Business Mathematics and Basic Statistics · Ch 11 — Determinants and Cramer's Rule

Solving a Two-Variable System by Cramer's Rule

The formula

Once $D$ , $D_x$ and $D_y$ are found, and provided $D\neq0$ , the solution of the system is

x=\frac{D_x}{D} \qquad\qquad y=\frac{D_y}{D}

Worked reasoning

Consider $2x+y=7$ and $x-y=2$ . Here $a_1=2,b_1=1,c_1=7$ and $a_2=1,b_2=-1,c_2=2$ .

D=\begin{vmatrix}2&1\\1&-1\end{vmatrix}=2(-1)-1(1)=-3

D_x=\begin{vmatrix}7&1\\2&-1\end{vmatrix}=7(-1)-1(2)=-9

D_y=\begin{vmatrix}2&7\\1&2\end{vmatrix}=2(2)-7(1)=-3

So $x=\dfrac{-9}{-3}=3$ and $y=\dfrac{-3}{-3}=1$ .

Note

Always verify by substitution

Once $x$ and $y$ are found, substituting both values back into BOTH original equations is the fastest way to catch an arithmetic slip in $D$ , $D_x$ or $D_y$ — here, $2(3)+1=7$ ✓ and $3-1=2$ ✓, confirming the solution.

Word problems …

Definition 1Cramer's Rule (Two Variables)

For $a_1x+b_1y=c_1$ , $a_2x+b_2y=c_2$ with $D\neq0$ : $x=D_x/D$ an …

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