Cramer’s Rule in R2 Through Elimination #
Adrian Jose Hernandez Vega
Elimination #
Recall the elimination method of solving a system of equations: negating either \(x\) or \(y\) by manipulating the equations and then solving. With enough algebra, any system can be solved via elimination.
Lemma #
Suppose there is a system of two equations:
\[ \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \]To isolate a variable, multiply the first equation by \(a_2\) or \(b_2\), the second by \(a_1\) or \(b_1\), and then subtract the equations:
\[ \begin{cases} a_2(a_1 x + b_1 y = c_1) \\ a_1(a_2 x + b_2 y = c_2) \end{cases} \]Proof: Appending \(a_2\) to the first equation and \(a_1\) to the second aligns the \(x\)-terms so that they cancel when the equations are subtracted. The same logic applies using \(b_1\) and \(b_2\) to cancel \(y\)-terms.
Following this lemma, one can solve explicitly for \(x\) and \(y\):
\[ \begin{aligned} x &= \frac{b_2 c_1 - b_1 c_2}{b_2 a_1 - b_1 a_2} \\ y &= \frac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1} \end{aligned} \]Theorem 1 #
Given neither denominator is zero, the solution to
\[ \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \]is explicitly
\[ (x,y) = \left( \frac{b_2 c_1 - b_1 c_2}{b_2 a_1 - b_1 a_2}, \frac{a_1 c_2 - a_2 c_1}{b_2 a_1 - b_1 a_2} \right) \]Theorem 2 (Cramer’s Rule) #
For an arbitrary system
\[ \begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \]the solution can also be obtained via determinants:
\[ (x,y) = \left( \frac{\det(D_x)}{\det(D)}, \frac{\det(D_y)}{\det(D)} \right) \]where explicitly:
\[ (x,y) = \left( \frac{\det \begin{bmatrix} c_1 & b_1 \\ c_2 & b_2 \end{bmatrix}}{\det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}}, \frac{\det \begin{bmatrix} a_1 & c_1 \\ a_2 & c_2 \end{bmatrix}}{\det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}} \right) \]Proof: Recall that the determinant of a \(2 \times 2\) matrix is the difference of the products of its diagonals. Cramer’s Rule is simply a restatement of the previous theorem using determinants:
\[ \left( \frac{\det \begin{bmatrix} c_1 & b_1 \\ c_2 & b_2 \end{bmatrix}}{\det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}}, \frac{\det \begin{bmatrix} a_1 & c_1 \\ a_2 & c_2 \end{bmatrix}}{\det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}} \right)\]\[= \left( \frac{b_2 c_1 - b_1 c_2}{b_2 a_1 - b_1 a_2}, \frac{a_1 c_2 - a_2 c_1}{b_2 a_1 - b_1 a_2} \right) \]