Cramer’s Rule: Solution of a System of Linear Equations

Cramer’s Rule is used to solve the system of linear equations in two variables using determinants.

Analytical Geometry and Cramer’s Rule

Let us consider the following two linear equations:

i) $a_1x + b_1y + c_1 = 0$

ii) $a_2x + b_2y + c_2 = 0$

The values of x and y which satisfy both the equations can be obtained as follows:

$\text{(x,y)} = \left( \dfrac{ b_1c_2 – b_2c_1 }{ a_1b_2 – a_2 b_1 } , \dfrac{ c_1a_2 – c_2a_1 }{ a_1b_2 – a_2 b_1 } \right) $ – (iii)

provided that $a_1b_2 – a_2 b_1 \neq 0$.

This relation is obtained from Analytical Geometry. It can also be interpreted in the form of determinants of the matrices formed from two linear equations.

Cramer’s Rule suggests to solve a system of linear equation when the constant terms $c_1$ and $c_2$ are on the right-hand side of the equation separately.

Thus, when we put constant terms on the right-hand side of the equation, we observe that $c_1$ changes to $-c_1$ and $c_2$ changes to $-c_2$. Now, re-writing the equation (iii), we get,

$\text{(x,y)} = \left( \dfrac{b_1(-c_2) – b_2(-c_1)}{a_1b_2 – a_2 b_1} , \dfrac{(-c_1)a_2 – (c_2)a_1}{a_1b_2 – a_2 b_1} \right)$

$\text{(x,y)} = \left( \dfrac{c_1b_2 – c_2b_1}{a_1b_2 – a_2 b_1} , \dfrac{a_1c_2 – a_2c_1}{a_1b_2 – a_2 b_1} \right)$ – (iv)

How to form matrix D_x, D_y, and D?

Clearly in equation (iv), we have the determinant value of three matrices. The matrices are as follows:

$(c_1b_2 – c_2b_1) \text{= D}_\text{x =} \left | \displaylines{c_1 & b_1 \\ c_2 & b_2} \right |$

$(a_1c_2 – a_2c_1) \text{= D}_\text{y =} \left | \displaylines{a_1 & c_1 \\ a_2 & c_2} \right |$

$(a_1b_2 – a_2 b_1) \text{= D =} \left | \displaylines{a_1 & b_1 \\ a_2 & b_2} \right |$

These matrices are formed by the following coefficients:

Coeffficent of x	Coefficient of y	Coefficent of Constant Term
$a_1$	$b_1$	$c_1$
$a_2$	$b_2$	$c_2$

The matrix D_x is obtained by taking the coefficients of constant term and y, in order.

The matrix D_y is obtained by taking the coefficients of x and constant term, in order.

The matrix D is obtained by taking the coefficients of x and y, in order.

Since in the beginning, we had a condition

$a_1b_2 – a_2 b_1 \neq 0$

$\implies \text{Matrix D should be non-singular.}$

Cramer’s Rule in Solving a System of Linear Equations

Let us consider the following two linear equations:

i) $a_1x + b_1y = c_1$

ii) $a_2x + b_2y = c_2$

Remember, the equations must always be wrriten in the form of $ax + by = c$ in order to apply Cramer’s rule.

STEP 1: Create a table and write down the coefficients of x, y, and constant terms.

Coeffficent of x	Coefficient of y	Coefficent of Constant Term
$a_1$	$b_1$	$c_1$
$a_2$	$b_2$	$c_2$

STEP 2: Form three matrices: D_x, D_y, and D as follows:

$\text{D}_\text{x =} \left ( \displaylines{c_1 & b_1 \\ c_2 & b_2} \right )$

$\text{D}_\text{y =} \left ( \displaylines{a_1 & c_1 \\ a_2 & c} \right )$

$\text{D =} \left ( \displaylines{a_1 & b_1 \\ a_2 & b_2} \right )$

STEP 3: Check whether matrix D is singular or non-singular. If the matrix is singular, the solution of system of linear equations does not exist. If the matrix is non-singular, the solution of system of linear equations exists.

For the solution to exist:

$\text{D =} \left | \displaylines{a_1 & b_1 \\ a_2 & b_2} \right | = (a_1b_2 – a_2 b_1) \neq 0$

STEP 4: Find the determinant of D_x and D_y as follows:

$ \text{D}_\text{x =} \left | \displaylines{c_1 & b_1 \\ c_2 & b_2} \right | = (c_1b_2 – c_2b_1)$

$ \text{D}_\text{y =} \left | \displaylines{a_1 & c_1 \\ a_2 & c_2} \right | = (a_1c_2 – a_2c_1)$

STEP 5: Solve for x and y as follows:

$(x,y) = \left ( \dfrac{D_x}{D}, \dfrac{D_y}{D} \right)$

Solved Examples

Example 1: If $D = \left | \displaylines{ 1&1\\1&2} \right |$, $D_x = \left | \displaylines{2&1 \\3&2} \right |$ and $D_y = \left | \displaylines{1&2\\1&3} \right |$ then, find the values of x and y.

Solution:

$D = \left | \displaylines{1 & 1 \\ 1& 2} \right|$

$= 1×2- 1×1$

$=  2-1$

$= 1$

$D_x = \left |\displaylines{2&1\\3&2} \right |$

$= 2×2 – 1×3$

$= 4 – 3$

$= 1$

$D_y = \left | \displaylines{1&2\\1&3} \right |$

$= 1×3 – 1×2$

$= 3 – 2$

$= 1$

We know,

$x = \dfrac{ D_x}{D}$

$or, x = \frac{1}{1}$

$\therefore x = 1$

And,

$y = \dfrac{D_y }{D}$

$or, y = \frac{1}{1}$

$\therefore y = 1$

Hence, the required values of x and y are 1 and 1, respectively.

Example 2: Use Cramer’s rule to solve the following equations if the solution exists: $x + y = 7$ , $x – y = 3$.

Solution:

Given,

$x + y = 7$ and $x – y = 3$

Arranging the coefficients of given equations in table,

Coeffficent of x	Coefficient of y	Coefficent of Constant Term
$1$	$1$	$7$
$1$	$-1$	$3$

We know,

The actual determinant can be obtained by the values of coefficient of x and y,

D = $\left | \displaylines{ 1 &1\\1&-1} \right |$

$or, D = 1(-1) – (1)(1)$

$or, D = -1 – 1$

$\therefore D = -2 \neq 0$

Since, the value of D is not equal to 0, the solution of above linear equations exist.

And,

Determinant value of x can be obtained by the values of constant terms c and coefficient of y

$D_x = \left | \displaylines{7&1\\3&-1} \right |$

$or, D_x = 7(-1) – (3)1$

$or, D_x = -7-3$

$\therefore D_x =-10$

Also,

Determinant value of y can be obtained by the values of coefficient of x and constant terms c ,

$D_y = \left | \displaylines{ 1&7 \\1&3} \right | $

$or, D_y = 1(3) – 1(7)$

$or, D_y = 3 – 7$

$or, D_y = -4$

Now,

$x = \dfrac{D_x}{D}$

$= \frac{-10}{-2}$

$= 5$

$y = \dfrac{D_y}{D}$

$= \frac{-4}{-2}$

$= 2$

Hence, the required values of x and y are 5 and 2, respectively.