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Difference between revisions of "Cramer's Rule"

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<cmath>x_1 = \frac{|M_1|}{|A|} = \frac{18}{18}=1 \qquad x_2 = \frac{|M_2|}{|A|} = \frac{36}{18} = 2 \qquad x_3 = \frac{|M_3|}{|A|} = \frac{54}{18} = 3</cmath>
 
<cmath>x_1 = \frac{|M_1|}{|A|} = \frac{18}{18}=1 \qquad x_2 = \frac{|M_2|}{|A|} = \frac{36}{18} = 2 \qquad x_3 = \frac{|M_3|}{|A|} = \frac{54}{18} = 3</cmath>
  
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Revision as of 12:18, 14 July 2021

Cramer's Rule is a method of solving systems of equations using matrices.

General Form for n variables

Cramer's Rule employs the matrix determinant to solve a system of n linear equations in n variables.

We wish to solve the general linear system $A \mathbf{x}= \mathbf{b}$ for the vector $\mathbf{x} = \left( \begin{array}{c} x_1 \\ \vdots \\ x_n \end{array} \right)$. Here, $A$ is the coefficient matrix, $\mathbf{b}$ is a column vector.

Let $M_j$ be the matrix formed by replacing the jth column of $A$ with $\mathbf{b}$.

Then, Cramer's Rule states that the general solution is $x_j = \frac{|M_j|}{|A|} \; \; \; \forall j \in \mathbb{N}^{\leq n}$

General Solution for 2 Variables

Consider the following system of linear equations in $x$ and $y$, with constants $a, b, c, d, r, s$:

\begin{eqnarray*} ax + cy &=& r\\ bx + dy &=& s \end{eqnarray*}

By Cramer's Rule, the solution to this system is:

$x = \frac{\begin{vmatrix} r & c \\ s & d \end{vmatrix}} {\begin{vmatrix} a & c \\ b & d \end{vmatrix}} = \frac{rd - sc}{ad - bc} \qquad y = \frac{\begin{vmatrix} a & r \\ b & s \end{vmatrix}} {\begin{vmatrix} a & c \\ b & d \end{vmatrix}} = \frac{sa - rb}{ad - cb}$

Example in 3 Variables

\begin{eqnarray*} x_1+2x_2+3x_3&=&14\\ 3x_1+x_2+2x_3&=&11\\ 2x_1+3x_2+x_3&=&11 \end{eqnarray*}

Here, $A = \left( \begin{array}{ccc} 1 & 2 & 3\\ 3 & 1 & 2\\ 2 & 3 & 1 \end{array} \right) \qquad \mathbf{b} = \left( \begin{array}{c} 14\\ 11\\ 11 \end{array} \right)$

Thus, \[M_1 = \left( \begin{array}{ccc} 14 & 2 & 3\\ 11 & 1 & 2\\ 11 & 3 & 1 \end{array} \right) \qquad M_2 = \left( \begin{array}{ccc} 1 & 14 & 3\\ 3 & 11 & 2\\ 2 & 11 & 1 \end{array} \right) \qquad M_3 = \left( \begin{array}{ccc} 1 & 2 & 14\\ 3 & 1 & 11\\ 2 & 3 & 11 \end{array} \right)\]

We calculate the determinants: \[|A| = 18 \qquad |M_1| = 18 \qquad |M_2| = 36 \qquad |M_3| = 54\]

Finally, we solve the system: \[x_1 = \frac{|M_1|}{|A|} = \frac{18}{18}=1 \qquad x_2 = \frac{|M_2|}{|A|} = \frac{36}{18} = 2 \qquad x_3 = \frac{|M_3|}{|A|} = \frac{54}{18} = 3\]

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