Two-Variable Systems with Cramer's Rule

Three Determinants for Two Unknowns

A two-variable linear system asks for values of \(x\) and \(y\) that satisfy two equations at once. When the coefficient matrix has a nonzero determinant, Cramer’s rule gives each unknown using a determinant in the denominator and a determinant formed by replacing one coefficient column with the right-hand side. The method is compact, but the column order and the condition on the denominator matter: if that determinant is zero, the formulas do not give a unique solution.

Write a system over \(\mathbb{R}\) as

Its coefficient matrix is \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), and its right-hand-side column is \(\begin{bmatrix}e\\f\end{bmatrix}\). The determinant of a \(2\times2\) matrix \(\begin{bmatrix}r&s\\t&u\end{bmatrix}\) is \(ru-st\). For the coefficient matrix, this gives \(ad-bc\).

The order of entries follows the columns: the first column contains the coefficients of \(x\), and the second contains the coefficients of \(y\). In \(\Delta_x\), the right-hand side goes in the first column; in \(\Delta_y\), it goes in the second. Reversing a column changes the sign of a \(2\times2\) determinant, so keeping that order is essential.

Why the Quotient Formulas Work

Proof. First derive equations that any solution must satisfy. Multiply \(ax+by=e\) by \(d\), multiply \(cx+dy=f\) by \(b\), and subtract the second resulting equation from the first. The \(y\)-terms cancel, leaving $$ (ad-bc)x=ed-bf, $$ so \(\Delta x=\Delta_x\). Next, multiply \(cx+dy=f\) by \(a\), multiply \(ax+by=e\) by \(c\), and subtract the second resulting equation from the first. The \(x\)-terms cancel, leaving $$ (ad-bc)y=af-ec, $$ so \(\Delta y=\Delta_y\). If \(\Delta\ne0\), division gives the displayed formulas, and these equations show that any solution must have those values.

It remains to verify that the values do satisfy both original equations. Using the definitions, $$ a\Delta_x+b\Delta_y =a(ed-bf)+b(af-ec) =ead-abf+abf-bec =e(ad-bc)=e\Delta, $$ and $$ c\Delta_x+d\Delta_y =c(ed-bf)+d(af-ec) =ced-cbf+adf-cde =f(ad-bc)=f\Delta. $$ Therefore, for \(x=\Delta_x/\Delta\) and \(y=\Delta_y/\Delta\), the left sides of the original equations are \(e\) and \(f\), respectively. Existence is verified, and the earlier derivation shows that any solution must equal these values. Thus the solution is unique. \(\square\)

The theorem gives a precise condition for using the quotient formulas: the coefficient determinant must be nonzero. Each determinant in a numerator is computed directly from its displayed matrix; it is not obtained by replacing a row or by changing the order of the equations.

Solving a System Step by Step

What Happens When the Coefficient Determinant Is Zero?

If \(\Delta=0\), the quotients in Cramer’s rule are undefined. This does not by itself tell us whether the system has no solution or many solutions. The following result describes the possibilities for any two-variable system over \(\mathbb{R}\).

Proof. Regard a pair \((u,v)\) as a column of two numbers. We show that when \(\Delta=0\), there is a nonzero column \(\mathbf{z}\) for which \(\mathbf{A}\mathbf{z}=\mathbf{0}\). If the first coefficient row \((a,b)\) is nonzero, choose \(\mathbf{z}=\begin{bmatrix}-b\\a\end{bmatrix}\). This column is nonzero, and $$ \mathbf{A}\mathbf{z} =\begin{bmatrix}-ab+ab\\-bc+ad\end{bmatrix} =\begin{bmatrix}0\\\Delta\end{bmatrix} =\mathbf{0}. $$ If the first row is zero but the second row \((c,d)\) is nonzero, choose \(\mathbf{z}=\begin{bmatrix}-d\\c\end{bmatrix}\). It is nonzero, and both rows of \(\mathbf{A}\mathbf{z}\) are zero. If both rows are zero, \(\mathbf{z}=\begin{bmatrix}1\\0\end{bmatrix}\) is a nonzero column with \(\mathbf{A}\mathbf{z}=\mathbf{0}\).

If the system has a solution \(\mathbf{u}\), then for every real number \(t\), $$ \mathbf{A}(\mathbf{u}+t\mathbf{z}) =\mathbf{A}\mathbf{u}+t\mathbf{A}\mathbf{z} =\begin{bmatrix}e\\f\end{bmatrix}. $$ Because \(\mathbf{z}\ne\mathbf{0}\), distinct values of \(t\) give distinct solutions, so there are infinitely many. If the system has no solution, it has zero solutions. These cases exhaust the possibilities, proving that there cannot be exactly one solution. \(\square\)

Common Mistake: Reversing a Replacement Determinant

A frequent sign error is to calculate the determinant for \(y\) in the wrong order. In the first worked example, the correct replacement matrix is \(\begin{bmatrix}2&8\\1&1\end{bmatrix}\), so \(\Delta_y=2(1)-8(1)=-6\). A mistaken reversal gives \(8(1)-2(1)=6\), producing the incorrect value \(y=6/(-5)=-6/5\). Keeping the replacement column in the second position fixes the sign.

Correction check: Combining the mistaken \(y=-6/5\) with the correct \(x=11/5\) gives \(2(11/5)+3(-6/5)=22/5-18/5=4/5\), not \(8\). The error is caught by substituting into the original equations. Checking both equations is part of solving the system, not an optional extra.

A Reliable Checklist

Cramer’s rule is especially transparent for two variables because each numerator and denominator is a single \(2\times2\) determinant. Its condition is just as important as its formulas: a nonzero coefficient determinant guarantees exactly one solution, while a zero determinant means the quotient method cannot decide the system by division. In the singular case, checking the equations themselves distinguishes inconsistency from infinitely many solutions.