Why is "Claiming $\det A = 0$ automatically means 'no solution'" a common mistake in Consistency of 3×3 Linear Systems and Geometric Interpretation?

Why it happens: Confusing 'no inverse' with 'no solution'; not realising a line of solutions can exist. How to avoid it: A = 0 only rules out a UNIQUE solution. Row-reduce: 0=0 row → infinitely many; 0=k row → none. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Giving the algebra but no geometric description (or vice versa)" a common mistake in Consistency of 3×3 Linear Systems and Geometric Interpretation?

Why it happens: Answering only the part of the question that feels mathematical. How to avoid it: Always pair the conclusion with the geometry: single point / common line (sheaf) / triangular prism / parallel planes. Source: 9231 Paper 2 mark schemes

Why is "Forgetting to apply row operations to the constant column" a common mistake in Consistency of 3×3 Linear Systems and Geometric Interpretation?

Why it happens: Focusing on zeroing the coefficients and neglecting the right-hand side of [Ab]. How to avoid it: Treat the augmented matrix as one block — every row operation acts on the constant too. The bottom-row constant is what decides consistency. Source: 9231 Paper 2 Examiner Reports

Why is "Stating 'infinitely many solutions' without writing the line" a common mistake in Consistency of 3×3 Linear Systems and Geometric Interpretation?

Why it happens: Stopping at the 0=0 row without parametrising. How to avoid it: Set the free variable z = t and express x, y in terms of t, then give the vector form r = a + td. Source: 9231 Paper 2 Examiner Reports

Why is "Describing the no-solution case vaguely instead of naming the prism" a common mistake in Consistency of 3×3 Linear Systems and Geometric Interpretation?

Why it happens: Not knowing the specific mark-scheme term for three planes with no common point. How to avoid it: Use the keyword 'triangular prism' (or 'two/three parallel planes' if applicable) — vague answers like 'they don't all meet' lose the description mark. Source: 9231 Paper 2 mark schemes

Why is "Sign slip in the $3\times3$ cofactor expansion gives a wrong $\det A$" a common mistake in Consistency of 3×3 Linear Systems and Geometric Interpretation?

Why it happens: Forgetting the middle term's minus sign (-b()) in the +,-,+ pattern. How to avoid it: Write the signs +\,-\,+ above the top row before expanding; double-check brackets with negative entries. Source: 9231 Paper 2 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Consistency of 3×3 Linear Systems and Geometric Interpretation

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Detailed Study Notes

Detailed notes on Matrices (paper 2) for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Consistency of 3×3 Linear Systems and Geometric Interpretation — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (2026-2027 syllabus)

Three linear equations in $x, y, z$ are three planes. They meet at a single point, along a common line, or nowhere. Learn to classify the outcome from $\det A$ and the reduced augmented matrix — and to picture the geometry.

At a glance

Each equation is a plane in 3-D space; the solution set is where the planes intersect.
$\det A \neq 0$ : unique solution — three planes meet at a single point.
$\det A = 0$ : either NO solution (inconsistent) or INFINITELY many — row-reduce to tell which.
Common line (sheaf/pencil): planes share a whole line → infinitely many solutions.
Triangular prism / parallel: no common point → no solution (inconsistent).

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.2d — Determine whether a $3\times3$ linear system has a unique solution, no solution, or infinitely many solutions, using $\det A$ and row reduction.
2.2e — Interpret the three cases geometrically as the intersection of three planes (point, line, or no common point).
2.2f — Identify the row pattern of the reduced augmented matrix that signals each case ( $0=0$ vs $0=\text{nonzero}$ ).

Three planes, three possible outcomes

Unique point, common line, or no common point — decided by det A and the reduced rows.

The key idea. Each equation $ax + by + cz = d$ is the equation of a plane. Three equations are three planes, and the solution of the system is exactly the set of points lying on all three at once.

There are only three possibilities:

1. Unique solution — the planes meet at a single point. This happens precisely when $\det A \neq 0$ . The coefficient matrix is non-singular, $A^{-1}$ exists, and $\mathbf{x} = A^{-1}\mathbf{b}$ is the one and only solution.

