Cramer's Rule

A method for solving a system of $n$ linear equations in $n$ variables using determinants. Named after Swiss mathematician Gabriel Cramer (1704–1752), who published it in 1750.

General Statement

For a system $AX = B$ where $A$ is an $n \times n$ non-singular coefficient matrix ($|A| \neq 0$), the unique solution is:

$$x_i = \frac{|A_i|}{|A|} \quad \text{for } i = 1, 2, \ldots, n$$

where $A_i$ is the matrix formed by replacing the $i$-th column of $A$ with the constants column $B$.

$$A_i = \begin{bmatrix} a_{11} & \cdots & b_1 & \cdots & a_{1n} \ a_{21} & \cdots & b_2 & \cdots & a_{2n} \ \vdots & & \vdots & & \vdots \ a_{n1} & \cdots & b_n & \cdots & a_{nn} \end{bmatrix} \quad \text{(column $i$ replaced by $B$)}$$

Why Cramer's Rule Works

Cramer's Rule follows directly from the matrix inverse formula $X = A^{-1}B$ and the adjoint method:

$$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{|A|} C^T$$

where $C$ is the cofactor matrix. Multiplying both sides by $B$:

$$X = \frac{1}{|A|} C^T B$$

For the $i$-th component $x_i$, this equals $\frac{1}{|A|}$ times the dot product of the $i$-th row of $C^T$ (i.e., the $i$-th column of $C$) with $B$. But by the cofactor expansion of a determinant, this is exactly $|A_i|$ — the determinant of $A$ with the $i$-th column replaced by $B$.

In short: $x_i = \frac{\sum b_j C_{ji}}{|A|} = \frac{|A_i|}{|A|}$.

2×2 Cramer's Rule

For the system: $$\begin{aligned} a_1 x + b_1 y &= c_1 \ a_2 x + b_2 y &= c_2 \end{aligned}$$

Step 1: Form coefficient matrix $A = \begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix}$, constant matrix $B = \begin{pmatrix} c_1 \ c_2 \end{pmatrix}$

Step 2: Compute $|A| = \begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1$

Step 3: Compute $|A_x|$ (column 1 replaced by $B$): $$|A_x| = \begin{vmatrix} c_1 & b_1 \ c_2 & b_2 \end{vmatrix} = c_1 b_2 - c_2 b_1$$

Step 4: Compute $|A_y|$ (column 2 replaced by $B$): $$|A_y| = \begin{vmatrix} a_1 & c_1 \ a_2 & c_2 \end{vmatrix} = a_1 c_2 - a_2 c_1$$

Step 5: Solve: $$x = \frac{|A_x|}{|A|}, \quad y = \frac{|A_y|}{|A|}$$

Worked Example (2×2)

Solve: $2x + 3y = 8$, $x - y = -1$

$$\begin{aligned} |A| &= \begin{vmatrix} 2 & 3 \ 1 & -1 \end{vmatrix} = 2(-1) - 1(3) = -2 - 3 = -5 \ |A_x| &= \begin{vmatrix} 8 & 3 \ -1 & -1 \end{vmatrix} = 8(-1) - (-1)(3) = -8 + 3 = -5 \ |A_y| &= \begin{vmatrix} 2 & 8 \ 1 & -1 \end{vmatrix} = 2(-1) - 1(8) = -2 - 8 = -10 \end{aligned}$$

$$x = \frac{-5}{-5} = 1, \quad y = \frac{-10}{-5} = 2$$

Check: $2(1) + 3(2) = 8$ ✓, $1 - 2 = -1$ ✓

3×3 Cramer's Rule

For the system: $$\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \ a_2 x + b_2 y + c_2 z &= d_2 \ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$$

Components:

$$|A| = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix}$$

$$|A_x| = \begin{vmatrix} d_1 & b_1 & c_1 \ d_2 & b_2 & c_2 \ d_3 & b_3 & c_3 \end{vmatrix}, \quad |A_y| = \begin{vmatrix} a_1 & d_1 & c_1 \ a_2 & d_2 & c_2 \ a_3 & d_3 & c_3 \end{vmatrix}, \quad |A_z| = \begin{vmatrix} a_1 & b_1 & d_1 \ a_2 & b_2 & d_2 \ a_3 & b_3 & d_3 \end{vmatrix}$$

