Recipe

Solve a system by Cramer's rule

Use when: you have a square system $A\mathbf x=\mathbf b$ with $\det A\neq0$ and want one variable in isolation, as a ratio of two determinants, without eliminating the whole system

Goal: every $x_i$ , each computed as $\det A_i(\mathbf b)/\det A$ , with $A_i(\mathbf b)$ built by replacing exactly one column of $A$ with $\mathbf b$

Taught in Cramer's rule and the adjugate

Quick recipe

Write $A$ and $\mathbf b$ , and confirm $\det A\neq0$
Build $A_i(\mathbf b)$ by replacing column $i$ with $\mathbf b$
Compute $\det A_i(\mathbf b)$ for each variable
Divide each by $\det A$ to get that variable
Check by substituting back into the original equations

1
Set up the coefficient matrix and the right-hand-side vector, then compute $\det A$ .

Example

For $x+2y=5,\ 3x-y=1$ : $A=\begin{pmatrix}1&2\\3&-1\end{pmatrix}$ , $\mathbf b=\begin{pmatrix}5\\1\end{pmatrix}$ , $\det A=(1)(-1)-(2)(3)=-1-6=-7\neq0$ .
- $\det A\neq0$ — continue; a unique solution exists and Cramer's rule can find it.
- $\det A=0$ — stop; Cramer's rule doesn't apply, since every ratio below would divide by zero.
Why: every later step divides by $\det A$ , so confirming it's nonzero first is what guarantees Cramer's rule has an answer to give at all.
2
For the variable $x_i$ , copy $A$ but replace its column $i$ with $\mathbf b$ , leaving every other column exactly as it was in $A$ .

Example

Replacing column $1$ (for $x$ ) with $\mathbf b$ : $A_1(\mathbf b)=\begin{pmatrix}5&2\\1&-1\end{pmatrix}$ — column $2$ is untouched. Replacing column $2$ (for $y$ ) with $\mathbf b$ instead: $A_2(\mathbf b)=\begin{pmatrix}1&5\\3&1\end{pmatrix}$ — column $1$ is untouched.

Why: replacing exactly one column with $\mathbf b$ is what isolates that one variable's contribution — every other column still carries the coefficients that pin the other variables in place.
3
Evaluate each column-replaced determinant the same way you computed $\det A$ .

Example

$\det A_1(\mathbf b) = (5)(-1)-(2)(1) = -5-2=-7, \qquad \det A_2(\mathbf b) = (1)(1)-(5)(3) = 1-15=-14$

Why: these are the only two extra numbers, besides $\det A$ itself, that Cramer's rule needs — no elimination on the system ever happens.
4
Compute $x_i = \det A_i(\mathbf b)/\det A$ for each $i$ .

Example

$x=\frac{\det A_1(\mathbf b)}{\det A}=\frac{-7}{-7}=1,\qquad y=\frac{\det A_2(\mathbf b)}{\det A}=\frac{-14}{-7}=2$

Why: this ratio is exactly what the theorem behind Cramer's rule guarantees equals $x_i$ — nothing further to solve for once both determinants are known.
5
Plug every computed value back into every original equation, not just one of them.

Example

$x+2y = 1+2(2) = 5$ ✓, and $3x-y = 3(1)-2 = 1$ ✓ — both original equations hold.

Why: this catches an arithmetic slip in any determinant before it gets reported as the final answer.

Common mistakes

Replacing a ROW of $A$ instead of a column — Cramer's rule always swaps a column for $\mathbf b$ , never a row.
Forgetting to check $\det A\neq0$ before dividing by it.
Building $A_2(\mathbf b)$ by editing the already-modified $A_1(\mathbf b)$ instead of starting fresh from $A$ each time, accidentally leaving $\mathbf b$ sitting in two columns at once.
Dividing $\det A$ by $\det A_i(\mathbf b)$ instead of the other way around.