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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.

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to handle matrices larger than 2×22\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=A1b\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 3×33\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,

detA=a1b2c2b3c3b1a2c2a3c3+c1a2b2a3b3.\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 (+,,++, -, +) are essential. The determinant decides invertibility: AA has an inverse if and only if detA0\det A \neq 0.

Inverse and solving

When detA0\det A \neq 0 the system has a unique solution

x=A1b.\mathbf{x} = A^{-1}\mathbf{b}.

In practice the inverse of a 3×33\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 (detA0\det A \neq 0): the three planes meet at a single point.
  • No solution (detA=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 (detA=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.

Solving by elimination (row reduction)

When the inverse is not used, Gaussian elimination is the systematic alternative. Use one equation to eliminate a variable from the others, then repeat, reducing the system to an upper-triangular form that can be back-substituted. If elimination produces a contradictory row such as 0=50 = 5, the system is inconsistent and has no solution. If it produces an all-zero row such as 0=00 = 0, one equation was dependent on the others, leaving a free variable and infinitely many solutions. This procedure works whether or not the determinant is zero, and it is the tool that distinguishes the two degenerate cases.

Parametrising an infinite solution set

When there are infinitely many solutions, express the answer with a parameter. Solve for two variables in terms of the third, set the free variable equal to tt, and write the solution as x=x0+td\mathbf{x} = \mathbf{x}_0 + t\mathbf{d}. Geometrically this is exactly the line of intersection of the three planes: x0\mathbf{x}_0 is one point on the line and d\mathbf{d} is its direction. Recognising the parametric solution as a line ties this dot point back to vector equations of lines, and QCAA rewards stating the geometric interpretation alongside the algebra.

Reading the geometry of three planes

It helps to picture the configurations. Three planes meeting at a single point give a unique solution. Three planes sharing a common line give a one-parameter family. Three planes with no common point arise either when two are parallel and distinct, or when each pair meets in a line but the three lines are parallel (a triangular-prism arrangement). Coincident planes (all the same plane) give a two-parameter family. Matching the algebraic outcome to one of these pictures is the kind of justification an extended-response item asks for.

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.

QCAA 20244 marksPaper 2 (complex familiar). A system is given by x2y2z=6x - 2y - 2z = -6, 3xy+z=2-3x - y + z = 2 and 2x+3y5z=102x + 3y - 5z = 10. (a) Express the system as a matrix equation Ax=bA\mathbf{x} = \mathbf{b}. (b) Express x\mathbf{x} in terms of AA and b\mathbf{b}. (c) Determine the solution. (d) Verify using one of the equations.
Show worked answer →

(a) [1 mark] A=(122311235)A = \begin{pmatrix} 1 & -2 & -2 \\ -3 & -1 & 1 \\ 2 & 3 & -5 \end{pmatrix}, x=(x,y,z)T\mathbf{x} = (x, y, z)^{\mathsf T}, b=(6,2,10)T\mathbf{b} = (-6, 2, 10)^{\mathsf T}, so Ax=bA\mathbf{x} = \mathbf{b}.

(b) [1 mark] Since detA0\det A \neq 0, pre-multiply by A1A^{-1}: x=A1b\mathbf{x} = A^{-1}\mathbf{b}.

(c) [1 mark] Evaluating A1bA^{-1}\mathbf{b} (by calculator on this technology-active paper) gives x=(2,3,1)T\mathbf{x} = (-2, 3, -1)^{\mathsf T}, that is x=2x = -2, y=3y = 3, z=1z = -1.

(d) [1 mark] In equation one: (2)2(3)2(1)=26+2=6(-2) - 2(3) - 2(-1) = -2 - 6 + 2 = -6, matching the right-hand side, so the solution is verified.

QCAA 20224 marksPaper 1 (technique). The matrix A=(211112301)A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 0 & 1 \end{pmatrix} is the coefficient matrix of a system Ax=bA\mathbf{x} = \mathbf{b}. (a) Determine detA\det A by cofactor expansion along the first row. (b) State, with reason, whether the system has a unique solution for every b\mathbf{b}.
Show worked answer →

(a) detA=2[(1)(1)(2)(0)]1[(1)(1)(2)(3)]+(1)[(1)(0)(1)(3)]\det A = 2\big[(-1)(1) - (2)(0)\big] - 1\big[(1)(1) - (2)(3)\big] + (-1)\big[(1)(0) - (-1)(3)\big] =2(1)1(16)1(0+3)=2+53=0.= 2(-1) - 1(1 - 6) - 1(0 + 3) = -2 + 5 - 3 = 0.

(b) Since detA=0\det A = 0, AA is not invertible, so the system does not have a unique solution for every b\mathbf{b}; depending on b\mathbf{b} it has either no solution or infinitely many.

Markers reward the cofactor expansion with the correct alternating signs and the conclusion that a zero determinant rules out a unique solution.

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