Systems of linear equations

Matrix form AX = B, the consistency test by rank, and four solution methods — Cramer's rule, the inverse method, Gauss elimination and Gauss–Jordan — including homogeneous systems.

Module contents

This is what the whole module was building towards.

Matrix form

A system of $m$ equations in $n$ unknowns

$a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{mn} x_{n} = b_{1} = b_{2} ⋮ = b_{m}$

collapses to a single matrix equation

$A X = B$

where $A$ is the coefficient matrix ( $m \times n$ ), $X$ the column of unknowns ( $n \times 1$ ), and $B$ the column of constants ( $m \times 1$ ).

The augmented matrix is $A$ with $B$ attached as an extra column:

$[A ∣ B]$

A system with $B = O$ is homogeneous; otherwise non-homogeneous.

Consistency — will it even have a solution?

Compare the rank of the coefficient matrix with the rank of the augmented matrix.

$ρ (A) \neq = ρ ([A ∣ B]) ρ (A) = ρ ([A ∣ B]) = n ρ (A) = ρ ([A ∣ B]) = r < n ⟹ inconsistent — no solution ⟹ consistent, unique solution ⟹ consistent, infinitely many solutions$

where $n$ is the number of unknowns. In the third case, $n - r$ unknowns can be chosen freely; they are the free variables or parameters.

Inconsistency shows up in the reduced augmented matrix as a row of the form

$[00 \dots 0 ∣ c], c \neq = 0$

which says $0 = c$ . That is impossible, so no solution exists.

Method 1 — Cramer's rule

For a system of $n$ equations in $n$ unknowns with $Δ = ∣ A ∣ \neq = 0$ :

$x_{i} = \frac{Δ _{i}}{Δ}$

where $Δ_{i}$ is $∣ A ∣$ with the $i$ -th column replaced by $B$ .

For three unknowns:

$x = \frac{Δ _{1}}{Δ}, y = \frac{Δ _{2}}{Δ}, z = \frac{Δ _{3}}{Δ}$

When $Δ = 0$ , Cramer's rule gives no answer directly:

If $Δ = 0$ and at least one $Δ_{i} \neq = 0$ — inconsistent, no solution.
If $Δ = 0$ and every $Δ_{i} = 0$ — consistent, infinitely many solutions (analyse by rank).

Cramer's rule is clean for $2 \times 2$ and $3 \times 3$ but costs $n + 1$ determinants, so it is unusable for large systems.

Method 2 — Matrix inversion

If $A$ is square and non-singular:

$A X = B ⟹ X = A^{- 1} B$

Compute $A^{- 1}$ (usually via adjoint), then multiply. Useful when several different $B$ vectors must be solved with the same $A$ — invert once, reuse.

Method 3 — Gauss elimination

Form $[A ∣ B]$ .
Reduce the left block to upper triangular / echelon form by row operations.
Read the last equation, solve for the last unknown, and back-substitute upwards.

This is the method actual software uses. It has the lowest operation count of the three and never requires a determinant.

Method 4 — Gauss–Jordan

Same start, but continue until the left block is $I$ :

$[A ∣ B] ⟶ [I ∣ X]$

The solution is read off directly — no back-substitution. Slightly more arithmetic than Gauss elimination, but no second phase.

Homogeneous systems $A X = O$

These are always consistent, because $X = O$ always works. That solution is called the trivial solution. The real question is whether anything else exists.

For $n$ equations in $n$ unknowns:

$∣ A ∣ \neq = 0 ⟹ only the trivial solution X = O$
$∣ A ∣ = 0 ⟹ infinitely many non-trivial solutions$

In general, $A X = O$ has non-trivial solutions if and only if $ρ (A) < n$ , and the solution set then has $n - ρ (A)$ free parameters.

A homogeneous system with fewer equations than unknowns ( $m < n$ ) always has non-trivial solutions, since $ρ (A) \leq m < n$ .