2. Infinitely many solutions — the planes share a common line. Here $\det A = 0$ and the system is consistent. Reducing the augmented matrix produces a row of the form $0 = 0$ (a redundant equation). The solution is a line, described with a parameter $t$ .

3. No solution — the planes have no common point. Here $\det A = 0$ and the system is inconsistent. Reduction produces an impossible row such as $0 = 5$ — a contradiction. The planes might form a triangular prism, or be parallel.

Left: three planes cut at one point (unique). Middle: they share a line (infinitely many). Right: a triangular prism — each pair meets in a line, but no point is on all three (no solution).

Cambridge tip. $\det A = 0$ on its own does not mean "no solution". It only tells you the solution is not unique — you must row-reduce to decide between the line case and the no-solution case.

Unique point $\Leftrightarrow \det A \neq 0$ .
$\det A = 0$ splits into a common line OR no common point.
Reduce the augmented matrix to tell the two $\det A=0$ cases apart.

See the full worked example for consistency of 3×3 linear systems and geometric interpretation →

Reading the case off the reduced augmented matrix

A 0 = 0 bottom row means a line; a 0 = nonzero row means no solution.

When $\det A = 0$ , row-reduce $[A\mid\mathbf{b}]$ to upper-triangular form and look at the bottom row — its pattern tells you everything.

Signature A — infinitely many solutions (common line). The bottom row collapses to all zeros, including the constant: $\left(\begin{array}{ccc|c} \ast & \ast & \ast & \ast \\ 0 & \ast & \ast & \ast \\ 0 & 0 & 0 & 0 \end{array}\right)$ The row says $0 = 0$ — true but useless (one equation was redundant). You have two genuine equations in three unknowns, so introduce a parameter $z = t$ and back-substitute to get $x$ and $y$ in terms of $t$ . The solution is a line.

Signature B — no solution (inconsistent). The coefficient part of the bottom row is zero but the constant is not: $\left(\begin{array}{ccc|c} \ast & \ast & \ast & \ast \\ 0 & \ast & \ast & \ast \\ 0 & 0 & 0 & k \end{array}\right),\quad k \neq 0$ This reads $0 = k$ with $k \neq 0$ — a contradiction. No point can satisfy it, so the system has no solution.

The parameter form of a line. Once you reach Signature A, write (for example) $z = t$ , then $y = (\ldots) + (\ldots)t$ and $x = (\ldots) + (\ldots)t$ . In vector form: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{a} + t\,\mathbf{d}$ where $\mathbf{a}$ is a point on the line and $\mathbf{d}$ is its direction.

Cambridge tip. State your conclusion in words and give the geometry: " $0=0$ row ⇒ the system is consistent with infinitely many solutions; the three planes meet in a common line." Both the algebra and the interpretation carry marks.

$0 = 0$ bottom row → infinitely many solutions (a line).
$0 = k$ ( $k\neq0$ ) bottom row → no solution (inconsistent).
For the line, set $z = t$ and express $x, y$ in terms of $t$ .

See the full worked example for consistency of 3×3 linear systems and geometric interpretation →

The geometry of three planes in full

Point, sheaf (line), prism, parallel/coincident — match each to its algebra.

Examiners love asking you to describe the geometric configuration. Here is the complete catalogue.

Unique point ( $\det A \neq 0$ ). The three planes are like three walls of a room corner — they cross at exactly one point.

Sheaf / pencil — common line ( $\det A = 0$ , consistent, $0=0$ row). All three planes pass through one common line, like pages of an open book sharing the spine. Infinitely many solutions, parametrised by $t$ .

Triangular prism — no common point ( $\det A = 0$ , inconsistent, $0=k$ row). Each pair of planes meets in a line, but the three lines are parallel and distinct — they enclose a triangular tube. No point lies on all three, so no solution.

