Cramer’s rule is a method of solving $n$ simultaneous equations for $n$ unknowns.
Any system of equations of this kind can be written in the form
\[\begin{array}{c} a_{11} \, x_1 + a_{12} \, x_2 + ... + a_{1n} \, x_n = b_{1} \\ a_{21} \, x_1 + a_{22} \, x_2 + ... + a_{2n} \, x_n = b_{2} \\ \vdots \\ a_{n1} \, x_1 + a_{n2} \, x_n + ... + a_{nn} \, x_n = b_{n} \end{array} \label{eq:neq}\]The coefficients on the left side of Eq.’s 1 can be written as a matrix,
\[A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots &\vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \end{array} \right)\]The value of any of the $x_i$ can be found via
\[x_i = \frac{|B_i|}{|A|} \label{eq:sols}\]where the notation $|A|$ and $|B_i|$ denote the determinants of matrices $A$ and $B_i$, and the matrix $B_i$ is obtained by replacing the $i^{th}$ column of matrix $A$ with the coefficients on the left side of Eq.’s 1 For example,
\[B_2 = \left( \begin{array}{cccc} a_{11} & b_1 & ... & a_{1n} \\ a_{21} & b_2 & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & b_n & ... & a_{nn} \end{array} \right)\]Two equations involving two unknowns can be written in the form
\[\begin{array}{c} a_{11} \, x_1 + a_{12} \, x_2 = b_{1} \\ a_{21} \, x_1 + a_{22} \, x_2 = b_{2} \end{array}\]which yields
\[A = \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)\]Eq. 3 gives
\[x_1 = \frac{ \left| \begin{array}{cc} b_1 & a_{12} \\ b_2 & a_{22} \end{array} \right| }{ \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| } = \frac{b_1 \, a_{22} - a_{12} \, b_2} {a_{11} \, a_{22} - a_{12} \, a_{21}}\]and
\[x_2 = \frac{ \left| \begin{array}{cc} a_{11} & b_1 \\ a_{21} & b_2 \end{array} \right| }{ \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| } = \frac{a_{11} \, b_2 - b_1 \, a_{21}} {a_{11} \, a_{22} - a_{12} \, a_{21}}\]The system of equations
\[\begin{array}{c} 2 \, x_1 - \, x_2 = 1 \\ 5 \, x_1 + 3 \, x_2 = 2 \end{array} \label{eq:2eq}\]has solutions
\[x_1 = \frac{ \left| \begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array} \right| }{ \left| \begin{array}{cc} 2 & -1 \\ 5 & 3 \end{array} \right| } = \frac{5}{11}\]and
\[x_2 = \frac{ \left| \begin{array}{cc} 2 & 1 \\ 5 & 2 \end{array} \right| }{ \left| \begin{array}{cc} 2 & -1 \\ 5 & 3 \end{array} \right| } = -\frac{1}{11}\]The results can be verified by substituting them into either of Eq.’s 9,
\[\frac{10}{11} + \frac{1}{11} = \frac{11}{11} = 1 \hspace{24pt}\checkmark\]The system of equations
\[\begin{array}{c} 2 \, x_1 - ~ \, x_2 + 4 \, x_3 = 2 \\ 5 \, x_1 + 3 \, x_2 + 2 \, x_3 = 1 \\ ~~~ \, x_1 + 6 \, x_2 + ~~ \, x_3 = -3 \\ \end{array} \label{eq:3eq}\]has solutions
\[\nonumber x_1 = \frac{ \left| \begin{array}{ccc} 2 & -1 & 4 \\ 1 & 3 & 2 \\ -3 & 6 & 1 \end{array} \right| }{ \left| \begin{array}{ccc} 2 & -1 & 4 \\ 5 & 3 & 2 \\ 1 & 6 & 1 \end{array} \right| }\] \[x_1 = \frac{ 2 (3 \cdot 1 - 2 \cdot 6) - (-1)(1 \cdot 1 - 2 \cdot -3) + 4 (1 \cdot 6 - 3 \cdot -3) }{ 2 (3 \cdot 1 - 2 \cdot 6) - (-1)(5 \cdot 1 - 2 \cdot 1) + 4 (5 \cdot 6 - 3 \cdot 1) } = \frac{49}{93}\] \[x_2 = \frac{ \left| \begin{array}{ccc} 2 & 2 & 4 \\ 5 & 1 & 2 \\ 1 & -3 & 1 \end{array} \right| }{ \left| \begin{array}{ccc} 2 & -1 & 4 \\ 5 & 3 & 2 \\ 1 & 6 & 1 \end{array} \right| }\] \[= \frac{ 2 (1 \cdot 1 - 2 \cdot -3) - 2 (5 \cdot 1 - 2 \cdot 1) + 4 (5 \cdot -3 - 1 \cdot 1) }{93} = -\frac{56}{93}\]and
\[x_3 = \frac{ \left| \begin{array}{ccc} 2 & -1 & 2 \\ 5 & 3 & 1 \\ 1 & 6 & -3 \end{array} \right| }{ \left| \begin{array}{ccc} 2 & -1 & 4 \\ 5 & 3 & 2 \\ 1 & 6 & 1 \end{array} \right| }\] \[= \frac{ 2 (3 \cdot -3 - 1 \cdot 6) - (-1)(5 \cdot -3 - 1 \cdot 1) + 2 (5 \cdot 6 - 3 \cdot 1) } {93} = \frac{8}{93}\]The results can be verified by substituting them into any of Eq.’s 13,
\[\frac{2 \cdot 49 - (-56) + 4 \cdot 8}{93} = \frac{186}{93} = 2 \hspace{24pt}\checkmark\]This work is licensed under the Creative Commons Attribution-ShareAlike
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L.A. Riley (lriley@ursinus.edu), updated June 2021