Solution: $$x = \frac{|A_x|}{|A|}, \quad y = \frac{|A_y|}{|A|}, \quad z = \frac{|A_z|}{|A|}$$

Worked Example (3×3)

Solve: $$\begin{aligned} x + y - z &= 0 \ 2x - 3y + z &= 1 \ 2x + y + 2z &= 7 \end{aligned}$$

Step 1: $|A|$ — compute by cofactor expansion along row 1: $$\begin{aligned} |A| &= 1 \cdot \begin{vmatrix} -3 & 1 \ 1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 1 \ 2 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -3 \ 2 & 1 \end{vmatrix} \ &= 1(-6 - 1) - 1(4 - 2) - 1(2 + 6) \ &= 1(-7) - 1(2) - 1(8) \ &= -7 - 2 - 8 = -17 \end{aligned}$$

Step 2: $|A_x|$ (column 1 replaced by $d$): $$\begin{aligned} |A_x| &= \begin{vmatrix} 0 & 1 & -1 \ 1 & -3 & 1 \ 7 & 1 & 2 \end{vmatrix} \ &= 0 \cdot \begin{vmatrix} -3 & 1 \ 1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \ 7 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -3 \ 7 & 1 \end{vmatrix} \ &= 0 - 1(2 - 7) - 1(1 + 21) \ &= 0 - 1(-5) - 1(22) \ &= 5 - 22 = -17 \end{aligned}$$

Step 3: $|A_y|$ (column 2 replaced by $d$): $$\begin{aligned} |A_y| &= \begin{vmatrix} 1 & 0 & -1 \ 2 & 1 & 1 \ 2 & 7 & 2 \end{vmatrix} \ &= 1 \cdot \begin{vmatrix} 1 & 1 \ 7 & 2 \end{vmatrix} - 0 \cdot \begin{vmatrix} 2 & 1 \ 2 & 2 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & 1 \ 2 & 7 \end{vmatrix} \ &= 1(2 - 7) - 0 - 1(14 - 2) \ &= -5 - 12 = -17 \end{aligned}$$

Step 4: $|A_z|$ (column 3 replaced by $d$): $$\begin{aligned} |A_z| &= \begin{vmatrix} 1 & 1 & 0 \ 2 & -3 & 1 \ 2 & 1 & 7 \end{vmatrix} \ &= 1 \cdot \begin{vmatrix} -3 & 1 \ 1 & 7 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 1 \ 2 & 7 \end{vmatrix} + 0 \cdot \begin{vmatrix} 2 & -3 \ 2 & 1 \end{vmatrix} \ &= 1(-21 - 1) - 1(14 - 2) + 0 \ &= -22 - 12 = -34 \end{aligned}$$

Step 5: Final answers: $$x = \frac{-17}{-17} = 1, \quad y = \frac{-17}{-17} = 1, \quad z = \frac{-34}{-17} = 2$$

Check all 3 equations:

  • $1 + 1 - 2 = 0$ ✓
  • $2(1) - 3(1) + 2 = 1$ ✓
  • $2(1) + 1 + 2(2) = 7$ ✓

Cramer's Rule for $n \times n$ Systems (General Case)

For a system of $n$ equations in $n$ unknowns:

$$\begin{aligned} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \ &\ \vdots \ a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n &= b_n \end{aligned}$$

Each variable $x_i$ is given by:

$$x_i = \frac{|A_i|}{|A|} \quad \text{for } i = 1, 2, \ldots, n$$

where $A_i$ is the $n \times n$ matrix formed by replacing column $i$ of $A$ with the constants column $B = (b_1, b_2, \ldots, b_n)^T$.

Procedure Summary

  1. Write the system in matrix form: $AX = B$
  2. Compute $|A|$ — if $|A| = 0$, stop (Cramer's Rule cannot be used; the system has either no solution or infinitely many)
  3. For each variable $x_i$: form $A_i$ by replacing column $i$ of $A$ with $B$, compute $|A_i|$
  4. Read off the solution: $x_i = |A_i| / |A|$

Solvability and Solution Types

Condition Result Interpretation
$ A \neq 0$
$ A = 0$ and $(\text{adj } A)B = 0$
$ A = 0$ and $(\text{adj } A)B \neq 0$

Important: Cramer's Rule only applies to square systems ($n$ equations, $n$ unknowns) with $|A| \neq 0$.

