): the planes share a common line, or all coincide, giving a one-parameter family of solutions.

Why the determinant matters

A zero determinant signals that the rows (or the normal vectors of the planes) are linearly dependent, so the inverse method fails. You then test consistency directly by elimination to decide between no solution and infinitely many.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA4 marksA system of linear equations is given by x - 2y - 2z = -6, -3x - y + z = 2, 2x + 3y - 5z = 10. a) Express the system of equations as a matrix equation of the form AX = B, where A is a 3 x 3 matrix and both X and B are 3 x 1 column vectors. b) Use matrix algebra to express X in terms of A and B. c) Use your result from part b) to determine the solution of the system of equations. d) Verify your result from part c) using one of the given linear equations.
Show worked answer β†’

a) [1 mark] A = [[1, -2, -2], [-3, -1, 1], [2, 3, -5]], X = [x, y, z]^T (column), B = [-6, 2, 10]^T (column), so AX = B.

b) [1 mark] Since A is invertible (det A is non-zero), pre-multiply both sides by A inverse: X = A^(-1) B.

c) [1 mark] Evaluating A^(-1) B (by calculator on this technology-active paper) gives X = [-2, 3, -1]^T, that is x = -2, y = 3, z = -1.

d) [1 mark] Substitute into the first equation x - 2y - 2z = -6: (-2) - 2(3) - 2(-1) = -2 - 6 + 2 = -6, which matches the right-hand side, so the solution is verified.