Solving Systems of Equations with Matrices

Matrices turn a system of simultaneous linear equations into a single equation $A\mathbf{x} = \mathbf{b}$Ax=b, which — when $A$A is invertible — is solved in one stroke by $\mathbf{x} = A^{-1}\mathbf{b}$x=A−1b. Just as powerfully, the determinant of $A$A classifies the geometry: a non-zero determinant means the planes meet at exactly one point (a unique solution), while a zero determinant means they fail to do so — either meeting in a line or sheet (infinitely many solutions) or not meeting at all (no solution). This lesson writes systems in matrix form, solves them with the inverse, and decodes every geometric configuration of two lines or three planes. It is where determinants, inverses and geometry all come together. The single most important habit to build is this: always compute $\det A$detA first. It instantly separates the easy "unique solution" world (where you simply apply $\mathbf{x} = A^{-1}\mathbf{d}$x=A−1d) from the more interesting singular world, where a second test — consistency — decides between infinitely many solutions and none at all. Examiners deliberately set systems in that singular world precisely because navigating it demands genuine understanding rather than rote inversion.

---

1. Where this sits in AQA 7367

Solving systems is compulsory pure content for Papers 1 and 2, and it is the most AO3-rich topic in the matrices strand: it asks you to interpret — to translate between algebra (the value of $\det A$detA, the consistency of the equations) and geometry (how lines or planes are arranged in space). The mechanics ($\mathbf{x} = A^{-1}\mathbf{b}$x=A−1b) draw directly on the inverse-matrix lesson, and the classification draws on the determinant lesson. Expect questions that combine a computation with a "interpret your answer geometrically" demand.

---

2. Writing a system in matrix form

A system of linear equations such as

$$\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \\ a_2 x + b_2 y + c_2 z &= d_2 \\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$$a1​x+b1​y+c1​za2​x+b2​y+c2​za3​x+b3​y+c3​z​=d1​=d2​=d3​​

is captured by the single matrix equation $A\mathbf{x} = \mathbf{d}$Ax=d, where

$$A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}, \qquad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \qquad \mathbf{d} = \begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}.$$A=​a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​​,x=​xyz​​,d=​d1​d2​d3​​​.

$A$A is the coefficient matrix, $\mathbf{x}$x the vector of unknowns, and $\mathbf{d}$d the constants. The single equation $A\mathbf{x} = \mathbf{d}$Ax=d is exactly equivalent to the whole system — multiplying out $A\mathbf{x}$Ax reproduces the three left-hand sides.

---

3. Solving with the inverse

If $\det(A) \neq 0$det(A)=0, then $A^{-1}$A−1 exists and left-multiplying gives the unique solution

$A\mathbf{x} = \mathbf{d} \implies A^{-1}A\mathbf{x} = A^{-1}\mathbf{d} \implies \mathbf{x} = A^{-1}\mathbf{d}.$Ax=d⟹A−1Ax=A−1d⟹x=A−1d.

Worked Example 1 — a 2×2 system (with mark scheme)

Solve $2x + y = 5$2x+y=5, $3x + 2y = 8$3x+2y=8.

Matrix form: $\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}$(23​12​)(xy​)=(58​). Here $\det A = 4 - 3 = 1$detA=4−3=1, so $A^{-1} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}$A−1=(2−3​−12​) and

$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 8 \end{pmatrix} = \begin{pmatrix} 10 - 8 \\ -15 + 16 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}.$$(xy​)=(2−3​−12​)(58​)=(10−8−15+16​)=(21​).

So $x = 2$x=2, $y = 1$y=1. (Check in the original: $2(2)+1 = 5$2(2)+1=5 ✓, $3(2)+2 = 8$3(2)+2=8 ✓.)

Mark scheme: M1 write in matrix form and find $\det A = 1$detA=1; M1 correct $A^{-1}$A−1; M1 compute $A^{-1}\mathbf{d}$A−1d; A1 $x = 2, y = 1$x=2,y=1. A substitution check is good practice and protects the final mark.

Worked Example 2 — a 3×3 system (with mark scheme)

Solve $x + y + z = 6$x+y+z=6, $x - y + z = 2$x−y+z=2, $2x + y - z = 1$2x+y−z=1.

Coefficient matrix $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{pmatrix}$A=​112​1−11​11−1​​. Expanding along row 1,

$$\det A = 1\begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + 1\begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1(0) - 1(-3) + 1(3) = 6 \neq 0,$$detA=1​−11​1−1​​−1​12​1−1​​+1​12​−11​​=1(0)−1(−3)+1(3)=6=0,

so there is a unique solution. Rather than invert the whole matrix, eliminate: subtracting equation (2) from equation (1) gives $2y = 4$2y=4, so $y = 2$y=2. Then (1) becomes $x + z = 4$x+z=4 and (3) becomes $2x - z = 1 - y = -1$2x−z=1−y=−1; adding these, $3x = 3$3x=3, so $x = 1$x=1 and $z = 3$z=3.

Thus $x = 1, y = 2, z = 3$x=1,y=2,z=3. (Check (2): $1 - 2 + 3 = 2$1−2+3=2 ✓; (3): $2 + 2 - 3 = 1$2+2−3=1 ✓.)

Mark scheme: M1 form the coefficient matrix and compute $\det A = 6$detA=6 (establishing a unique solution); M1 a valid elimination or inverse method; A1 two of the three values; A1 all three correct with a check. Either the inverse route or systematic elimination earns full marks; elimination is usually faster for a clean $3\times3$3×3.