Word Problem (3×3 Application)

Aishah bought a marker pen, two examination pads and a file for RM36. Balqis bought three marker pens, two examination pads and a file for RM43. Camelia bought three marker pens, an examination pad and four files for RM74.

Step 1 — Define variables: Let $x =$ price of marker pen, $y =$ price of exam pad, $z =$ price of file.

Step 2 — Write equations: $$\begin{aligned} x + 2y + z &= 36 \ 3x + 2y + z &= 43 \ 3x + y + 4z &= 74 \end{aligned}$$

Step 3 — Matrix form: $$\begin{pmatrix} 1 & 2 & 1 \ 3 & 2 & 1 \ 3 & 1 & 4 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 36 \ 43 \ 74 \end{pmatrix}$$

Step 4 — Compute $|A|$: $$\begin{aligned} |A| &= 1(2 \cdot 4 - 1 \cdot 1) - 2(3 \cdot 4 - 1 \cdot 3) + 1(3 \cdot 1 - 2 \cdot 3) \ &= 1(8 - 1) - 2(12 - 3) + 1(3 - 6) \ &= 7 - 18 - 3 = -14 \end{aligned}$$

$|A| \neq 0$, so a unique solution exists.

Step 5 — Compute $|A_x|, |A_y|, |A_z|$: $$\begin{aligned} |A_x| &= \begin{vmatrix} 36 & 2 & 1 \ 43 & 2 & 1 \ 74 & 1 & 4 \end{vmatrix} = 36(8-1) - 2(172-74) + 1(43-148) \ &= 36(7) - 2(98) + 1(-105) = 252 - 196 - 105 = -49 \[4pt] |A_y| &= \begin{vmatrix} 1 & 36 & 1 \ 3 & 43 & 1 \ 3 & 74 & 4 \end{vmatrix} = 1(172-74) - 36(12-3) + 1(222-129) \ &= 98 - 36(9) + 93 = 98 - 324 + 93 = -133 \[4pt] |A_z| &= \begin{vmatrix} 1 & 2 & 36 \ 3 & 2 & 43 \ 3 & 1 & 74 \end{vmatrix} = 1(148-43) - 2(222-129) + 36(3-6) \ &= 105 - 2(93) + 36(-3) = 105 - 186 - 108 = -189 \end{aligned}$$

Step 6 — Solutions: $$x = \frac{-49}{-14} = 3.5, \quad y = \frac{-133}{-14} = 9.5, \quad z = \frac{-189}{-14} = 13.5$$

Answer: Marker pen = RM 3.50, Exam pad = RM 9.50, File = RM 13.50.

Comparison with Other Methods

Method Used When Advantages Disadvantages
Cramer's Rule $ A \neq 0$, small $n$
Inverse Matrix $ A \neq 0$
Gauss-Jordan Elimination Any size Efficient, handles all solution types More steps, row operations

For $n \le 3$, Cramer's Rule is competitive with other methods. For $n > 3$, Gaussian elimination is preferred because computing determinants becomes expensive: an $n \times n$ determinant requires $\sim n!$ operations if computed naively (though cofactor expansion reduces this).

Computational Note

The number of operations to compute an $n \times n$ determinant by cofactor expansion is $O(n!)$. For Cramer's Rule, you need $n+1$ such determinants, making the total $O((n+1)!)$ ≈ $O(n!)$ — impractical for large $n$.

In practice:

  • $n = 2$: very fast (2 multiplications, 1 subtraction per determinant)
  • $n = 3$: manageable (~9 terms per determinant)
  • $n \ge 4$: use Gaussian elimination or inverse matrix method instead

Key Formulas to Memorize

2×2 Determinant: $$\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$$

3×3 Determinant (diagonal expansion for checking): $$|A| = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33}$$

3×3 Cofactor Sign Pattern: $$\begin{pmatrix} + & - & + \ - & + & - \ + & - & + \end{pmatrix}$$

Related Sources

Related Concepts

  • Matrices — matrix algebra fundamentals
  • Systems of Linear Equations — $AX = B$ formulation and solution types
  • Determinant — |A| computation and properties
  • Gauss-Jordan Elimination — alternative row-reduction method
  • Matrix Inverse — $A^{-1}$ method for solving systems
  • Adjoint Matrix — $\text{adj}(A) = C^T$, used in deriving Cramer's Rule

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