Cramer's Rule — Solve Systems Using Determinants

Cramer's Rule solves a square system of linear equations using determinants. The whole method rests on one formula: for the variable $x_i$,

$$x_i = \frac{\det(A_i)}{\det(A)}$$

where $A$ is the coefficient matrix and $A_i$ is the matrix formed by replacing the $i$th column of $A$ with the constants from $b$. If the system is written as $Ax = b$ with $A$ square and $\det(A) \ne 0$, the system has a unique solution and this formula finds each variable directly.

Why The Formula Holds

The symbols mean: $\det(A)$ is the determinant of the original coefficient matrix, and each $\det(A_i)$ is the determinant after one column has been replaced by the constants column $b$. The condition $\det(A) \ne 0$ is essential, because the formula divides by $\det(A)$.

This is why the determinant doubles as the test for solvability. If $\det(A) \ne 0$, the system has one unique solution. If $\det(A) = 0$, the denominator is zero, so the rule does not apply: there may be no solution or infinitely many. Cramer's Rule is learned precisely because it connects three ideas cleanly: solving linear systems, determinants, and the condition for a unique solution. Use it only when the system has the same number of equations and unknowns, the coefficient matrix is square, and $\det(A) \ne 0$.

Worked Example: A $2 \times 2$ System Step By Step

Solve

$$\begin{cases} 2x + y = 5 \\ x - y = 1 \end{cases}$$

First identify the coefficient matrix and the constants column:

$$A = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}, \qquad b = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$$

Compute the determinant of $A$:

$$\det(A) = \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = 2(-1) - 1(1) = -3$$

Because $\det(A) = -3 \ne 0$, the system has a unique solution, so Cramer's Rule applies.

Find $x$

Replace the first column of $A$ with $b$:

$$A_x = \begin{bmatrix} 5 & 1 \\ 1 & -1 \end{bmatrix}$$

Then

$$\det(A_x) = \begin{vmatrix} 5 & 1 \\ 1 & -1 \end{vmatrix} = 5(-1) - 1(1) = -6$$

Divide by the original determinant:

$$x = \frac{\det(A_x)}{\det(A)} = \frac{-6}{-3} = 2$$

Find $y$

Replace the second column of $A$ with $b$:

$$A_y = \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix}$$

Then

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

Divide by $\det(A)$:

$$y = \frac{\det(A_y)}{\det(A)} = \frac{-3}{-3} = 1$$

So the solution is

$$(x,y) = (2,1)$$

That is the full pattern: one determinant for the original matrix, then one more determinant for each variable.

Practice And The Traps To Avoid

Try the same routine on

$$\begin{cases} 3x + 2y = 8 \\ x - y = 0 \end{cases}$$

Compute $\det(A)$ first; if it is nonzero, replace one column at a time and solve for $x$ and $y$. After finishing by hand, compare your setup with a matrix solver to check the determinants as well as the final answer.

Three calculation traps recur:

• Using it when $\det(A) = 0$. This is the main check. The method divides by $\det(A)$, so a zero determinant means it does not apply for a unique solution.
• Replacing the wrong column. To solve for $x$, replace the $x$-column; for $y$, replace the $y$-column. The constants column replaces one column at a time, it is not appended.
• Treating it as the best method for every system. For larger systems, row reduction or numerical methods are usually more practical.

When Cramer's Rule Is Used

You will see it in algebra and linear algebra courses when the goal is understanding rather than speed, especially when you want to show how each variable depends on the coefficients and constants. It is most comfortable for $2 \times 2$ and sometimes $3 \times 3$ systems. Beyond that, the determinant work grows quickly, so it stops being the default.