Choosing a method

Situation	Method
$2$ or $3$ unknowns, $Δ \neq = 0$ , question says "Cramer"	Cramer's rule
Same $A$ , several different $B$	Inverse method
Any size, method free	Gauss elimination
Consistency to be discussed	Rank of $[A ∣ B]$ vs rank of $A$
Homogeneous	Check $∣ A ∣$ ; if zero, row-reduce and parametrise

Worked examples

5 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Three unknowns by Cramer's rule

Question

Solve by Cramer's rule:
$x + y + z x - y + z 2 x + y - z = 6 = 2 = 1$

Write the coefficient determinant
$Δ = 112 1 - 1 1 11 - 1$

Expanding along row 1:

$= 1 [(- 1) (- 1) - (1) (1)] - 1 [(1) (- 1) - (1) (2)] + 1 [(1) (1) - (- 1) (2)]$
$= 1 (1 - 1) - 1 (- 1 - 2) + 1 (1 + 2) = 0 + 3 + 3 = 6$

$Δ = 6 \neq = 0$ , so a unique solution exists.
$Δ_{1}$ — replace column 1 with $B = (6, 2, 1)$
$Δ_{1} = 621 1 - 1 1 11 - 1 = 6 (1 - 1) - 1 (- 2 - 1) + 1 (2 + 1) = 0 + 3 + 3 = 6$
$Δ_{2}$ — replace column 2
$Δ_{2} = 112621 11 - 1 = 1 (- 2 - 1) - 6 (- 1 - 2) + 1 (1 - 4) = - 3 + 18 - 3 = 12$
$Δ_{3}$ — replace column 3
$Δ_{3} = 112 1 - 1 1 621 = 1 (- 1 - 2) - 1 (1 - 4) + 6 (1 + 2) = - 3 + 3 + 18 = 18$
Divide
$x = \frac{6}{6} = 1, y = \frac{12}{6} = 2, z = \frac{18}{6} = 3$
Check in all three equations
$1 + 2 + 3 = 6$ ✓ $1 - 2 + 3 = 2$ ✓ $2 + 2 - 3 = 1$ ✓

Answer

$x = 1, y = 2, z = 3$

Example 2

The same system by Gauss elimination

Question

Solve by Gauss elimination:
$x + y + z x - y + z 2 x + y - z = 6 = 2 = 1$

Form the augmented matrix
$112 1 - 1 1 11 - 1 621$
Clear column 1
$R_{2} \to R_{2} - R_{1}$ : $(0, - 2, 0 ∣ - 4)$

$R_{3} \to R_{3} - 2 R_{1}$ : $(0, - 1, - 3 ∣ - 11)$

$100 1 - 2 - 1 10 - 3 6 - 4 - 11$
Clear column 2
$R_{3} \to 2 R_{3} - R_{2}$ : $(0, - 2 + 2, - 6 - 0 ∣ - 22 + 4) = (0, 0, - 6 ∣ - 18)$

$100 1 - 2 0 10 - 6 6 - 4 - 18$

Upper triangular — elimination is complete.
Back-substitute
From row 3: $- 6 z = - 18 \Rightarrow z = 3$

From row 2: $- 2 y = - 4 \Rightarrow y = 2$

From row 1: $x + y + z = 6 \Rightarrow x + 2 + 3 = 6 \Rightarrow x = 1$
Note the ranks
$ρ (A) = 3$ and $ρ ([A ∣ B]) = 3$ , with $n = 3$ unknowns. Equal ranks, both equal to $n$ — consistent with a unique solution, exactly as found.

Answer

$x = 1, y = 2, z = 3$ — the same as by Cramer's rule, with far less arithmetic.

NoteScaling R₃ by 2 before subtracting avoids fractions. Legal, and worth doing every time.