---

4. The determinant classifies the solution

The value of $\det A$detA tells you, before any solving, how many solutions to expect:

Determinant	Number of solutions	Geometric meaning
$\det A \neq 0$detA=0	exactly one (unique)	lines/planes meet at a single point
$\det A = 0$detA=0, consistent	infinitely many	lines coincide; planes meet in a line or sheet
$\det A = 0$detA=0, inconsistent	none	lines/planes are parallel or form a prism

The key subtlety: $\det A = 0$detA=0 by itself does not decide between "infinitely many" and "none" — you must check consistency (whether the equations are compatible) to tell them apart. That extra step is exactly what the harder exam questions probe.

A special, always-consistent case is the homogeneous system $A\mathbf{x} = \mathbf{0}$Ax=0. It can never be inconsistent, because $\mathbf{x} = \mathbf{0}$x=0 (the trivial solution) always works. So for a homogeneous system the only question is whether there are other solutions: if $\det A \neq 0$detA=0 the trivial solution is the unique one, while if $\det A = 0$detA=0 there are infinitely many (a line or plane of solutions through the origin). This "$\det A = 0 \iff$detA=0⟺ non-trivial solutions exist" principle is precisely what defines eigenvalues in lesson 8, where the homogeneous system is $(A - \lambda I)\mathbf{x} = \mathbf{0}$(A−λI)x=0.

---

5. Geometric interpretation: three planes in 3D

Each equation $ax + by + cz = d$ax+by+cz=d is a plane. The solution set is the intersection of the three planes, and there is a finite menu of configurations:

Configuration	Description	Solutions	$\det A$detA
Single point	three planes cross at one point	one	$\neq 0$=0
Sheaf (common line)	three planes share a common line	infinitely many (1 parameter)	$0$0, consistent
Coincident planes	two or more planes are identical	infinitely many	$0$0, consistent
Triangular prism	the three pairwise intersection lines are parallel but distinct	none	$0$0, inconsistent
Parallel planes	two or more planes are parallel and distinct	none	$0$0, inconsistent

When $\det A \neq 0$detA=0 the only possibility is the single point. When $\det A = 0$detA=0 the planes are in "special position" and you decode which case by attempting to solve: a contradiction (e.g. $0 = 5$0=5) means inconsistent (prism or parallel, no solution); a free parameter means consistent (sheaf or coincident, infinitely many).

It helps to picture the two "no solution" cases distinctly. Parallel planes are the obvious one: like sheets of paper held apart, two (or three) planes with the same orientation but different positions never share a point. The triangular prism is subtler and a favourite of examiners: here no two planes are parallel — each pair meets in a line — but the three lines of intersection are themselves parallel to one another, so the three planes enclose an infinite triangular tube and have no point common to all three. The determinant is zero in both cases, and only by solving (reaching a contradiction) do you confirm there is no solution. Conversely the two "infinitely many" cases are the sheaf (three distinct planes pivoting about one shared line, like pages of an open book meeting at the spine) and coincident planes (literally the same plane written two or three ways).

graph TD
  A["3×3 system A x = d"] --> B{"det A = 0 ?"}
  B -->|"no"| C["Unique solution<br/>planes meet at a point"]
  B -->|"yes"| D{"Consistent?"}
  D -->|"yes"| E["Infinitely many<br/>sheaf / coincident planes"]
  D -->|"no"| F["No solution<br/>prism / parallel planes"]

---

6. Worked Example 3 — the singular, consistent case

Solve $x + y + z = 6$x+y+z=6, $2x + 2y + 2z = 12$2x+2y+2z=12, $x + y + z = 6$x+y+z=6.

Here the second equation is exactly twice the first, and the third repeats the first: there is really only one independent equation, $x + y + z = 6$x+y+z=6. The coefficient matrix has three identical (up to scaling) rows, so $\det A = 0$detA=0. The system is consistent (the equations agree), so there are infinitely many solutions: choosing $x$x and $y$y freely as parameters, $z = 6 - x - y$z=6−x−y. Geometrically, all three planes coincide.

7. Worked Example 4 — the singular, inconsistent case

Solve $x + y = 3$x+y=3, $2x + 2y = 7$2x+2y=7.

The second equation says $x + y = 3.5$x+y=3.5, contradicting the first ($x + y = 3$x+y=3). The coefficient matrix $\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$(12​12​) has $\det = 0$det=0, and the system is inconsistent, so there is no solution. Geometrically, the two lines are parallel and distinct — they never meet.

These two examples are the heart of the topic: same $\det A = 0$detA=0, opposite conclusions, separated only by consistency. The lesson to carry into the exam is that a zero determinant is never the end of the analysis — it is the signal to switch from "apply the inverse" to "investigate the geometry", and the quality of your answer is judged on how clearly you then resolve which singular case you are in.

---

8. Specimen-style exam question

(Specimen-style — not a real past paper.) The system $x + 2y - z = 1$x+2y−z=1, $2x - y + 3z = 4$2x−y+3z=4, $x + 7y - 6z = k$x+7y−6z=k is given. (a) Show that the coefficient matrix is singular. (b) Find the value of $k$k for which the system is consistent, and describe the solution set geometrically. (7 marks)

Model solution. (a) $\det\begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 1 & 7 & -6 \end{pmatrix} = 1(6 - 21) - 2(-12 - 3) + (-1)(14 + 1) = -15 + 30 - 15 = 0$det​121​2−17​−13−6​​=1(6−21)−2(−12−3)+(−1)(14+1)=−15+30−15=0, so the matrix is singular.