Solve the system of linear equations using Cramer’s rule:
\[3x + 2y = 1\] and \[2x - 3y = 5\]
A.\[x = 1\], \[y = - 1\]
B.\[x = 2\], \[y = - 1\]
C.\[x = 1\], \[y = - 2\]
D.\[x = 1\], \[y = 2\]

Hint: Here we need to solve the system of linear equations using the Cramer’s rule. Using the Cramer’s rule, we will find the determinant of the coefficient matrix and the determinant of the \[x\] coefficient matrix. Then we will find the determinant of the \[y\] coefficient matrix. Finally, we will use the formula of Cramer’s rule to get the value of the variables.

Complete step-by-step answer:
Here we need to solve the system of linear equations using the Cramer’s rule.
The given simultaneous equations are:
\[3x + 2y = 1\] ……… \[\left( 1 \right)\]
 \[2x - 3y = 5\] ………. \[\left( 2 \right)\]
The given equations are in \[ax + by + c = 0\] form.
We will compare the given equations with \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\].
\[\begin{array}{l}{a_1} = 3\\{b_1} = 2\\{c_1} = 1\\{a_2} = 2\\{b_2} = - 3\\{c_2} = 5\end{array}\]
Now, we will use the Cramer’s rule to find the determinant of the coefficient matrix which is given by
\[D = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}\\{{a_2}}&{{b_2}}\end{array}} \right|\]
Now, we will substitute the values of all the coefficients here.
\[ \Rightarrow D = \left| {\begin{array}{*{20}{c}}3&2\\2&{ - 3}\end{array}} \right|\]
Now, we will find the determinant of the coefficient matrix.
\[ \Rightarrow D = 3 \times - 3 - 2 \times 2\]
On multiplying the terms, we get
\[ \Rightarrow D = - 9 - 4\]
On subtracting the terms, we get
\[ \Rightarrow D = - 13\]
Now, we will use the Cramer’s rule to find the determinant of the \[x\] coefficient matrix which is given by
\[{D_x} = \left| {\begin{array}{*{20}{c}}{{c_1}}&{{b_1}}\\{{c_2}}&{{b_2}}\end{array}} \right|\]
Now, we will substitute the values of all the coefficients and the constants here.
\[ \Rightarrow {D_x} = \left| {\begin{array}{*{20}{c}}1&2\\5&{ - 3}\end{array}} \right|\]
Now, we will find the determinant of the \[x\] coefficient matrix.
\[ \Rightarrow {D_x} = 1 \times - 3 - 5 \times 2\]
On multiplying the terms, we get
\[ \Rightarrow {D_x} = - 3 - 10\]
On subtracting the terms, we get
\[ \Rightarrow {D_x} = - 13\]
Now, we will use the Cramer’s rule to find the determinant of the \[y\] coefficient matrix which is given by\[{D_y} = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{c_1}}\\{{a_2}}&{{c_2}}\end{array}} \right|\] .
Now, we will substitute the values of all the coefficients and the constants here.
\[ \Rightarrow {D_y} = \left| {\begin{array}{*{20}{c}}3&1\\2&5\end{array}} \right|\]
Now, we will find the determinant of the \[y\] coefficient matrix.
\[ \Rightarrow {D_x} = 3 \times 5 - 1 \times 2\]
On multiplying the terms, we get
\[ \Rightarrow {D_y} = 15 - 2\]
On subtracting the terms, we get
\[ \Rightarrow {D_y} = 13\]
We know the formula to find the value of the variables using the Cramer’s rule is given by \[x = \dfrac{{{D_x}}}{D}\] and \[y = \dfrac{{{D_y}}}{D}\].
Now, we will substitute the values of the determinants in the formula.
\[x = \dfrac{{ - 13}}{{ - 13}}\]
On further simplification, we get
\[ \Rightarrow x = 1\]
Also, we have
\[y = \dfrac{{13}}{{ - 13}}\]
On further simplification, we get
\[ \Rightarrow y = - 1\]
Hence, the value of the variable \[x\] is 1 and the value of the variable \[x\] is -1.
Hence, the correct option is option A.

Note: Here, we have been given linear equations in two variables. Linear equations are the equations that have the highest degree of and have only one solution. We have used Cramer’s rule to find the solution. Cramer’s rule is defined as a method to solve the system of equations in which the number of variables and the number equation is the same. This method is used to get unique solutions if they exist. If the value of the determinant of the coefficient matrix is zero then Cramer’s rule can’t be used there.