Two (or three) planes parallel ( $\det A = 0$ , inconsistent). At least two planes are parallel but not identical, so they never meet. No solution. (If two are coincident — the same plane — and the third cuts them, you get a line again.)

Parallel planes share no point — the algebraic signal is a $0 = k$ contradiction row after reduction.

The decision flow:

Compute $\det A$ . If $\neq 0$ → unique point. Done.
If $\det A = 0$ , reduce $[A\mid\mathbf{b}]$ .
$0=0$ row → consistent → common line (infinitely many).
$0=k$ row ( $k\neq0$ ) → inconsistent → no solution (prism or parallel).

Cambridge tip. When asked "describe the geometrical relationship", the safe full answer names the case: meet at a single point / meet in a common line (sheaf) / form a triangular prism (no common point) / include parallel planes. Vague answers like "they don't meet" lose the description mark.

Unique point ↔ three planes cross once ( $\det A \neq 0$ ).
Common line (sheaf) ↔ $0=0$ row, infinitely many solutions.
Triangular prism or parallel planes ↔ $0=k$ row, no solution.

See the full worked example for consistency of 3×3 linear systems and geometric interpretation →

Quick recap

Each linear equation is a plane; the solution set is the planes' intersection.
$\det A \neq 0$ : unique solution — a single point.
$\det A = 0$ : not unique — must row-reduce to decide which case.
$0=0$ bottom row → infinitely many solutions → a common line (sheaf).
$0=k$ ( $k\neq0$ ) bottom row → no solution → triangular prism or parallel planes.
Always give BOTH the algebraic conclusion and the geometric description.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\det A \neq 0 \Rightarrow$ unique solution (single point).
$\det A = 0 \Rightarrow$ not unique — reduce to decide line vs none.
$0=0$ row $\Rightarrow$ consistent, infinitely many (line/sheaf).
$0=k,\,k\neq0$ row $\Rightarrow$ inconsistent, no solution (prism/parallel).
Line solution: set $z=t$ , write $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ .

How it’s examined

Topic 2.2 consistency questions appear in Paper 2 (Further Pure Mathematics 2), often 8-12 marks. A typical structure: 'For what value of $\lambda$ does the system have (i) a unique solution, (ii) no solution, (iii) infinitely many solutions? Describe the geometry in each case.' Marks are lost by treating $\det A = 0$ as automatically 'no solution', by not row-reducing to separate the two $\det A = 0$ cases, and by giving the algebra without the geometric description (or vice versa). The phrase 'triangular prism' is a specific mark-scheme keyword for the inconsistent $\det A = 0$ case.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2); 9231/21 & 9231/23 Further Pure Mathematics 2 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Consistency of 3×3 Linear Systems and Geometric Interpretation

Step-by-step solutions to past-paper-style questions on consistency of 3×3 linear systems and geometric interpretation, written exactly the way a tutor would explain them at the board.

Question type:

