"Cramer's Rule" is another way to solve a system of linear equations with matrices.  It uses a formula to calculate the solution to the system utilizing the definition of determinants.

Cramer's Rule:  Definition

Cramer's Rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, i.e. a square matrix, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the vector of right hand sides of the equations.  

Cramer's Rule:  Formula

Rules for a $2\times 2$ Matrix

Consider the linear system:

$\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{\color{Red}e}\\{\color{Red}f}\end{bmatrix}$

Assume the determinant is non-zero. Then, $x$ and $y$ can be found by Cramer's rule: 

$\displaystyle x=\frac{\begin{vmatrix}{\color{Red}e}&b\\{\color{Red}f}&d\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}=\frac{{\color{Red}e}d-b{\color{Red}f}}{ad-bc}$

And:

$\displaystyle y=\frac{\begin{vmatrix}a&{\color{Red}e}\\c&{\color{Red}f}\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}=\frac{a{\color{Red}f}-{\color{Red}e}c}{ad-bc}$

Rules for a $3 \times 3$ Matrix 

Given:

$\displaystyle \begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}{\color{Red}j}\\{\color{Red}k}\\{\color{Red}l}\end{bmatrix}$

Then the values of $x$, $y$ and $z$ can be found as follows:

$\displaystyle x=\frac{\begin{vmatrix}{\color{Red}j}&b&c\\{\color{Red}k}&e&f\\{\color{Red}l}&h&i\end{vmatrix}}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}} \quad y=\frac{\begin{vmatrix}a&{\color{Red}j}&c\\d&{\color{Red}k}&f\\g&{\color{Red}l}&i\end{vmatrix}}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}} \quad z=\frac{\begin{vmatrix}a&b&{\color{Red}j}\\d&e&{\color{Red}k}\\g&h&{\color{Red}l}\end{vmatrix}}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}$

Using Cramer's Rule

Example 1:  Solve the system using Cramer's Rule:

$\displaystyle \left\{\begin{matrix} 3x+2y & = 10\\ -6x+4y & = 4 \end{matrix}\right.$

In matrix format:

$\displaystyle \begin{bmatrix}3&2\\-6&4\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}10\\4\end{bmatrix}$

$\displaystyle \begin{aligned} x&=\frac{\begin{vmatrix}{\color{Red}e}&b\\{\color{Red}f}&d\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}\\&=\frac{{\color{Red}e}d-b{\color{Red}f}}{ad-bc}\end{aligned} $

$\displaystyle \begin{aligned} x&=\frac{\begin{vmatrix}10&2\\4&4\end{vmatrix}}{\begin{vmatrix}3&2\\-6&4\end{vmatrix}}\\&=\frac{10\cdot 4-2 \cdot 4}{(3 \cdot 4) -[2 \cdot (-6)]}\\&=\frac{32}{24}=\frac{4}{3}\end{aligned}$ 

$\displaystyle \begin{aligned} y&=\frac{\begin{vmatrix}a&{\color{Red}e}\\c&{\color{Red}f}\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}\\ &=\frac{a{\color{Red}f}-{\color{Red}e}c}{ad-bc} \end{aligned}$       

$\displaystyle \begin{aligned} y&=\frac{\begin{vmatrix}3&10\\-6&4\end{vmatrix}}{\begin{vmatrix}3&2\\-6&4\end{vmatrix}}\\ &=\frac{(3 \cdot 4)-[10 \cdot(-6)]}{(3 \cdot 4)-[2 \cdot (-6)]}\\ &=\frac{72}{24}=3 \end{aligned}$

The solution to the system is $(\frac{4}{3}, 3)$.