Solve linear systems using determinants. Enter coefficients and constants; get solution via Cramer’s rule with clear steps, previews, and checks.
x.
Enter any real number.
y.
Decimals allowed.
z.
Negative values OK.
x+y+z=6.
x.
Use 0 if term missing.
y.
Can be negative.
z.
Enter any real number.
2x-y+3z=9.
x.
Zero allowed.
y.
Decimals allowed.
z.
Negative values OK.
-x+4y+z=10.
1x + 1y + 1z = 6
2x + (-1)y + 3z = 9
-1x + 4y + 1z = 10
Δ = det(A), Δx = det(A with col x → b), etc. x = Δx/Δ, y = Δy/Δ, z = Δz/Δ
A Cramer’s Rule Calculator solves a square linear system \(A\mathbf{x}=\mathbf{b}\) by expressing each unknown as a ratio of determinants. When the coefficient matrix \(A\in\mathbb{R}^{n\times n}\) has nonzero determinant, the system has a unique solution and Cramer’s formula provides it directly: replace the \(i\)-th column of \(A\) by \(\mathbf{b}\) to form \(A_i\), compute \(\det(A_i)\) and \(\det(A)\), and take \(x_i=\det(A_i)/\det(A)\). This method is exact and transparent—ideal for hand checks, teaching, and small systems (2×2 or 3×3). The calculator automates determinant evaluation, shows each column replacement, simplifies rational numbers, and verifies the solution by substituting back into \(A\mathbf{x}\). If \(\det(A)=0\), it diagnoses singularity and indicates whether the system is inconsistent or has infinitely many solutions (requiring rank analysis or parameterization).
Determinants summarize linear dependence and scaling. The calculator computes \(\det(A)\) and, when feasible, symbolic cofactors to preserve exactness with integers or fractions. For 2×2 and 3×3 systems it uses closed-form determinant formulas; for larger but still modest sizes it uses cofactor expansion or efficient algorithms. It then assembles solution components \(x_i=\det(A_i)/\det(A)\), reports simplified fractions and decimals, and includes a check that \(A\mathbf{x}=\mathbf{b}\). Because Cramer’s rule is numerically and computationally expensive for large \(n\), the tool warns about performance and recommends Gaussian elimination for big systems. It also flags common input issues: non-square matrices, mismatched RHS size, and rows or columns that make \(\det(A)=0\) (e.g., proportional rows). For learning, it highlights the relationship to the adjugate formula \(\mathbf{x}=\operatorname{adj}(A)\mathbf{b}/\det(A)\).
Cramer’s rule (general): \[ A\mathbf{x}=\mathbf{b},\quad \det(A)\ne0,\quad x_i=\frac{\det(A_i)}{\det(A)},\ \ A_i:\ \text{replace column }i\text{ of }A\text{ by }\mathbf{b}. \]
2×2 system: For \(A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\), \(\mathbf{b}=\begin{bmatrix}e\\ f\end{bmatrix}\), \[ \Delta=\det(A)=ad-bc,\quad x=\frac{\det\!\begin{bmatrix}e&b\\ f&d\end{bmatrix}}{\Delta}=\frac{ed-bf}{\Delta},\quad y=\frac{\det\!\begin{bmatrix}a&e\\ c&f\end{bmatrix}}{\Delta}=\frac{af-ec}{\Delta}. \]
3×3 determinant (cofactor form): \[ \det(A)=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13},\quad M_{ij}=\det(\text{minor of }a_{ij}). \]
Adjugate relation: \[ A^{-1}=\frac{\operatorname{adj}(A)}{\det(A)},\qquad \mathbf{x}=A^{-1}\mathbf{b}=\frac{\operatorname{adj}(A)\mathbf{b}}{\det(A)}. \]
\(2x+3y=7,\ -x+4y=5\). \(\Delta=2\cdot4-3(-1)=11\). \(x=\dfrac{\det\!\begin{bmatrix}7&3\\ 5&4\end{bmatrix}}{11}=\dfrac{28-15}{11}=\dfrac{13}{11}\), \quad \(y=\dfrac{\det\!\begin{bmatrix}2&7\\ -1&5\end{bmatrix}}{11}=\dfrac{10+7}{11}=\dfrac{17}{11}\).
\(\begin{cases} x+y+z=6\\ 2x-y+3z=14\\ -x+4y+z=9 \end{cases}\), \(A=\begin{bmatrix}1&1&1\\2&-1&3\\-1&4&1\end{bmatrix}\), \(\det(A)=-11\). \(\det(A_x)=0,\ \det(A_y)=-11,\ \det(A_z)=-55\Rightarrow (x,y,z)=\big(0,\ 1,\ 5\big).\)
\(2x+4y=8,\ x+2y=4\Rightarrow A=\begin{bmatrix}2&4\\1&2\end{bmatrix}\) with \(\det(A)=0\). Columns are proportional \(\Rightarrow\) infinitely many solutions (or none if RHS inconsistent). Cramer’s rule not applicable.
Only for square systems with \(\det(A)\ne0\). If \(\det(A)=0\), use row reduction to analyze solutions.
No. It’s best for small systems (2×2, 3×3). Use elimination or LU for larger \(n\).
That’s normal. The calculator returns exact rational results and decimal approximations.
Substitute \(\mathbf{x}\) back into \(A\mathbf{x}\) and confirm it equals \(\mathbf{b}\); the tool shows this check.
Yes—the same formulas hold using complex determinants and conjugation where appropriate.
Linearly dependent rows/columns (e.g., proportional rows) make \(A\) singular, eliminating uniqueness.