1Classifying by the determinant
Getting startedIdentify & classify• determinant, easy
Question
For the system with $A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & -1 & 3 \\ 1 & 1 & 2 \end{pmatrix}$ , decide whether a unique solution is guaranteed.
Step-by-step solution
1. Step 1
  Evaluate $\det A$ by expanding along row 1.
  $\det A = 1((-1)(2)-(3)(1)) - 2((2)(2)-(3)(1)) + 1((2)(1)-(-1)(1))$
2. Step 2
  Simplify each bracket.
  $\det A = 1(-5) - 2(1) + 1(3) = -5 - 2 + 3 = -4$
Answer
$\det A = -4 \neq 0$ , so the system has a unique solution — the three planes meet at a single point.
Examiner tip
A non-zero determinant alone settles it: unique solution. No row reduction needed. Always state the geometric meaning too.
2Spotting det A = 0
Getting startedIdentify & classify• determinant, easy
Question
Show that $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix}$ is singular, and say what this means for the number of solutions.
Step-by-step solution
1. Step 1
  Notice row 3 = row 1 + row 2: $(2,3,4) = (1,1,1)+(1,2,3)$ . A dependent row forces $\det A = 0$ , but compute to confirm.
  $\det A = 1(2\cdot4 - 3\cdot3) - 1(1\cdot4 - 3\cdot2) + 1(1\cdot3 - 2\cdot2)$
2. Step 2
  Evaluate.
  $\det A = 1(-1) - 1(-2) + 1(-1) = -1 + 2 - 1 = 0$
Answer
$\det A = 0$ , so $A$ is singular: the system does not have a unique solution — it has either no solution or infinitely many (decide by row-reducing the augmented matrix).
Examiner tip
Crucial point: $\det A = 0$ rules out a unique solution but does NOT by itself mean 'no solution'. You must look at the constants $\mathbf{b}$ .
3A system with infinitely many solutions
Building confidenceMulti-step problem• infinite-solutions, medium
Question
Solve $x + y + z = 2$ , $x + 2y + 3z = 5$ , $2x + 3y + 4z = 7$ , giving the solution in terms of a parameter.
Step-by-step solution
1. Step 1
  Augment and eliminate $R_2 \to R_2 - R_1$ , $R_3 \to R_3 - 2R_1$ .
  $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{array}\right)$
2. Step 2
  $R_3 \to R_3 - R_2$ gives an all-zero bottom row.
  $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right)$
3. Step 3
  $0=0$ row ⇒ infinitely many solutions. Set $z = t$ . From row 2: $y + 2t = 3 \Rightarrow y = 3 - 2t$ . From row 1: $x + (3-2t) + t = 2 \Rightarrow x = t - 1$ .
  $x = t - 1,\quad y = 3 - 2t,\quad z = t$
Answer
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ — the three planes meet in a common line.
Examiner tip
Check in eqn 3: $2(t-1)+3(3-2t)+4t = 2t-2+9-6t+4t = 7$ ✓ for all $t$ . The vector form names a point and a direction — that is the line.
4A system with no solution
Building confidenceMulti-step problem• no-solution, medium
Question
Show that $x + y + z = 2$ , $x + 2y + 3z = 5$ , $2x + 3y + 4z = 10$ is inconsistent.
Step-by-step solution
1. Step 1
  Eliminate $R_2 \to R_2 - R_1$ , $R_3 \to R_3 - 2R_1$ .
  $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 6 \end{array}\right)$
2. Step 2
  $R_3 \to R_3 - R_2$ . The coefficients vanish but the constant does not.
  $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 3 \end{array}\right)$
3. Step 3
  Read the bottom row: $0 = 3$ , a contradiction.
  $0x + 0y + 0z = 3 \;\text{(impossible)}$
Answer
The bottom row gives $0 = 3$ , which is impossible, so the system has no solution — the planes form a triangular prism (no common point).
Examiner tip
Same coefficient matrix as the previous example (so $\det A = 0$ ), but the different constant turns a line into a contradiction. Geometry: triangular prism.
5Finding the critical parameter value
StretchMulti-step problem• parameter, hard
Question
For $x + y + 2z = 3$ , $2x - y + z = 0$ , $x + 2y + \lambda z = b$ , find the value of $\lambda$ for which the system fails to have a unique solution.
Step-by-step solution
1. Step 1
  Unique solution fails when $\det A = 0$ . Set up the determinant.
  $\det A = \begin{vmatrix} 1 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 2 & \lambda \end{vmatrix}$
2. Step 2
  Expand along row 1.
  $\det A = 1(-\lambda - 2) - 1(2\lambda - 1) + 2(4 + 1) = -\lambda - 2 - 2\lambda + 1 + 10$
3. Step 3
  Simplify and set to zero.
  $\det A = -3\lambda + 9 = 0 \;\Rightarrow\; \lambda = 3$
Answer
$\lambda = 3$ makes the system singular; for any $\lambda \neq 3$ there is a unique solution.
Examiner tip
When $\lambda = 3$ you would then test the constant $b$ to decide between no solution and infinitely many. Finding the critical $\lambda$ from $\det A = 0$ is the first move.
6Describing the geometry in each case
StretchShow that / prove• geometry, hard
Question
A system has $\det A = 0$ . After reduction the augmented matrix is $\left(\begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right)$ . Describe the solution set and the geometry.
Step-by-step solution
1. Step 1
  Identify the case. The bottom row is $0=0$ , so the system is consistent with infinitely many solutions.
2. Step 2
  Parametrise. Set $z = t$ . Row 2: $y + t = 1 \Rightarrow y = 1 - t$ . Row 1: $x - t = 2 \Rightarrow x = 2 + t$ .
  $x = 2 + t,\quad y = 1 - t,\quad z = t$
3. Step 3
  Write in vector form — a point plus a direction.
  $\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$
Answer
Infinitely many solutions lying on a line through $(2,1,0)$ with direction $\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ . The three planes form a sheaf, meeting in this common line.
Examiner tip
Full marks need BOTH the parametric/vector solution AND the geometric description ('three planes meet in a common line / sheaf').

