Cramer's Rule Calculator is an easy tool which solves the given linear equations using cramer's rule and gives you the answer instantly. Cramer's rule is used to solve the given set of linear equations. It is an easy method to solve instead of solving all these entire equations.

## Cramer's Rule Problems

Consider the two equations
a1x + b1y = c1

a2x + b2y = c2
To solve this using Cramer's rule you should follow the following steps:

Step 1: Read the given problem and observe whether the given matrix is a square matrix if so carry on.

Step 2 : Find the determinant of given matrix and check whether |A| $\neq$ 0, if so cramer's rule can be applied.
D = $\begin{vmatrix} a_{1}&b_{1} \\ a_{2}&b_{2} \end{vmatrix}$ $\neq$ 0
i.e. (a1 b2 - a2 b1) $\neq$ 0

Step 3 : Replace the first column by the elements present in the right and get the value of determinant D1
$\begin{vmatrix} c_{1}&b_{1} \\ c_{2}&b_{2} \end{vmatrix}$
Replace second column by the elements present in the right and get the value of determinant D2
$\begin{vmatrix} a_{1}&c_{1} \\ a_{2}&c_{2} \end{vmatrix}$

Step 4 : Find the value of variable using formula

x = $\frac{D_{1}}{D}$ and  y = $\frac{D_{2}}{D}$.
Below are given some problems based on cramer's rule which may be useful for you.

### Solved Examples

Question 1: Solve the following equations using cramer's rule:
5x + 6y = 2
2x - y = 5.
Solution:

The given equations are:
5x + 6y = 2
2x -y = 5

D = $\begin{vmatrix} 5 & 6\\2 & -1\end{vmatrix}$ =  - 5 - 12 = -17
D1 = $\begin{vmatrix} 2&6 \\ 5&-1 \end{vmatrix}$ = -2 - 30 = -32
D2 = $\begin{vmatrix} 5&2 \\ 2&5 \end{vmatrix}$ = 25 - 4 = 21

The value of the variables are
x = $\frac{D_{1}}{D}$ = $\frac{-32}{-17}$ = 1.882 and
y = $\frac{D_{2}}{D}$ = $\frac{21}{-17}$ = -1.235
The solution for the system are (1.882, -1.235).

Question 2: Find the cramer's rule for the following equations:
3x + 2y + z = 2
2x + y - z = 1
x + y + z = 2
Solution:

The given equations are
3x + 2y + z = 2
2x + y - z = 1
x + y + z = 2
D = $\begin{vmatrix} 3 & 2 & 1\\ 2 & 1 & -1\\ 1 & 1 & 1 \end{vmatrix}$
= 3 (1+1) -2 (2+1) + 1(2-1)
= 6 - 6 + 1
= 1
D1 = $\begin{vmatrix} 2 & 2 & 1\\ 1 & 1 & -1 \\ 2 & 1 & 1 \end{vmatrix}$
= 2(1+1) - 2 (1+2) + 1(1-2)
= 4 - 6 + 1
= 1
D2 = $\begin{vmatrix} 3& 2 & 1\\ 2& 1 & -1\\ 1& 2 & 1 \end{vmatrix}$
= 3(1+2) - 2(2+1) + 1(4-1)
= 9 - 6 + 3
= 6
D3 = $\begin{vmatrix} 3& 2 & 2\\ 2& 1 &1 \\ 1& 1 & 2 \end{vmatrix}$
= 3(2-1) - 2(4-1) + 2(2-1)
= 3 - 6 + 2
= -1
The variables are
x = $\frac{D_{1}}{D}$ = $\frac{1}{1}$ = 1
y = $\frac{D_{2}}{D}$ = $\frac{6}{1}$ = 6
z = $\frac{D_{3}}{D}$ = $\frac{-1}{1}$ = -1
The solution for the system are (1,6,-1).

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