Example 3

Solving by the inverse method

Question

Solve using $X = A^{- 1} B$ :
$2 x - 3 y 4 x + y = 5 = 3$

Write in matrix form
$A [24 - 3 1] X [x y] = B [53]$
Invert $A$
$∣ A ∣ = 2 (1) - (- 3) (4) = 2 + 12 = 14 \neq = 0$

$A^{- 1} = \frac{1}{14} [1 - 4 32]$
Multiply
$X = A^{- 1} B = \frac{1}{14} [1 - 4 32] [53] = \frac{1}{14} [5 + 9 - 20 + 6] = \frac{1}{14} [14 - 14] = [1 - 1]$
Check
$2 (1) - 3 (- 1) = 2 + 3 = 5$ ✓ $4 (1) + (- 1) = 3$ ✓

Answer

$x = 1, y = - 1$

NoteKeep the 1/14 outside until the very last step — it turns the whole computation into integer arithmetic.

Example 4

Testing consistency by rank

Question

Test for consistency and solve if possible:
$x + 2 y + z 2 x + 3 y + 2 z 3 x + 5 y + 3 z = 3 = 5 = 9$

Form the augmented matrix
$123235123359$
Reduce
$R_{2} \to R_{2} - 2 R_{1}$ : $(0, - 1, 0 ∣ - 1)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(0, - 1, 0 ∣ 0)$

$100 2 - 1 - 1 100 3 - 1 0$

$R_{3} \to R_{3} - R_{2}$ : $(0, 0, 0 ∣ 1)$

$100 2 - 1 0 100 3 - 1 1$
Read the ranks
Coefficient block: two non-zero rows, so $ρ (A) = 2$ .

Augmented matrix: three non-zero rows, so $ρ ([A ∣ B]) = 3$ .
Conclude
$ρ (A) = 2 \neq = 3 = ρ ([A ∣ B])$

The system is inconsistent — no solution.

The last row says $0 x + 0 y + 0 z = 1$ , i.e. $0 = 1$ , which is impossible. Geometrically the three planes have no common point.

Answer

Inconsistent: $ρ (A) = 2 \neq = ρ ([A ∣ B]) = 3$ . No solution exists.

Example 5

A homogeneous system with non-trivial solutions

Question

Solve
$x + 2 y - z 2 x + y + z 3 x + 3 y = 0 = 0 = 0$

Check the determinant
$∣ A ∣ = 123213 - 1 10 = 1 (0 - 3) - 2 (0 - 3) + (- 1) (6 - 3) = - 3 + 6 - 3 = 0$

$∣ A ∣ = 0$ , so non-trivial solutions exist.
Row-reduce (no $B$ column needed — it stays zero)
$R_{2} \to R_{2} - 2 R_{1}$ : $(0, - 3, 3)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(0, - 3, 3)$

$100 2 - 3 - 3 - 1 33$

$R_{3} \to R_{3} - R_{2}$ :

$100 2 - 3 0 - 1 30$
Count the free variables
$ρ (A) = 2$ , $n = 3$ unknowns, so there are $n - ρ (A) = 1$ free variable — a one-parameter family of solutions.
Parametrise and back-substitute
Let $z = k$ .

Row 2: $- 3 y + 3 z = 0 \Rightarrow y = z = k$

Row 1: $x + 2 y - z = 0 \Rightarrow x + 2 k - k = 0 \Rightarrow x = - k$
Write the solution
$x y z = k - 1 11, k \in R$

Check with $k = 1$ , i.e. $(x, y, z) = (- 1, 1, 1)$ : $- 1 + 2 - 1 = 0$ ✓, $- 2 + 1 + 1 = 0$ ✓, $- 3 + 3 = 0$ ✓

Setting $k = 0$ recovers the trivial solution.

Answer

$(x, y, z) = k (- 1, 1, 1)$ for any real $k$ .

NoteAlways present a homogeneous answer as a multiple of a fixed vector. Giving one particular solution loses marks — the question asks for all of them.

Practice problems

7 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

Solve by Cramer's rule: $3 x + 2 y = 7, x - y = 1$ .

Show solution

$Δ = 31 2 - 1 = - 3 - 2 = - 5 \neq = 0$

$Δ_{1} = 71 2 - 1 = - 7 - 2 = - 9 Δ_{2} = 3171 = 3 - 7 = - 4$

$x = \frac{- 9}{- 5} = \frac{9}{5}, y = \frac{- 4}{- 5} = \frac{4}{5}$

Check: $3 (9/5) + 2 (4/5) = 27/5 + 8/5 = 35/5 = 7$ ✓ $9/5 - 4/5 = 1$ ✓

Answer

x = 9/5, y = 4/5

Problem 2Basic

Write $2 x - y + 3 z = 9, x + y + z = 6, x - y + z = 2$ in the form $A X = B$ and state the augmented matrix.