Model Answers — Consistency of 3×3 Linear Systems and Geometric Interpretation

High-scoring sample answers for consistency of 3×3 linear systems and geometric interpretation on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
Determine whether the system $2x + y - z = 1$ , $x + 3y + 2z = 4$ , $3x - y + z = 2$ has a unique solution, justifying your answer.
Model answer
Evaluate $\det A$ for $A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & 3 & 2 \\ 3 & -1 & 1 \end{pmatrix}$ , expanding along row 1: $\det A = 2(3\cdot1 - 2\cdot(-1)) - 1(1\cdot1 - 2\cdot3) + (-1)(1\cdot(-1) - 3\cdot3)$ $= 2(3 + 2) - 1(1 - 6) - 1(-1 - 9) = 2(5) - 1(-5) - 1(-10) = 10 + 5 + 10 = 25$

Since $\det A = 25 \neq 0$ , the matrix is non-singular, so the system has a unique solution: the three planes meet at a single point.
Why this scores
M1 for the determinant expansion, A1 for $\det A = 25$ , A1 for 'unique solution', B1 for the geometric statement (single point). Watch the signs in the cofactor expansion.
Question 2
4 marks
(a) Show that the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 1 \end{pmatrix}$ is singular. (b) State what this tells you, and what it does NOT tell you, about the solutions of $A\mathbf{x} = \mathbf{b}$ .
Model answer
(a) Row 2 is exactly twice row 1: $(2,4,6) = 2(1,2,3)$ . Two proportional rows force a zero determinant. Confirming by expansion along column 1: $\det A = 1(4\cdot1 - 6\cdot0) - 2(2\cdot1 - 3\cdot0) + 1(2\cdot6 - 3\cdot4) = 4 - 4 + 0 = 0$

So $A$ is singular.

(b) This tells us the system does not have a unique solution. It does not tell us which of the two remaining cases applies — there could be no solution (inconsistent) or infinitely many (a common line). To decide, row-reduce the augmented matrix $[A\mid\mathbf{b}]$ and inspect the bottom row.
Why this scores
(a) M1 reasoning/expansion, A1 $\det A = 0$ . (b) B1 'no unique solution', B1 'cannot distinguish no-solution from infinitely-many without $\mathbf{b}$ '. The common error is claiming $\det A = 0 \Rightarrow$ no solution.
Question 3
6 marks
Solve $x + 2y - z = 1$ , $2x + y + z = 5$ , $3x + 3y = 6$ , expressing your answer in terms of a parameter, and describe the geometry.
Model answer
Augment and eliminate $R_2 \to R_2 - 2R_1$ , $R_3 \to R_3 - 3R_1$ : $\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & -3 & 3 & 3 \\ 0 & -3 & 3 & 3 \end{array}\right)$

$R_3 \to R_3 - R_2$ : $\left(\begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & -3 & 3 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right)$

The $0 = 0$ row shows the system is consistent with infinitely many solutions.

