Cramer’s Rule: Definition, Properties, Formula and Examples

In linear algebra, Cramer's rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column with the column vector of the right sides of the equations. In real life, we use Cramer's Rule to solve the system of linear equations which helps us to solve age-related problems and time-related problems.

In this article, we will cover the concept of Cramer's Rule. This category falls under the broader category of Matrices, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Questions based on this topic have been asked frequently in JEE Mains

System of Linear Equation

A system of linear equations are group of $n$ linear equations containing $n$ number of variables.

1. System of 2 Linear Equations:

It is a pair of linear equations in two variables. It is usually of the form

$a_1x +b_1y + c_1 = 0$

$a_2x +b_2y + c_2 = 0$

Finding a solution for this system means finding the values of $x$ and $y$ that satisfy both equations.

2. System of 3 Linear Equations:

It is a group of 3 linear equations in three variables. It is usually of the form

$a_1x +b_1y + +c_1z + d_1 = 0$

$a_2x +b_2y + +c_2z + d_2 = 0$

$a_3x +b_3y + +c_3z + d_3 = 0$

Finding a solution for this system means finding the values of $x, y$, and $z$ that satisfy all three equations.

The system of equations is broadly classified into two types:

Methods to solve Systems of Linear Equations

We use the following method to solve a System of linear equations in three variables

• Cramers Rule
• Inverse method
• Gaussian elimination method
• Gaussian Jordan method
• LU Decomposition method

Cramer’s law

For the system of equations in two variables:

Let $a_1 x+b_1 y=c_1$ and $a_2 x+b_2 y=c_2$, where

$
\frac{\mathrm{a}_1}{\mathrm{a}_2} \neq \frac{\mathrm{b}_1}{\mathrm{~b}_2}
$

On solving this equation by cross multiplication, we get

$
\begin{aligned}
& \frac{x}{b_2 c_1-b_1 c_2}=\frac{y}{a_1 c_2-a_2 c_1}=\frac{1}{a_1 b_2-a_2 b_1} \\
& \text { or } \frac{\mathrm{x}}{\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|}=\frac{\mathrm{y}}{\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|}=\frac{1}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|} \\
& \text { or } \mathrm{x}=\frac{\left|\begin{array}{ll}
c_1 & b_1 \\
c_2 & b_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}, \mathrm{y}=\frac{\left|\begin{array}{ll}
a_1 & c_1 \\
a_2 & c_2
\end{array}\right|}{\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|}
\end{aligned}
$

We can observe that the first column in the numerator of x is of constants and 2nd column in the numerator of $y$ is of constants, and the denominator is of the coefficient of variables.

We can follow this analogy for the system of equations of 3 variables where the third column in the numerator of the value of $z$ will be constant and the denominator will be formed by the value of coefficients of the variables.

For the system of equations in three variables:

Let us consider the system of equations

$
\begin{aligned}
& \mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_{1 \mathrm{z}}=\mathrm{d}_1 \\
& \mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2 \mathrm{z}=\mathrm{d}_2 \\
& \mathrm{a}_3 \mathrm{x}+\mathrm{b}_3 \mathrm{y}+\mathrm{c}_3 \mathrm{z}=\mathrm{d}_3
\end{aligned}
$

then $\Delta$, which will be the determinant of the coefficient of variables, will be $\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$
$\Delta_1$ numerator of $x$ is :
$\Delta_1=\left|\begin{array}{lll}d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3\end{array}\right|$
Similarly $\Delta_2=\left|\begin{array}{lll}a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3\end{array}\right|$ and $\Delta_3=\left|\begin{array}{lll}a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3\end{array}\right|$

i) If $\Delta \neq0$, then the system of equations has a unique finite solution and so equations are consistent, and solutions are

$\\\mathrm{x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

ii) If $\Delta =0$ , and any of

$\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

The system of equations is consistent and it has an infinite number of solutions (except when all three equations represent parallel planes, in which case there is no solution)

Solved Examples Based on Cramer’s Rule

Example 1: If the system of linear equations

Solution:

For infinitely many solutions :

$\Delta = 0= \Delta _{1}= \Delta _{2}= \Delta _{3}$

$\Delta = 0\Rightarrow \begin{vmatrix} 2 & 1 & -1\\ 1& -1 & -1\\ 3& 3 & \beta \end{vmatrix}= 0$
$\Rightarrow 2\left ( -\beta +3 \right )-1\left ( \beta +3 \right )-1\left ( 3+3 \right )= 0$

$\Rightarrow -3\beta -3= 0\Rightarrow \beta = -1$

$\Delta_{1} = 0\Rightarrow \begin{vmatrix} 3 & 1 & -1\\ \alpha & -1 & -1\\ 3& 3 & -1 \end{vmatrix}= 0$

$\Rightarrow 3\left ( 1+3 \right )-1\left ( -\alpha +3 \right )-1\left ( 3\alpha +3 \right )= 0$

$\Rightarrow 12+\alpha -3-3\alpha -3= 0$

$\Rightarrow 2\alpha= 6\Rightarrow \alpha = 3$

$\alpha = 3,\beta = -1$

$\alpha +\beta -\alpha \beta -3-1+3= 5$

Example 2: Two pairs of dice are thrown. The number on them is taken as $\lambda \, and\, \mu ,$ and a system of linear equations

Solution:

For unique solution $\Delta \neq 0$

$\Rightarrow \begin{vmatrix} 1& 1 &1 \\ 1 &2 &3 \\ 1& 3& \lambda \end{vmatrix}\neq0$

$\Rightarrow \lambda\neq 5$
$\therefore p= \frac{5}{6}$

For no solution

$\Delta = 0 \Rightarrow \lambda= 5 \: \: and$

$\Delta_{1} \neq 0 \Rightarrow \begin{vmatrix} 1 & 1 &5 \\ 1& 2 &\mu \\ 1& 3 & 1 \end{vmatrix} \neq 0\Rightarrow \mu\neq3$
$\therefore q= \frac{1}{6}\cdot \frac{5}{6}= \frac{5}{36}$

Example 3: If the following system of linear equations

$\begin{aligned} &2 x+y+z=5 \\ &x-y+z=3 \\ &x+y+a z=b \end{aligned}$
has no solution, then : [JEE MAINS 2021]