Show solution

$A = 211 - 1 1 - 1 311, X = x y z, B = 962$

$A X = B : 211 - 1 1 - 1 311 x y z = 962$

Augmented matrix:

$[A ∣ B] = 211 - 1 1 - 1 311962$

Answer

[A ∣ B] = 211 - 1 1 - 1 311962

Problem 3Exam level

Solve by Gauss–Jordan:
$x + y + z 2 x + 5 y + 7 z 2 x + y - z = 9 = 52 = 0$

Show solution

$122151 17 - 1 9520$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 2 R_{1}$ :

$100 13 - 1 15 - 3 934 - 18$

$R_{2} \leftrightarrow R_{3}$ , then $R_{2} \to - R_{2}$ to get a leading $1$ :

$10011313591834$

$R_{1} \to R_{1} - R_{2}$ , $R_{3} \to R_{3} - 3 R_{2}$ :

$100010 - 2 3 - 4 - 9 18 - 20$

$R_{3} \to - \frac{1}{4} R_{3}$ :

$100010 - 2 31 - 9 185$

$R_{1} \to R_{1} + 2 R_{3}$ , $R_{2} \to R_{2} - 3 R_{3}$ :

$100010001135$

Read off directly:

$x = 1, y = 3, z = 5$

Check: $1 + 3 + 5 = 9$ ✓ $2 + 15 + 35 = 52$ ✓ $2 + 3 - 5 = 0$ ✓

Answer

x = 1, y = 3, z = 5

Problem 4Exam level

Test for consistency and solve:
$x + y + z x + 2 y + 3 z x + 4 y + 7 z = 6 = 14 = 30$

Show solution

$11112413761430$

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ :

$1001131266824$

$R_{3} \to R_{3} - 3 R_{2}$ :

$100110120680$

Ranks. $ρ (A) = 2$ and $ρ ([A ∣ B]) = 2$ . Equal, so the system is consistent. But $n = 3$ unknowns and $r = 2$ , so there are $n - r = 1$ free variable and infinitely many solutions.

Parametrise. Let $z = k$ .

Row 2: $y + 2 k = 8 \Rightarrow y = 8 - 2 k$

Row 1: $x + (8 - 2 k) + k = 6 \Rightarrow x = k - 2$

$x y z = - 2 80 + k 1 - 2 1, k \in R$

Check with $k = 2$ , giving $(0, 4, 2)$ : $0 + 4 + 2 = 6$ ✓, $0 + 8 + 6 = 14$ ✓, $0 + 16 + 14 = 30$ ✓

AnswerConsistent with infinitely many solutions:

x = k - 2, y = 8 - 2 k, z = k

Problem 5Exam level

Find the values of $λ$ for which the system has (i) a unique solution, (ii) no solution, (iii) infinitely many solutions:
$x + y + z x + 2 y + 4 z x + 4 y + 10 z = 1 = λ = λ^{2}$

Show solution

$1111241410 1 λ λ^{2}$

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ :

$100113139 1 λ - 1 λ^{2} - 1$

$R_{3} \to R_{3} - 3 R_{2}$ :

$100110130 1 λ - 1 λ^{2} - 3 λ + 2$

since $λ^{2} - 1 - 3 (λ - 1) = λ^{2} - 3 λ + 2 = (λ - 1) (λ - 2)$ .

(i) Unique solution. This would need $ρ (A) = 3$ , but the coefficient block has a zero row for every $λ$ , so $ρ (A) = 2$ always. A unique solution never occurs.