Parametrise: let $z = t$ . Row 2: $-3y + 3t = 3 \Rightarrow y = t - 1$ . Row 1: $x + 2(t-1) - t = 1 \Rightarrow x = 3 - t$ .

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

Geometry: the three planes meet in this common line (a sheaf/pencil of planes).
Why this scores
M1 elimination, A1 the $0=0$ row, M1 setting $z=t$ , A1 A1 for $x, y$ in terms of $t$ , B1 for the geometric description. Check eqn 2: $2(3-t)+(t-1)+t = 6-2t+t-1+t = 5$ ✓.
Question 4
6 marks
Show that $x + 2y + z = 3$ , $2x - y + 3z = 1$ , $3x + y + 4z = 7$ has no solution, and describe the configuration of the three planes.
Model answer
Augment and eliminate $R_2 \to R_2 - 2R_1$ , $R_3 \to R_3 - 3R_1$ : $\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & -5 & 1 & -5 \\ 0 & -5 & 1 & -2 \end{array}\right)$

$R_3 \to R_3 - R_2$ : $\left(\begin{array}{ccc|c} 1 & 2 & 1 & 3 \\ 0 & -5 & 1 & -5 \\ 0 & 0 & 0 & 3 \end{array}\right)$

The bottom row reads $0x + 0y + 0z = 3$ , i.e. $0 = 3$ , which is impossible. Hence the system is inconsistent — no solution.

Geometry: since each pair of planes meets in a line but there is no point common to all three, the planes form a triangular prism.
Why this scores
M1 elimination, A1 reaching a $0 = k$ ( $k\neq0$ ) row, A1 'no solution', M1 A1 justification, B1 'triangular prism'. The keyword 'triangular prism' is specifically credited for this $\det A = 0$ inconsistent case.
Question 5
8 marks
The system $x + y + z = 1$ , $x + 2y + 4z = \mu$ , $x + 4y + 10z = \mu^2$ depends on the constant $\mu$ . (a) Show the coefficient matrix is singular. (b) Find the values of $\mu$ for which the system is consistent, and describe the geometry for each.
Model answer
(a) $\det A = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{vmatrix}$ . Expand along row 1: $\det A = 1(20 - 16) - 1(10 - 4) + 1(4 - 2) = 4 - 6 + 2 = 0$ so $A$ is singular and there is no unique solution for any $\mu$ .

(b) Reduce $[A\mid\mathbf{b}]$ with $R_2 \to R_2 - R_1$ , $R_3 \to R_3 - R_1$ : $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & \mu - 1 \\ 0 & 3 & 9 & \mu^2 - 1 \end{array}\right)$ $R_3 \to R_3 - 3R_2$ : $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & \mu - 1 \\ 0 & 0 & 0 & \mu^2 - 1 - 3(\mu - 1) \end{array}\right)$ The bottom-right entry is $\mu^2 - 3\mu + 2 = (\mu - 1)(\mu - 2)$ .

Consistent ( $0 = 0$ row) requires $(\mu - 1)(\mu - 2) = 0$ , i.e. $\mu = 1$ or $\mu = 2$ . In each of these cases the system has infinitely many solutions and the three planes meet in a common line (a sheaf).

For all other $\mu$ the bottom row is $0 = \text{nonzero}$ , so the system is inconsistent (no solution) — the planes form a triangular prism.
Why this scores
(a) M1 A1 $\det A = 0$ . (b) M1 elimination, A1 the $(\mu-1)(\mu-2)$ entry, M1 setting it to $0$ , A1 $\mu = 1, 2$ , B1 B1 for the two geometric descriptions. Factorising the bottom-right entry is the key step.
Question 6
8 marks
For the system $2x + y + z = 1$ , $x - y + 2z = 3$ , $x + 2y - z = \lambda$ , find the value of $\lambda$ that makes the system consistent when $\det A = 0$ , and describe the geometry for (i) that $\lambda$ and (ii) all other $\lambda$ .
Model answer
Determinant: $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 2 & -1 \end{pmatrix}$ , expand along row 1: $\det A = 2(1 - 4) - 1(-1 - 2) + 1(2 + 1) = -6 + 3 + 3 = 0$ so $A$ is singular for all $\lambda$ — never a unique solution.

