Question

Use Cramer’s rule to solve this system of equations: \[x + y = 4\] and \[2x + y = 3\].
A). \[x = - 1,y = 2\]
B). \[x = 1,y = - 2\]
C). \[x = 2,y = 5\]
D). \[x = - 1,y = 5\]

Answer 1

Hint: We have to find the solutions of the given equation by the Cramer’s rule. So 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.

Complete step-by-step solution:
Given,
A pair of linear equation in two variable
\[x + y = 4\] and
\[2x + y = 3\]
To find,
The solution of the given equation by Cramer’s rule
To use Cramer’s rule we have to find the express in terms of matrix
\[\left( {\begin{array}{*{20}{c}}
  {{x_1}}&{{y_1}} \\
  {{x_2}}&{{y_2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {{c_1}} \\
  {{c_2}}
\end{array}} \right)\]
\[\left( D \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( x \right) = \,\left( c \right)\]
Matrix \[D\] is explained
Matrix \[{D_x}\] is the replacement of x terms with constant terms
So, now we are generating all the matrixes.
\[D = \left( {\begin{array}{*{20}{c}}
  1&1 \\
  2&1
\end{array}} \right)\]
\[{D_x} = \left( {\begin{array}{*{20}{c}}
  4&1 \\
  3&1
\end{array}} \right)\]
\[{D_y} = \left( {\begin{array}{*{20}{c}}
  1&4 \\
  2&3
\end{array}} \right)\]
The values of \[x\] and \[y\] are
\[x = \dfrac{{|{D_x}|}}{{|D|}}\] and \[y = \dfrac{{|{D_y}|}}{{|D|}}\]
On putting the values of \[D\] and \[{D_x}\] in order to find the value of \[x\]
\[x = \dfrac{{\left( {\begin{array}{*{20}{c}}
  4&1 \\
  3&1
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  1&1 \\
  2&1
\end{array}} \right)}}\]
On taking the determinant in numerator and denominator
\[x = \dfrac{{\left( {4 \times 1} \right) - \left( {3 \times 1} \right)}}{{(1 \times 1) - \left( {2 \times 1} \right)}}\]
On further solving
\[x = \dfrac{{4 - 3}}{{1 - 2}}\]
\[\Rightarrow x = \dfrac{1}{{ - 1}}\]
\[\Rightarrow x = - 1\]
The value of \[x\] is
\[ \Rightarrow x = - 1\]
On putting the values of \[D\] and \[{D_y}\] in order to find the value of \[y\]
\[y = \dfrac{{\left( {\begin{array}{*{20}{c}}
  1&4 \\
  2&3
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  1&1 \\
  2&1
\end{array}} \right)}}\]
On taking the determinant in numerator and denominator
\[y = \dfrac{{\left( {3 \times 1} \right) - \left( {4 \times 2} \right)}}{{(1 \times 1) - \left( {2 \times 1} \right)}}\]
On further solving
\[y = \dfrac{{3 - 8}}{{1 - 2}}\]
\[\Rightarrow y=\dfrac{-5}{-1}\]
\[\Rightarrow y = 5\]
The value of \[y\] is
\[ \Rightarrow y = 5\]
Final answer:
The solution of the pair of linear equation in two variable are:
\[ \Rightarrow x = - 1\]
\[ \Rightarrow y = 5\]

Note: To solve this type of question you must know the concept of Cramer’s rule. While applying the formula for finding the value of x and y remember that we take the determinant of the matrix, not the matrix. Make the determinant part carefully and take their determinant in order to find the solution of the linear equation in two variables. Cramer’s rule is also used to determine the solution of linear equations in three variables.