(ii) No solution. Need $(λ - 1) (λ - 2) \neq = 0$ , which makes $ρ ([A ∣ B]) = 3 \neq = 2 = ρ (A)$ :

$λ \neq = 1 and λ \neq = 2$

(iii) Infinitely many. Need $(λ - 1) (λ - 2) = 0$ , giving $ρ (A) = ρ ([A ∣ B]) = 2 < 3$ :

$λ = 1 or λ = 2$

Solution when $λ = 1$ : row 2 becomes $y + 3 z = 0$ . With $z = k$ : $y = - 3 k$ , and $x = 1 - y - z = 1 + 3 k - k = 1 + 2 k$ .

$(x, y, z) = (1 + 2 k, - 3 k, k)$

Solution when $λ = 2$ : row 2 becomes $y + 3 z = 1$ . With $z = k$ : $y = 1 - 3 k$ , and $x = 1 - (1 - 3 k) - k = 2 k$ .

$(x, y, z) = (2 k, 1 - 3 k, k)$

Answer(i) never; (ii)

λ \neq = 1, 2

; (iii)

λ = 1

λ = 2

Problem 6Exam level

Show that the system $x + 2 y + 3 z = 0, 3 x + 4 y + 4 z = 0, 7 x + 10 y + 12 z = 0$ has only the trivial solution.

Show solution

Homogeneous, so it is automatically consistent. Test $∣ A ∣$ .

$∣ A ∣ = 13724103412$

Expand along row 1:

$= 1 (48 - 40) - 2 (36 - 28) + 3 (30 - 28)$
$= 1 (8) - 2 (8) + 3 (2) = 8 - 16 + 6 = - 2$

$∣ A ∣ = - 2 \neq = 0$

Since the coefficient matrix is non-singular, $ρ (A) = 3 = n$ , and the only solution of $A X = O$ is

$X = A^{- 1} O = O$

that is, $x = y = z = 0$ — the trivial solution only. ∎

Answer

∣ A ∣ = - 2 \neq = 0

, so

x = y = z = 0

is the only solution.

Problem 7Challenge

Find $k$ so that $k x + 3 y - z = 0, 4 x + k y - 2 z = 0, 2 x + y - z = 0$ has a non-trivial solution, and find that solution for the smaller value of $k$ .

Show solution

A homogeneous system has non-trivial solutions exactly when $∣ A ∣ = 0$ .

$∣ A ∣ = k 42 3 k 1 - 1 - 2 - 1$

Expand along row 1:

$= k (- k + 2) - 3 (- 4 + 4) + (- 1) (4 - 2 k)$
$= - k^{2} + 2 k - 0 - 4 + 2 k$
$= - k^{2} + 4 k - 4 = - (k^{2} - 4 k + 4) = - (k - 2)^{2}$

$∣ A ∣ = 0 ⟺ (k - 2)^{2} = 0 ⟺ k = 2$

Only one value, $k = 2$ (a repeated root).

Solving at $k = 2$ .

$242321 - 1 - 2 - 1$

$R_{2} \to R_{2} - 2 R_{1}$ : $(0, - 4, 0)$
$R_{3} \to R_{3} - R_{1}$ : $(0, - 2, 0)$

$200 3 - 4 - 2 - 1 00 R_{3} \to 2 R_{3} - R_{2} 200 3 - 4 0 - 1 00$

$ρ (A) = 2$ , $n = 3$ , so one free parameter.

Row 2: $- 4 y = 0 \Rightarrow y = 0$ .

Row 1: $2 x + 0 - z = 0 \Rightarrow z = 2 x$ . Let $x = t$ .

$(x, y, z) = t (1, 0, 2), t \in R$

Check with $t = 1$ , i.e. $(1, 0, 2)$ : $2 + 0 - 2 = 0$ ✓, $4 + 0 - 4 = 0$ ✓, $2 + 0 - 2 = 0$ ✓

Answer

k = 2

; solution

(x, y, z) = t (1, 0, 2)

Matrix form

Consistency — will it even have a solution?

Method 1 — Cramer's rule

Method 2 — Matrix inversion

Method 3 — Gauss elimination

Method 4 — Gauss–Jordan

Homogeneous systems AX=O

Choosing a method

Worked examples

Practice problems

Homogeneous systems $A X = O$