Cramer’s Rule

Cramer’s Rule is a method in linear algebra used to solve systems of linear equations with the same number of equations and variables. It uses determinants of matrices to find the values of variables directly without substitution or elimination. The rule is applicable only when the coefficient matrix is square and has a non-zero determinant. Cramer’s Rule is particularly useful for solving small systems and is a foundational technique in mathematics and engineering.

1.0Cramer’s Rule Definition

In linear algebra, Cramer’s Rule is a powerful method used to solve systems of linear equations using determinants. It applies to square systems (number of equations = number of variables) and provides exact values for each variable without the need for elimination or substitution.

2.0What is Cramer’s Rule?

Cramer’s Rule is a technique that uses the determinant of matrices to solve a system of linear equations. It is applicable only when the coefficient matrix is square and has a non-zero determinant.

3.0What is the Cramer’s Rule Method?

The Cramer method involves replacing each column of the coefficient matrix with the constant matrix (RHS) and computing the determinant. Each variable is then found by:

$x_i=\frac{\text{ Determinant of matrix with column }i\text{ replaced by RHS }}{\text{ Determinant of coefficient matrix }}$

4.0Cramer’s Rule Formula

For a system of equations:

$\begin{array}{rl}& a_{1}x+b_{1}y=c_1\\ & a_{2}x+b_{2}y=c_2\end{array}$

The determinant of the coefficient matrix:

$D=\left|\begin{array}{ll}{a}_1& b_1\\ a_2& b_2\end{array}\right|=a_1{b}_2-a_2{b}_1$ 

Replace the respective columns for each variable:

• $D_x=\left|\begin{array}{ll}{c}_1& b_1\\ c_2& b_2\end{array}\right|,$
• $D_y=\left|\begin{array}{ll}{a}_1& c_1\\ a_2& c_2\end{array}\right|$

Then,

$Dx=\frac{D_x}{D},y=\frac{D_y}{D}$

This is the Cramer’s Rule formula for 2 variables.

5.0Why Use the Cramer Method?

• It's a direct algebraic method.
• Suitable for small systems (2 x 2 or 3 x 3).
• No need for row operations or substitution.
• Especially helpful in Maths and engineering problems with clear coefficient matrices.

6.0Applications of Cramer’s Rule

• Linear algebra problems
• Engineering circuits (KCL/KVL)
• Economics (linear optimization)
• Computer graphics (transformation matrices)
• Solving simultaneous equations with exact coefficients

7.0Solved Examples on Cramer’s Rule 

Example 1: 

$\begin{array}{rl}& 2x+3y=8\\ & 4x-y=2\end{array}$ 

Solution: 

Step 1: Find D (Determinant of coefficient matrix)

$D=\left|\begin{array}{cc}2& 3\\ 4& -1\end{array}\right|=(2)(-1)-(4)(3)=-2-12=-14$

Step 2: Find D_x (Replace x-column with constants)

$D_x=\left|\begin{array}{cc}8& 3\\ 2& -1\end{array}\right|=(8)(-1)-(2)(3)=-8-6=-14$

Step 3: Find $D_y$ (Replace y-column with constants)

$D_y=\left|\begin{array}{ll}2& 8\\ 4& 2\end{array}\right|=(2)(2)-(4)(8)=4-32=-28$ 

Step 4: Use Cramer’s Rule

$x=\frac{D_x}{D}=\frac{-14}{-14}=1,y=\frac{D_y}{D}=\frac{-28}{-14}=2$

Solution: x = 1, y = 2

Example 2: Solve using Cramer’s Rule:

$\{\begin{array}{l}2x+3y=8\\ 4x-y=2\end{array}$ 

Solution:

• Coefficient matrix:

$D=\left|\begin{array}{cc}2& 3\\ 4& -1\end{array}\right|=(2)(-1)-(4)(3)=-2-12=-14$

• Replace 1st column with constants for $D_x$:

$D_x=\left|\begin{array}{cc}8& 3\\ 2& -1\end{array}\right|=(8)(-1)-(2)(3)=-8-6=-14$

• Replace 2nd column with constants for $D_y$:

$D_y=\left|\begin{array}{ll}2& 8\\ 4& 2\end{array}\right|=(2)(2)-(4)(8)=4-32=-28$ 

Now,

$x=\frac{D_x}{D}=\frac{-14}{-14}=1,y=\frac{D_y}{D}=\frac{-28}{-14}=2$

Answer: x = 1, y = 2

Example 3: Solve using Cramer’s Rule:

$\{\begin{array}{l}x-\frac{1}{2}y=1\\ 3x+2y=5\end{array}$

Solution:

• Coefficient matrix:

$D=\left|\begin{array}{cc}1& -\frac{1}{2}\\ 3& 2\end{array}\right|=(1)(2)-(3)\left(-\frac{1}{2}\right)=2+\frac{3}{2}=\frac{7}{2}$

• $D_x=\left|\begin{array}{cc}1& -\frac{1}{2}\\ 5& 2\end{array}\right|=(1)(2)-(5)\left(-\frac{1}{2}\right)=2+\frac{5}{2}=\frac{9}{2}{D}_y=\left|\begin{array}{ll}1& 1\\ 3& 5\end{array}\right|=(1)(5)-(3)(1)=5-3=2$

$x=\frac{9/2}{7/2}=\frac{9}{7},y=\frac{2}{7/2}=\frac{4}{7}$

Answer: $x=\frac{9}{7},y=\frac{4}{7}$

Example 4: Solve the system using Cramer’s Rule:

$\{\begin{array}{l}x+y+z=6\\ 2x-y+3z=14\\ x+2y-z=-2\end{array}$ 

Solution:

