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Determinants and Cramer's Rule
Algebra Textbooks Boundless Algebra Matrices Determinants and Cramer's Rule
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Concept Version 11
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Cramer's Rule

Cramer's Rule uses determinants to solve for a solution to the equation $Ax=b$, when $A$ is a square matrix.

Learning Objective

  • Use Cramer's Rule to solve for a single variable in a system of linear equations


Key Points

    • Cramer's Rule only works on square matrices that have a non-zero determinant and a unique solution.
    • Consider the linear system $\left\{\begin{matrix} ax+by & ={\color{Red}e}\\ cx+dy & ={\color{Red}f} \end{matrix}\right.$, which in matrix format is $\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{\color{Red}e}\\{\color{Red}f}\end{bmatrix}$.   Assume the determinant is non-zero. Then, $x$ and $y$ and be found by Cramer's rule: $x=\frac{\begin{vmatrix}{\color{Red}e}&b\\{\color{Red}f}&d\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}=\frac{{\color{Red}e}d-b{\color{Red}f}}{ad-bc}$  and $y=\frac{\begin{vmatrix}a&{\color{Red}e}\\c&{\color{Red}f}\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}=\frac{a{\color{Red}f}-{\color{Red}e}c}{ad-bc}$.
    • Cramer's Rule is efficient for solving small systems and can be calculated quite quickly; however, as the system grows, calculating the new determinants can be tedious.

Terms

  • square matrix

    A matrix having the same number of rows as columns.

  • determinant

    The unique scalar function over square matrices which is distributive over matrix multiplication, multilinear in the rows and columns, and takes the value of $1$ for the unit matrix. Its abbreviation is "$\det$".


Full Text

"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 $(\frac{4}{3}, 3)$.

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