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Cramer's Rule

Cramer’s Rule

The Definition of Cramer’s Rule

Cramer's Rule

Let the n-variable linear system be:

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

If the determinant $\det(A)$ of the coefficient matrix $A = (a_{ij})$ satisfies $\det(A) \neq 0$ , then the system has a unique solution: $x_j = \frac{\det(A_j)}{\det(A)} \qquad (j=1,2,\dots,n)$

where $A_j$ is the matrix obtained from the coefficient matrix $A$ by replacing its $j$ -th column with the constant terms $\vec{b}$ .

The Conditions of Cramer’s Rule

Conditions:

The system must have n equations in n unknowns (a square matrix)
The determinant $|A|$ of the coefficient matrix must satisfy $|A| \neq 0$

Note: when $|A| = 0$ , Cramer’s rule does not apply and another method has to be used to solve the system.

The Proof of Cramer’s Rule

Proof idea:

Suppose $A\vec{x} = \vec{b}$ has the unique solution $\vec{x} = A^{-1}\vec{b}$
Use the property of the adjugate matrix: $A^{-1} = \frac{1}{|A|}A^*$
This gives $x_i = \frac{1}{|A|}\sum_{j=1}^n A_{ji}b_j$
Here $A_{ji}$ is the algebraic cofactor of $A$
One can prove that $\sum_{j=1}^n A_{ji}b_j = |A_i|$

Applications of Cramer’s Rule

The Solving Steps

Check the conditions: confirm that the system is n×n and that $|A| \neq 0$
Compute $|A|$ : compute the determinant of the coefficient matrix
Compute $|A_i|$ : replace the i-th column with the vector of constant terms and compute the determinant
Solve: $x_i = \frac{|A_i|}{|A|}$

Example

Example 1: solve the system by Cramer’s rule: $\begin{cases} x + 2y = 5 \\ 3x - y = 4 \end{cases}$

Solution:

The coefficient matrix: $A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$
$|A| = 1 \times (-1) - 2 \times 3 = -1 - 6 = -7 \neq 0$
$|A_1| = \begin{vmatrix} 5 & 2 \\ 4 & -1 \end{vmatrix} = 5 \times (-1) - 2 \times 4 = -5 - 8 = -13$
$|A_2| = \begin{vmatrix} 1 & 5 \\ 3 & 4 \end{vmatrix} = 1 \times 4 - 5 \times 3 = 4 - 15 = -11$
$x = \frac{-13}{-7} = \frac{13}{7}$ , $y = \frac{-11}{-7} = \frac{11}{7}$

The Advantages and Disadvantages of Cramer’s Rule

Advantages

Theoretical significance: it gives an explicit expression for the solution
Simple computation: for small systems the computation is relatively simple
Exact solutions: it yields exact fractional solutions

Disadvantages

Heavy computation: for large systems, several determinants have to be computed
Restricted applicability: it works only for square matrices with a nonzero determinant
Numerical stability: when the determinant is close to zero, the computational error is large

Extensions of Cramer’s Rule

Homogeneous Systems

The Necessary and Sufficient Condition for a Homogeneous System to Have a Nonzero Solution

The homogeneous linear system $A\vec{x} = \vec{0}$ has a nonzero solution if and only if $|A| = 0$ .

Proof

Proof:

If $|A| \neq 0$ , then the only solution is the zero solution
If $|A| = 0$ , then there are infinitely many solutions, including nonzero ones

The Non-Square Case

For a system of size m×n (m ≠ n), Cramer’s rule does not apply and another method has to be used to solve it.

Exercises

Exercise 1

Solve the system $\begin{cases} 2x + y = 7 \\ x - 3y = -4 \end{cases}$ by Cramer’s rule.

Reference Answer

Cramer's rulesystem of linear equationsdeterminantcoefficient matrix

Approach: Follow the steps of Cramer’s rule.

Steps:

The coefficient matrix: $A = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}$
$|A| = 2 \times (-3) - 1 \times 1 = -6 - 1 = -7 \neq 0$
$|A_1| = \begin{vmatrix} 7 & 1 \\ -4 & -3 \end{vmatrix} = 7 \times (-3) - 1 \times (-4) = -21 + 4 = -17$
$|A_2| = \begin{vmatrix} 2 & 7 \\ 1 & -4 \end{vmatrix} = 2 \times (-4) - 7 \times 1 = -8 - 7 = -15$
$x = \frac{-17}{-7} = \frac{17}{7}$ , $y = \frac{-15}{-7} = \frac{15}{7}$

Answer: $x = \frac{17}{7}$ , $y = \frac{15}{7}$

Exercise 2

Decide whether the system $\begin{cases} x + y = 1 \\ 2x + 2y = 2 \end{cases}$ can be solved by Cramer’s rule.

Reference Answer

Cramer's rulesystem of linear equationsdeterminantcoefficient matrix

Approach: Check the determinant of the coefficient matrix.

