2.4. Cramer’s rule#

Gabriel Cramer

Fig. 2.1 Gabriel Cramer (1704 - 1752)#

Cramer’s rule, named after Swiss mathematician Gabriel Cramer, is an explicit rule for calculating the solution to a system of linear equations using determinants. We saw in the section on determinants that the solution to the system of linear equations

\[\begin{split} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} e \\ f \end{pmatrix}\end{split}\]

is

\[ \begin{align*} x_1 &= \frac{de - bf}{ad - bc}, & \qquad x_2 &= \frac{af - ce}{ad - bc}. \end{align*} \]

We can recognise that the solution to both variables includes the determinant of the coefficient matrix, \(ad - bc\), in the denominator. But what about the numerator? If we consider the solution to \(x_1\) then in \(de - bf\) we can see that the constant values \(e\) and \(f\) are included and we have the subtraction of two products which is similar to the determinant of a \(2 \times 2\) matrix, i.e.,

\[\begin{split} \begin{align*} de - bf = \begin{vmatrix} e & b \\ f & d \end{vmatrix}. \end{align*} \end{split}\]

Doing similar for the solution to \(x_2\) we see that the numerator is

\[\begin{split} \begin{align*} af - ce = \begin{vmatrix} a & e \\ c & f \end{vmatrix}. \end{align*} \end{split}\]

These determinants are simply the coefficient matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) with columns 1 and 2 replaced by the constant vector \(\begin{pmatrix} e \\ f \end{pmatrix}\) for \(x_1\) and \(x_2\) respectively. This can be extended to larger systems to give us Cramer’s rule.

Theorem 2.2 (Cramer’s rule)

The solution to a non-singular linear system of equations of the form \(A\vec{x}=\vec{b}\) can be calculated using Cramer’s rule which is

(2.1)#\[ x_i = \frac{\det(A_i)}{\det(A)}, \]

where \(A_i\) is a matrix obtained by replacing column \(i\) of \(A\) with \(\vec{b}\).

Proof. The solution to a system of linear equations \(A \vec{x} = \vec{b}\) can be calculated using \(\vec{x} = A^{-1} \vec{b}\) where \(A^{-1}\) is the inverse of the coefficient matrix \(A\). The adjoint-determinant formula for calculating the inverse is

\[ A^{-1} = \frac{\operatorname{adj}(A)}{\det(A)}, \]

so

\[ \vec{x} = \frac{1}{\det(A)} \operatorname{adj}(A) \vec{b}. \]

Since \(\operatorname{adj}(A) = C^\mathsf{T}\) where \(C\) is the matrix of co-factors

\[\begin{split} C^\mathsf{T} = \begin{pmatrix} C_{11} & C_{21} & \cdots & C_{n1} \\ C_{12} & C_{22} & \cdots & C_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ C_{1n} & C_{2n} & \cdots & C_{nn} \end{pmatrix}. \end{split}\]

then for the \(i\)th element of \(\vec{x}\) using the definition of matrix multiplication we have

\[ x_i = \frac{1}{\det(A)} \sum_{j=1}^n C_{ji} b_j \]

In Cramer’s rule, \(A_i\) is the matrix formed by replacing the \(i\)th column of \(A\) with \(\vec{b}\)

\[\begin{split} A_i = \begin{pmatrix} a_{11} & \cdots & a_{1,i-1} & b_1 & a_{1,i+1} & \cdots & a_{1n} \\ a_{21} & \cdots & a_{2,i-1} & b_2 & a_{2,i+1} & \cdots & a_{2n} \\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{n,i-1} & b_n & a_{n,i+1} & \cdots & a_{nn} \end{pmatrix}. \end{split}\]

Since removing the \(i\)th column from both \(A\) and \(A_i\) (the one with the \(b\) values) results in the same matrix then the cofactors of \(A_i\) are the same as the cofactors of \(A\). If we calculate \(\det(A_i)\) by expanding along the \(i\)th column of \(A_i\) then

\[ \det(A_i) = b_1 C_{1i} + b_2 C_{2i} + \cdots + b_n C_{ni} = \sum_{j=1}^n C_{ji} b_j, \]

so \(x_i = \dfrac{\det(A_i)}{\det(A)}\).

