Cramer's Rule

Cramer’s Rule is a method for solving systems of linear equations using determinants. Instead of performing row operations or finding an inverse matrix, you compute several determinants and take their ratios. The formula gives each variable directly — no back-substitution needed.

Cramer’s Rule is named after Swiss mathematician Gabriel Cramer, who published it in 1750. While it is not the most efficient method for large systems, it is elegant, self-contained, and particularly useful for systems of 2 or 3 equations where you need only one of the variables.

Cramer’s Rule for 2x2 Systems

Consider the system:

$a_1 x + b_1 y = c_1$

$a_2 x + b_2 y = c_2$

The coefficient matrix is $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$ with determinant $D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$.

Cramer’s Rule states:

$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$

where:

• $D = \det(A)$ — the determinant of the coefficient matrix
• $D_x$ — replace the $x$-column (first column) of $A$ with the constants: $D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}$
• $D_y$ — replace the $y$-column (second column) of $A$ with the constants: $D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$

Memory aid: To find a variable, replace its column in the coefficient matrix with the constants column, compute the determinant, and divide by the original determinant $D$.

Example 1: Cramer’s Rule for a 2x2 System

Solve:

$3x + 2y = 16$

$x - y = 2$

Step 1 — Compute $D$ (determinant of the coefficient matrix):

$D = \begin{vmatrix} 3 & 2 \\ 1 & -1 \end{vmatrix} = 3(-1) - 2(1) = -3 - 2 = -5$

Since $D = -5 \neq 0$, the system has a unique solution and Cramer’s Rule applies.

Step 2 — Compute $D_x$ (replace the $x$-column with the constants):

$D_x = \begin{vmatrix} 16 & 2 \\ 2 & -1 \end{vmatrix} = 16(-1) - 2(2) = -16 - 4 = -20$

Step 3 — Compute $D_y$ (replace the $y$-column with the constants):

$D_y = \begin{vmatrix} 3 & 16 \\ 1 & 2 \end{vmatrix} = 3(2) - 16(1) = 6 - 16 = -10$

Step 4 — Divide:

$x = \frac{D_x}{D} = \frac{-20}{-5} = 4, \quad y = \frac{D_y}{D} = \frac{-10}{-5} = 2$

Answer: $x = 4$, $y = 2$

Verification: $3(4) + 2(2) = 12 + 4 = 16$ and $4 - 2 = 2$.

Example 2: Another 2x2 System

Solve:

$2x + 5y = 1$

$3x + 7y = 2$

Compute the determinants:

$D = \begin{vmatrix} 2 & 5 \\ 3 & 7 \end{vmatrix} = 14 - 15 = -1$

$D_x = \begin{vmatrix} 1 & 5 \\ 2 & 7 \end{vmatrix} = 7 - 10 = -3$

$D_y = \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 4 - 3 = 1$

$x = \frac{-3}{-1} = 3, \quad y = \frac{1}{-1} = -1$

Answer: $x = 3$, $y = -1$

Verification: $2(3) + 5(-1) = 6 - 5 = 1$ and $3(3) + 7(-1) = 9 - 7 = 2$.

Cramer’s Rule for 3x3 Systems

For a system of three equations in three unknowns:

$a_1 x + b_1 y + c_1 z = d_1$

$a_2 x + b_2 y + c_2 z = d_2$

$a_3 x + b_3 y + c_3 z = d_3$

Cramer’s Rule states:

$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}$

where:

• $D$ is the determinant of the $3 \times 3$ coefficient matrix
• $D_x$ replaces the first column with the constants $[d_1, d_2, d_3]^T$
• $D_y$ replaces the second column with the constants
• $D_z$ replaces the third column with the constants

Example 3: Cramer’s Rule for a 3x3 System

Solve:

$x + y + z = 6$

$2x - y + z = 3$

$x + 2y - z = 5$

Step 1 — Compute $D$:

$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix}$

Expand along the first row:

$D = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1) = -1 + 3 + 5 = 7$

Step 2 — Compute $D_x$ (replace column 1 with constants):

$D_x = \begin{vmatrix} 6 & 1 & 1 \\ 3 & -1 & 1 \\ 5 & 2 & -1 \end{vmatrix}$

$D_x = 6(1 - 2) - 1(-3 - 5) + 1(6 + 5) = 6(-1) - 1(-8) + 1(11) = -6 + 8 + 11 = 13$

Step 3 — Compute $D_y$ (replace column 2 with constants):

$D_y = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 3 & 1 \\ 1 & 5 & -1 \end{vmatrix}$

$D_y = 1(-3 - 5) - 6(-2 - 1) + 1(10 - 3) = -8 + 18 + 7 = 17$

Step 4 — Compute $D_z$ (replace column 3 with constants):

$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 2 & -1 & 3 \\ 1 & 2 & 5 \end{vmatrix}$

$D_z = 1(-5 - 6) - 1(10 - 3) + 6(4 + 1) = -11 - 7 + 30 = 12$

Step 5 — Divide:

$x = \frac{13}{7}, \quad y = \frac{17}{7}, \quad z = \frac{12}{7}$

Answer: $x = \frac{13}{7}$, $y = \frac{17}{7}$, $z = \frac{12}{7}$

Verification: $\frac{13}{7} + \frac{17}{7} + \frac{12}{7} = \frac{42}{7} = 6$. The other equations can be verified similarly.