Steps:

The coefficient matrix: $A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$
$|A| = 1 \times 2 - 1 \times 2 = 2 - 2 = 0$
The determinant is zero, so Cramer’s rule does not apply

Answer: it cannot be solved by Cramer’s rule

Exercise 3

Solve the system $\begin{cases} x + y + z = 6 \\ 2x + 3y + z = 13 \\ x - y + 2z = 4 \end{cases}$ by Cramer’s rule.

Reference Answer

Cramer's rulesystem of linear equationsdeterminantcoefficient matrix

Approach: Follow the steps of Cramer’s rule for the three-variable system.

Steps:

The coefficient matrix: $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 1 & -1 & 2 \end{pmatrix}$
$|A| = 1 \times (3 \times 2 - 1 \times (-1)) - 1 \times (2 \times 2 - 1 \times 1) + 1 \times (2 \times (-1) - 3 \times 1) = 1 \times 7 - 1 \times 3 + 1 \times (-5) = 7 - 3 - 5 = -1 \neq 0$
$|A_1| = \begin{vmatrix} 6 & 1 & 1 \\ 13 & 3 & 1 \\ 4 & -1 & 2 \end{vmatrix} = 6 \times (3 \times 2 - 1 \times (-1)) - 1 \times (13 \times 2 - 1 \times 4) + 1 \times (13 \times (-1) - 3 \times 4) = 6 \times 7 - 1 \times 22 + 1 \times (-25) = 42 - 22 - 25 = -5$
$|A_2| = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 13 & 1 \\ 1 & 4 & 2 \end{vmatrix} = 1 \times (13 \times 2 - 1 \times 4) - 6 \times (2 \times 2 - 1 \times 1) + 1 \times (2 \times 4 - 13 \times 1) = 1 \times 22 - 6 \times 3 + 1 \times (-5) = 22 - 18 - 5 = -1$
$|A_3| = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & 13 \\ 1 & -1 & 4 \end{vmatrix} = 1 \times (3 \times 4 - 13 \times (-1)) - 1 \times (2 \times 4 - 13 \times 1) + 6 \times (2 \times (-1) - 3 \times 1) = 1 \times 25 - 1 \times (-5) + 6 \times (-5) = 25 + 5 - 30 = 0$
$x = \frac{-5}{-1} = 5$ , $y = \frac{-1}{-1} = 1$ , $z = \frac{0}{-1} = 0$

Answer: $x = 5$ , $y = 1$ , $z = 0$

Exercise 4

Prove that: if the homogeneous linear system $A\vec{x} = \vec{0}$ has a nonzero solution, then $|A| = 0$ .

Reference Answer

Cramer's rulesystem of linear equationsdeterminantcriterion for solutions

Approach: Use Cramer’s rule together with a proof by contradiction.

Steps:

Assume $|A| \neq 0$
By Cramer’s rule, the system has a unique solution
Since the system is homogeneous, the unique solution is the zero solution
This contradicts the assumption that a nonzero solution exists
Therefore $|A| = 0$

Answer: the proof is complete

Exercise 5

Decide which case the system $\begin{cases} x + 2y = 5 \\ 2x + 4y = 10 \end{cases}$ falls into.

Reference Answer

Cramer's rulesystem of linear equationsdeterminantcoefficient matrix

Approach: First check the determinant, then analyse which case you are in.

Steps:

The coefficient matrix: $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$
$|A| = 1 \times 4 - 2 \times 2 = 4 - 4 = 0$
Cramer’s rule does not apply
Look at the two equations: the second equation is twice the first
The two equations are equivalent, so there are infinitely many solutions

Answer: there are infinitely many solutions

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$\det(A)$	Mathematical symbol	determinant of A	The determinant of the coefficient matrix $A$ ; Cramer’s rule requires $\det(A) \neq 0$
$x_j = \dfrac{\det(A_j)}{\det(A)}$	Mathematical symbol	x sub j equals A j over A	Cramer’s rule: the $j$ -th unknown equals the ratio of two determinants
$A_j$	Mathematical symbol	A sub j	The matrix obtained from $A$ by replacing its $j$ -th column with the constant terms $\vec{b}$
$A^*$	Mathematical symbol	A star	The adjugate matrix, used in $A^{-1} = \dfrac{1}{\det(A)}A^*$
$A_{ji}$	Mathematical symbol	A sub j i	The algebraic cofactor in the determinant
$A\vec{x} = \vec{0}$	Mathematical symbol	A x equals zero	The homogeneous system, which has a nonzero solution if and only if $\det(A) = 0$

Gaussian Elimination

The Structure of Solutions and Fundamental Systems of Solutions

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Cramer's Rule

Cramer’s Rule

The Definition of Cramer’s Rule

The Conditions of Cramer’s Rule

The Proof of Cramer’s Rule

Applications of Cramer’s Rule

The Solving Steps

Example

The Advantages and Disadvantages of Cramer’s Rule

Advantages

Disadvantages

Extensions of Cramer’s Rule

Homogeneous Systems

The Non-Square Case

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Summary

Symbols Used in This Article

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