Cramer's Rule and Consistency of Linear Equations | NIMCET

Quick answer: For a system of n linear equations in n unknowns with coefficient determinant D, Cramer's rule gives $x = \frac{D_x}{D}$, $y = \frac{D_y}{D}$, $z = \frac{D_z}{D}$, where $D_x$ replaces the x-column by the constants. If $D \ne 0$, the system has a unique solution. If $D = 0$ and some $D_x, D_y, D_z$ is non-zero, there is no solution; if all are 0, there are usually infinitely many (or none). A homogeneous system has a non-trivial solution only when $D = 0$.

This uses determinants and the inverse of a matrix.

Cramer's rule (3 variables)

For

$a_1x + b_1y + c_1z = d_1, \quad a_2x + b_2y + c_2z = d_2, \quad a_3x + b_3y + c_3z = d_3$

let D be the determinant of the coefficients. $D_x$, $D_y$ and $D_z$ are formed by replacing the x-, y- or z-column of D with $d_1, d_2, d_3$.

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

Consistency conditions

For two variables, the same rules apply with $2 \times 2$ determinants. Geometrically, $D = 0$ means the lines are parallel or coincident.

Matrix method

Write the system as $AX = B$.

Homogeneous systems

A system $AX = O$ always has the trivial solution $x = y = z = 0$.

Worked examples

Example 1: Cramer's rule in two variables

Solve $2x + 3y = 8$ and $x - y = -1$.

Step 1: $D = \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = -5$.

Step 2: $D_x = \begin{vmatrix} 8 & 3 \\ -1 & -1 \end{vmatrix} = -5$ and $D_y = \begin{vmatrix} 2 & 8 \\ 1 & -1 \end{vmatrix} = -10$.

Step 3: $x = 1$, $y = 2$.

Check: $2 + 6 = 8$ and $1 - 2 = -1$ ✓.

Example 2: three variables

Solve $x + y + z = 6$, $x - y + z = 2$, $2x + y - z = 1$.

Step 1:

$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2) = 6$

Step 2: $D_x = \begin{vmatrix} 6 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 6(0) - 1(-3) + 1(3) = 6$, so $x = 1$.

Step 3: Subtracting the first two equations gives $2y = 4$, so $y = 2$, and then $z = 3$.

Check: $2(1) + 2 - 3 = 1$ ✓.

Example 3: no solution

For what k does $x + 2y = 3$, $2x + ky = 5$ have no solution?

$D = k - 4 = 0$ gives $k = 4$. Then $D_x = 3k - 10 = 2 \ne 0$, so there is no solution at $k = 4$.

Example 4: non-trivial solution

For what λ does $x + y + z = 0$, $x + 2y + 3z = 0$, $x + 3y + \lambda z = 0$ have a non-trivial solution?

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = (2\lambda - 9) - (\lambda - 3) + (3 - 2) = \lambda - 5$

So $\lambda = 5$.

Summary table

Exam tips

• Most parameter questions only need $D = 0$ or $D \ne 0$.
• For homogeneous systems, "non-trivial solution" always means $D = 0$.
• After finding x by Cramer's rule, solve the remaining variables by elimination to save time.

See all topics in the NIMCET Algebra chapter.

Common mistakes

• Replacing the wrong column when forming $D_x$, $D_y$, $D_z$.
• Concluding "infinitely many solutions" from $D = 0$ alone.
• Forgetting that homogeneous systems are always consistent.
• Sign errors in 3 × 3 expansion.

Frequently Asked Questions

What is Cramer's rule?

For a system with coefficient determinant D ≠ 0, each variable equals Dᵢ/D, where Dᵢ is D with that variable's column replaced by the constants.

When does a system of linear equations have a unique solution?

When the determinant of the coefficients is not zero.

When is a system of linear equations inconsistent?

When D = 0 and at least one of Dₓ, D_y, D_z is non-zero.

When does a homogeneous system have a non-trivial solution?

When the determinant of the coefficient matrix is zero.

What is the matrix method for solving linear equations?

Write the system as AX = B. If |A| ≠ 0, the solution is X = A⁻¹B.