Reduce $[A\mid\mathbf{b}]$ . Swap to get a leading $1$ ( $R_1 \leftrightarrow R_2$ ): $\left(\begin{array}{ccc|c} 1 & -1 & 2 & 3 \\ 2 & 1 & 1 & 1 \\ 1 & 2 & -1 & \lambda \end{array}\right)$ $R_2 \to R_2 - 2R_1$ , $R_3 \to R_3 - R_1$ : $\left(\begin{array}{ccc|c} 1 & -1 & 2 & 3 \\ 0 & 3 & -3 & -5 \\ 0 & 3 & -3 & \lambda - 3 \end{array}\right)$ $R_3 \to R_3 - R_2$ : $\left(\begin{array}{ccc|c} 1 & -1 & 2 & 3 \\ 0 & 3 & -3 & -5 \\ 0 & 0 & 0 & \lambda - 3 - (-5) \end{array}\right) = \left(\begin{array}{ccc|c} 1 & -1 & 2 & 3 \\ 0 & 3 & -3 & -5 \\ 0 & 0 & 0 & \lambda + 2 \end{array}\right)$

(i) The bottom row is $0 = \lambda + 2$ , which is $0 = 0$ only when $\lambda = -2$ . Then the system is consistent with infinitely many solutions — the three planes meet in a common line (sheaf).

(ii) For $\lambda \neq -2$ the row is $0 = (\lambda + 2) \neq 0$ , a contradiction, so there is no solution — the planes form a triangular prism.
Why this scores
M1 A1 $\det A = 0$ , M1 elimination, A1 the $\lambda + 2$ bottom entry, M1 setting it to zero, A1 $\lambda = -2$ , B1 B1 for the two geometric descriptions. The whole question hinges on tracking the constant column accurately through the reduction.

Key Formulae — Consistency of 3×3 Linear Systems and Geometric Interpretation

The formulae you need to memorise for consistency of 3×3 linear systems and geometric interpretation on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Condition for a unique solution
$\det A \neq 0 \;\Longleftrightarrow\; \text{unique solution}$
$\det A$
determinant of the $3\times3$ coefficient matrix
When to use
First test in any consistency question — a non-zero determinant settles it (single point).
Singular case
$\det A = 0 \;\Rightarrow\; \text{no unique solution (reduce to decide)}$
$\det A = 0$
$A$ is singular; $A^{-1}$ does not exist
When to use
When the determinant vanishes — proceed to row-reduce the augmented matrix.
Bottom-row signatures
$0=0 \Rightarrow \infty\text{ solutions (line)};\qquad 0=k\,(k\neq0) \Rightarrow \text{no solution}$
$k$
the constant in a fully-eliminated bottom row
When to use
After reducing $[A\mid\mathbf{b}]$ with $\det A = 0$ — the bottom row decides the case.
Line solution (vector form)
$\mathbf{r} = \mathbf{a} + t\,\mathbf{d}$
$\mathbf{a}$
a particular point on the line
$\mathbf{d}$
direction vector of the line
$t$
free parameter
When to use
Writing the infinitely-many-solutions case as the common line of the three planes.
3×3 determinant (cofactor expansion)
$\det A = a(ei - fh) - b(di - fg) + c(dh - eg)$
$a,b,c$
entries of the top row of $A$
$d..i$
the remaining six entries, row by row
When to use
Computing $\det A$ to test for a unique solution. Mind the alternating $+,-,+$ signs.