Notice this is the same system from the Gaussian elimination lesson — both methods produce the same answer.

When Cramer’s Rule Fails

Cramer’s Rule requires $D \neq 0$. When $D = 0$:

• The system is either inconsistent (no solution) or dependent (infinitely many solutions)
• Cramer’s Rule cannot distinguish between these two cases — it simply fails
• You must use Gaussian elimination to determine which case applies

Example 4: $D = 0$ — Cramer’s Rule Does Not Apply

Solve:

$x + 2y = 3$

$2x + 4y = 6$

$D = \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = 4 - 4 = 0$

Cramer’s Rule fails. To determine why, note that the second equation is exactly twice the first — the system is dependent with infinitely many solutions. But if the second equation were $2x + 4y = 7$ instead, $D$ would still be 0, yet the system would be inconsistent (no solution). Cramer’s Rule cannot tell you which.

Comparison to Other Methods

When to choose Cramer’s Rule:

• The system is $2 \times 2$ or $3 \times 3$ and you want a direct formula
• You only need the value of one variable (compute just $D$ and that variable’s determinant)
• You are on a timed test and want to avoid row operations

When NOT to use Cramer’s Rule:

• The system has 4 or more variables (the number of determinants grows rapidly)
• $D = 0$ (the method fails entirely)
• You need to solve many systems with the same coefficient matrix (the inverse method is faster)

Real-World Application: Electrical Circuit with Three Loops

An electrician analyzing a three-loop circuit writes Kirchhoff’s voltage equations:

$5I_1 - 2I_2 = 10$

$-2I_1 + 8I_2 - 3I_3 = 0$

$-3I_2 + 6I_3 = -5$

To find the current $I_1$ without solving the entire system, compute only $D$ and $D_{I_1}$:

$D = \begin{vmatrix} 5 & -2 & 0 \\ -2 & 8 & -3 \\ 0 & -3 & 6 \end{vmatrix}$

Expanding along the first row (taking advantage of the zero in position $(1,3)$):

$D = 5(48 - 9) - (-2)(-12 - 0) + 0 = 5(39) + (-2)(12) = 195 - 24 = 171$

$D_{I_1} = \begin{vmatrix} 10 & -2 & 0 \\ 0 & 8 & -3 \\ -5 & -3 & 6 \end{vmatrix}$

$D_{I_1} = 10(48 - 9) - (-2)(0 - 15) + 0 = 10(39) + 2(-15) = 390 - 30 = 360$

$I_1 = \frac{360}{171} = \frac{40}{19} \approx 2.11 \text{ amps}$

The electrician found $I_1$ without ever computing $I_2$ or $I_3$. This is the key advantage of Cramer’s Rule — targeted answers.

Common Mistakes

1. Replacing the wrong column. To find $x$, replace the first column (the $x$-coefficients) with the constants. To find $y$, replace the second column. Replacing the wrong column gives the wrong variable’s value.
2. Forgetting to check $D \neq 0$. Always compute the main determinant $D$ first. If $D = 0$, stop — Cramer’s Rule does not apply.
3. Sign errors in 3x3 determinants. Each cofactor expansion involves a sign from the checkerboard pattern. One wrong sign propagates through the entire calculation.
4. Using Cramer’s Rule for large systems. A $4 \times 4$ system requires five $4 \times 4$ determinants. A $5 \times 5$ system requires six $5 \times 5$ determinants. The work grows factorially — use Gaussian elimination instead.

Practice Problems

Key Takeaways

• Cramer’s Rule solves a system by computing determinants: each variable equals the ratio of a modified determinant to the original determinant $D$
• To find a specific variable, replace that variable’s column in the coefficient matrix with the constants column, compute the determinant, and divide by $D$
• Cramer’s Rule works only when $D \neq 0$ — when $D = 0$, it fails and cannot distinguish between inconsistent and dependent systems
• It is most practical for 2x2 and 3x3 systems and particularly useful when you need only one variable
• For larger systems, Gaussian elimination or the inverse method is more efficient
• Always compute $D$ first — if it is zero, switch to another method

Return to College Algebra for more topics in this section.