Example 2.2

Solve the following systems of linear equations using Cramer’s rule

(i)   \(\begin{array}{rl} 3x_1 - 2x_2 \!\!\!\! &= -4, \\ x_1 - 3x_2 \!\!\!\! &= 1. \end{array}\)

Solution

Here \(A = \begin{pmatrix} 3 & -2 \\ 1 & -3 \end{pmatrix}\) and \(\vec{b} = \begin{pmatrix} -4 \\ 1 \end{pmatrix}\)

\[\begin{split} \begin{align*} x_1 &= \frac{ \begin{vmatrix} -4 & -2 \\ 1 & -3 \end{vmatrix}} {\begin{vmatrix} 3 & -2 \\ 1 & -3 \end{vmatrix}} = \frac{14}{-7} = -2, \\ \\ x_2 &= \frac{\begin{vmatrix} 3 & -4 \\ 1 & 1 \end{vmatrix}}{-7} = \frac{7}{-7} = -1. \end{align*} \end{split}\]

Checking the solution

\[\begin{split} \begin{align*} A\vec{x} = \begin{pmatrix} 3 & -2 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} = \vec{b} \qquad \checkmark \end{align*} \end{split}\]

(ii)   \(\begin{array}{rl} -2x_1 - 3x_2 - x_3 \!\!\!\! &= -5, \\ -4x_1 + 4x_2 + 3x_3 \!\!\!\! &= -20, \\ -3x_1 \!\!\!\! &= -12. \end{array}\)

Solution

Here \(A = \begin{pmatrix} -2 & -3 & -1 \\ -4 & 4 & 3 \\ -3 & 0 & 0 \end{pmatrix}\) and \(\vec{b} = \begin{pmatrix} -5 \\ -20 \\ -12 \end{pmatrix}\)

\[\begin{split} \begin{align*} x_1 &= \frac{ \begin{vmatrix}-5 & -3 & -1 \\ -20 & 4 & 3 \\ -12 & 0 & 0 \end{vmatrix}} {\begin{vmatrix} -2 & -3 & -1 \\ -4 & 4 & 3 \\ -3 & 0 & 0 \end{vmatrix}} = \frac{-12 \begin{vmatrix} -3 & -1 \\ 4 & 3 \end{vmatrix}} {-3\begin{vmatrix} -3 & -1 \\ 4 & 3 \end{vmatrix}} \\ &= \frac{-12(-5)}{-3(-5)} = \frac{60}{15} = 4,\\ \\ x_2 &= \frac{ \begin{vmatrix} -2 & -5 & -1 \\ -4 & -20 & 3 \\ -3 & -12 & 0 \end{vmatrix}}{15} = \frac{-\begin{vmatrix} -4 & -20 \\ -3 & -12 \end{vmatrix} - 3\begin{vmatrix}-2 & -5 \\ -3 & -12 \end{vmatrix}}{15} \\ &= \frac{12 - 3(9)}{15} = \frac{-15}{15} = -1, \\ \\ x_3 &= \frac{ \begin{vmatrix} -2 & -3 & - 5 \\ -4 & 4 & -20 \\ -3 & 0 & -12 \end{vmatrix}}{15} =\frac{-3\begin{vmatrix} -3 & -5 \\ 4 & -20 \end{vmatrix} - 12\begin{vmatrix} -2 & -3 \\ -4 & 4 \end{vmatrix}}{15} \\ & = \frac{-3(80) - 12(-20)}{15} = \frac{0}{15} = 0. \end{align*} \end{split}\]

Checking the solution

\[\begin{split} \begin{align*} A\vec{x} = \begin{pmatrix} -2 & -3 & -1 \\ -4 & 4 & 3 \\ -3 & 0 & 0 \end{pmatrix} \begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -5 \\ -20 \\ -12 \end{pmatrix} = \vec{b} \qquad \checkmark \end{align*} \end{split}\]