Key Definitions and Keywords — Consistency of 3×3 Linear Systems and Geometric Interpretation

Definitions to memorise and the exact keywords mark schemes credit for consistency of 3×3 linear systems and geometric interpretation answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Consistent system
Examiner keyword
A linear system that has at least one solution (a unique solution OR infinitely many). An inconsistent system has none.
Inconsistent system
Examiner keyword
A system with no solution. After reduction it produces a contradictory row of the form $0 = k$ with $k \neq 0$ .
Singular matrix
Examiner keyword
A square matrix with $\det A = 0$ ; it has no inverse. For $A\mathbf{x}=\mathbf{b}$ this means there is no unique solution.
Sheaf (pencil) of planes
Examiner keyword
A family of planes all passing through one common line. The infinitely-many-solutions case: three planes meeting in a common line.
Triangular prism configuration
Examiner keyword
Three planes where each pair meets in a line but no point lies on all three, enclosing a triangular tube. The no-solution case.
Free parameter
A variable (e.g. $z = t$ ) left undetermined when there are fewer independent equations than unknowns, generating infinitely many solutions along a line.

Common Mistakes and Misconceptions — Consistency of 3×3 Linear Systems and Geometric Interpretation

The traps other students keep falling into on consistency of 3×3 linear systems and geometric interpretation questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Claiming $\det A = 0$ automatically means 'no solution'
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
Confusing 'no inverse' with 'no solution'; not realising a line of solutions can exist.
How to avoid it
$\det A = 0$ only rules out a UNIQUE solution. Row-reduce: $0=0$ row → infinitely many; $0=k$ row → none.
✕Giving the algebra but no geometric description (or vice versa)
9231 Paper 2 mark schemes
Why it happens
Answering only the part of the question that feels mathematical.
How to avoid it
Always pair the conclusion with the geometry: single point / common line (sheaf) / triangular prism / parallel planes.
✕Forgetting to apply row operations to the constant column
9231 Paper 2 Examiner Reports
Why it happens
Focusing on zeroing the coefficients and neglecting the right-hand side of $[A\mid\mathbf{b}]$ .
How to avoid it
Treat the augmented matrix as one block — every row operation acts on the constant too. The bottom-row constant is what decides consistency.
✕Stating 'infinitely many solutions' without writing the line
9231 Paper 2 Examiner Reports
Why it happens
Stopping at the $0=0$ row without parametrising.
How to avoid it
Set the free variable $z = t$ and express $x, y$ in terms of $t$ , then give the vector form $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ .
✕Describing the no-solution case vaguely instead of naming the prism
9231 Paper 2 mark schemes
Why it happens
Not knowing the specific mark-scheme term for three planes with no common point.
How to avoid it
Use the keyword 'triangular prism' (or 'two/three parallel planes' if applicable) — vague answers like 'they don't all meet' lose the description mark.
✕Sign slip in the $3\times3$ cofactor expansion gives a wrong $\det A$
9231 Paper 2 Examiner Reports
Why it happens
Forgetting the middle term's minus sign ( $-b(\ldots)$ ) in the $+,-,+$ pattern.
How to avoid it
Write the signs $+\,-\,+$ above the top row before expanding; double-check brackets with negative entries.

Consistency of 3×3 Linear Systems and Geometric Interpretation

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Classifying by the determinant

2Spotting det A = 0

3A system with infinitely many solutions

4A system with no solution

5Finding the critical parameter value

6Describing the geometry in each case

Condition for a unique solution

Singular case

Bottom-row signatures

Line solution (vector form)

3×3 determinant (cofactor expansion)

Consistent system

Inconsistent system

Singular matrix

Sheaf (pencil) of planes

Triangular prism configuration

Free parameter

✕Claiming det⁡A=0\det A = 0detA=0 automatically means 'no solution'

✕Giving the algebra but no geometric description (or vice versa)

✕Forgetting to apply row operations to the constant column

✕Stating 'infinitely many solutions' without writing the line

✕Describing the no-solution case vaguely instead of naming the prism

✕Sign slip in the 3×33\times33×3 cofactor expansion gives a wrong det⁡A\det AdetA

✕Claiming $\det A = 0$ automatically means 'no solution'

✕Sign slip in the $3\times3$ cofactor expansion gives a wrong $\det A$