Step 1: Coefficient Matrix Determinant D

$D=\left|\begin{array}{ccc}1& 1& 1\\ 2& -1& 3\\ 1& 2& -1\end{array}\right|$

Using cofactor expansion:

$\begin{array}{rl}& D=1\cdot \left|\begin{array}{cc}-1& 3\\ 2& -1\end{array}\right|-1\cdot \left|\begin{array}{cc}2& 3\\ 1& -1\end{array}\right|+1\cdot \left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|\\ & =1((-1)(-1)-(3)(2))-1((2)(-1)-(3)(1))+1((2)(2)-(-1)(1))\\ & =1(1-6)-1(-2-3)+1(4+1)\\ & =1(-5)-1(-5)+1(5)\\ & =-5+5+5=5\\ & D=5\end{array}$

Step 2: D_x — Replace 1st column with constants (6, 14, -2)

$\begin{array}{rl}& D_x=\left|\begin{array}{ccc}6& 1& 1\\ 14& -1& 3\\ -2& 2& -1\end{array}\right|\\ & D_x=6\cdot \left|\begin{array}{cc}-1& 3\\ 2& -1\end{array}\right|-1\cdot \left|\begin{array}{cc}14& 3\\ -2& -1\end{array}\right|+1\cdot \left|\begin{array}{cc}14& -1\\ -2& 2\end{array}\right|\\ & D_x=6((-1)(-1)-(3)(2))-(14\cdot -1-3\cdot -2)+(14\cdot 2-(-1)(-2))=6(1-6)-(-14+6)+(28-2)\\ & D_x=6(-5)-(-8)+26\\ & D_x=-30+8+26=4\\ & D_{-}x=4\end{array}$

Step 3: D_y — Replace 2nd column with constants

$\begin{array}{rl}{D}_y& =\left|\begin{array}{ccc}1& 6& 1\\ 2& 14& 3\\ 1& -2& -1\end{array}\right|\\ D_y& =1\cdot \left|\begin{array}{cc}14& 3\\ -2& -1\end{array}\right|-6\cdot \left|\begin{array}{cc}2& 3\\ 1& -1\end{array}\right|+1\cdot \left|\begin{array}{cc}2& 14\\ 1& -2\end{array}\right|\end{array}$

$\begin{array}{rl}& D_y=1(14\cdot -1-3\cdot -2)-6(2\cdot -1-3\cdot 1)+1(2\cdot -2-14\cdot 1)\\ & D_y=1(-14+6)-6(-2-3)+1(-4-14)\\ & D_y=-8+30-18=4\\ & {\mathrm{D}}_{\mathrm{y}}=4\end{array}$

Step 4: D_z — Replace 3rd column with constants

$\begin{array}{rl}& D_z=\left|\begin{array}{ccc}1& 1& 6\\ 2& -1& 14\\ 1& 2& -2\end{array}\right|\\ & D_z=1\cdot \left|\begin{array}{cc}-1& 14\\ 2& -2\end{array}\right|-1\cdot \left|\begin{array}{cc}2& 14\\ 1& -2\end{array}\right|+6\cdot \left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|\\ & D_z=1((-1)(-2)-(14)(2))-1((2)(-2)-(14)(1))+6((2)(2)-(-1)(1))\\ & D_z=1(2-28)-(-4-14)+6(4+1)\\ & D_z=-26+18+30=22\\ & D_z=22\end{array}$

Final Step: Solve for x, y, z

$x=\frac{D_x}{D}=\frac{4}{5},y=\frac{D_y}{D}=\frac{4}{5},z=\frac{D_z}{D}=\frac{22}{5}$

Final Answer:

$x=\frac{4}{5},y=\frac{4}{5},z=\frac{22}{5}$

Example 5: Use Cramer’s Rule to check if this system has a unique solution:

$\{\begin{array}{l}x+2y=4\\ 2x+4y=8\end{array}$ 

Solution:

Coefficient matrix:

$D=\left|\begin{array}{ll}1& 2\\ 2& 4\end{array}\right|=1(4)-2(2)=4-4=0$

Since D = 0, Cramer’s Rule is not applicable. The system is dependent or inconsistent.

Answer: No unique solution (Cramer’s Rule fails)

Example 6: Solve using Cramer’s Rule:

$\{\begin{array}{l}3x-4y=10\\ 5x+2y=8\end{array}$

Solution:

• $D=\left|\begin{array}{cc}3& -4\\ 5& 2\end{array}\right|=(3)(2)-(5)(-4)=6+20=26$
• $D_x=\left|\begin{array}{cc}10& -4\\ 8& 2\end{array}\right|=(10)(2)-(8)(-4)=20+32=52$
• $D_y=\left|\begin{array}{cc}3& 10\\ 5& 8\end{array}\right|=(3)(8)-(5)(10)=24-50=-26$

$x=\frac{52}{26}=2,y=\frac{-26}{26}=-1$

Answer: x = 2, y = -1

Q7: What is the Cramer formula?

Ans: $x_i=\frac{D_i}{D}$ 

Where $D_i$ is the determinant after replacing column ii with the constant column.

8.0Practice Questions on Cramer’s Rule

1. Solve: 3x + 4y = 10, 5x − 2y = 8  
2. Use Cramer’s Rule to solve: 2x + y + z = 1, x + 3y – z = 4, 3x – y + 2z = 5  
3. Check if the following system can be solved using Cramer’s Rule:

x + 2y = 3, 2x + 4y = 6   

4. Find the determinant of:

$\left|\begin{array}{ccc}1& -2& 3\\ 4& 0& 5\\ 2& 1& 6\end{array}\right|$