
Find the values of $D_x$ and $D_y$ to solve the simultaneous equations by Cramer's method.
$3x - 4y = 10$, $4x + 3y = 5$
Answer
563.1k+ views
Hint: The concept of the Cramer Rule is a technique that incorporates determinants to solve systems of equations that have a similar number of equations as parameters. It is a feasible and useful approach for identifying solutions to systems with an unspecified range of unknowns, only when you have a similar set of equations as unknowns.
Here, initially we will check whether the system of linear equations and the number of unknown variables are the same or not. If it is the same, then we will proceed further for the Cramer’s Rule. In this question, both the value of the system of equations and the number of unknown variables are the same (that is $2 $).
Now, we will define the coefficients of the given sets of linear equations for the coefficient matrix ‘A’. Then, using the coefficient we will find the determinant of the coefficient matrix. It will give you the value of determinant $D$. Then, we will find the value of ${D_x} $ by using Cramer’s Rule that will eliminate the coefficient of one variable (that is $x$) from the coefficient matrix ‘$A$’, and solve the determinant value of the new matrix and calculate for value of ‘x’.
Similarly, we will find the value of ${D_y} $ by using Cramer’s Rule that will eliminate the coefficient of one variable (that is $y$) from the coefficient matrix ‘$A$’, and solve the determinant value of the new matrix and calculate for value of ‘y’.
Formula Used:
$\Rightarrow$ $D = \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right) = (a \cdot d) - (b \cdot c)$
$\Rightarrow$ $x = \dfrac{{{D_x}}}{D}$
$\Rightarrow$ $y = \dfrac{{{D_y}}}{D}$
Complete step-by-step answer:
As per the question, the system of linear equations are given as:
$\Rightarrow$ $3x - 4y = 10$ - (1)
$\Rightarrow$ $4x + 3y = 5$ - (2)
Let the coefficient matrix is given by:
$\Rightarrow$ $A = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right)$
Using equation (1) and (2), define the coefficient value:
$\Rightarrow$ ${a_{11}} = 3$ ,
$\Rightarrow$ ${a_{12}} = - 4$ ,
$\Rightarrow$ ${b_1} = 10$,
$\Rightarrow$ ${a_{21}} = 4$ ,
$\Rightarrow$ ${a_{22}} = 3$ ,
$\Rightarrow$ ${b_2} = 5$.
So, the coefficient matrix is given by:
$\Rightarrow$ $A = \left( {\begin{array}{*{20}{c}}
3&{ - 4} \\
4&3
\end{array}} \right)$
Now, finding the value of determinant of matrix ‘$A$’, we get:
$\Rightarrow$ $D = \left| {\begin{array}{*{20}{c}}
3&{ - 4} \\
4&3
\end{array}} \right| = (3 \times 3) - ( - 4 \times 4) = 9 + 16 = 25$
To find the value of ${D_x}$, eliminate the coefficient of $x$ variable from the matrix ‘$A$’, and we get a new matrix:
$\Rightarrow$ ${D_x} = \left| {\begin{array}{*{20}{c}}
{{b_1}}&{{a_{12}}} \\
{{b_2}}&{{a_{22}}}
\end{array}} \right|$
$\Rightarrow$ $ {D_x} = \left| {\begin{array}{*{20}{c}}
{10}&{ - 4} \\
5&3
\end{array}} \right| = (10 \times 3) - ( - 4 \times 5) = 30 + 20 = 50$
Similarly, to find the value of ${D_y}$, eliminate the coefficient of $y$ variable from the matrix ‘$A$’, and we get a new matrix:
$\Rightarrow$ ${D_y} = \left| {\begin{array}{*{20}{c}}
{{a_{11}}}&{{b_1}} \\
{{a_{21}}}&{{b_2}}
\end{array}} \right|$
$\Rightarrow$ $ {D_y} = \left| {\begin{array}{*{20}{c}}
3&{10} \\
4&5
\end{array}} \right| = (3 \times 5) - (10 \times 4) = 15 - 40 = - 25$
Now, by using Cramer’s Rule, the value of $x$ and $y$ are:
$\Rightarrow$ $x = \dfrac{{50}}{{25}} = 2$
$\Rightarrow$ $y = \dfrac{{ - 25}}{{25}} = - 1$
Note:
Whenever in the question, it was asked to find the values of ${D_x}$ and ${D_y}$, or to value of $x$ and $y$ for any system of equations or number of variables, always remember to check for the similar number of system of linear equations and the number of unknown variables. Then, apply Cramer’s Rule to find out the result.
$D = \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right) = (a \cdot d) - (b \cdot c)$
$x = \dfrac{{{D_x}}}{D}$
$y = \dfrac{{{D_y}}}{D}$
Try to remember the above formulas for solving the simultaneous equations by Cramer’s Rule.
Here, initially we will check whether the system of linear equations and the number of unknown variables are the same or not. If it is the same, then we will proceed further for the Cramer’s Rule. In this question, both the value of the system of equations and the number of unknown variables are the same (that is $2 $).
Now, we will define the coefficients of the given sets of linear equations for the coefficient matrix ‘A’. Then, using the coefficient we will find the determinant of the coefficient matrix. It will give you the value of determinant $D$. Then, we will find the value of ${D_x} $ by using Cramer’s Rule that will eliminate the coefficient of one variable (that is $x$) from the coefficient matrix ‘$A$’, and solve the determinant value of the new matrix and calculate for value of ‘x’.
Similarly, we will find the value of ${D_y} $ by using Cramer’s Rule that will eliminate the coefficient of one variable (that is $y$) from the coefficient matrix ‘$A$’, and solve the determinant value of the new matrix and calculate for value of ‘y’.
Formula Used:
$\Rightarrow$ $D = \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right) = (a \cdot d) - (b \cdot c)$
$\Rightarrow$ $x = \dfrac{{{D_x}}}{D}$
$\Rightarrow$ $y = \dfrac{{{D_y}}}{D}$
Complete step-by-step answer:
As per the question, the system of linear equations are given as:
$\Rightarrow$ $3x - 4y = 10$ - (1)
$\Rightarrow$ $4x + 3y = 5$ - (2)
Let the coefficient matrix is given by:
$\Rightarrow$ $A = \left( {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}} \\
{{a_{21}}}&{{a_{22}}}
\end{array}} \right)$
Using equation (1) and (2), define the coefficient value:
$\Rightarrow$ ${a_{11}} = 3$ ,
$\Rightarrow$ ${a_{12}} = - 4$ ,
$\Rightarrow$ ${b_1} = 10$,
$\Rightarrow$ ${a_{21}} = 4$ ,
$\Rightarrow$ ${a_{22}} = 3$ ,
$\Rightarrow$ ${b_2} = 5$.
So, the coefficient matrix is given by:
$\Rightarrow$ $A = \left( {\begin{array}{*{20}{c}}
3&{ - 4} \\
4&3
\end{array}} \right)$
Now, finding the value of determinant of matrix ‘$A$’, we get:
$\Rightarrow$ $D = \left| {\begin{array}{*{20}{c}}
3&{ - 4} \\
4&3
\end{array}} \right| = (3 \times 3) - ( - 4 \times 4) = 9 + 16 = 25$
To find the value of ${D_x}$, eliminate the coefficient of $x$ variable from the matrix ‘$A$’, and we get a new matrix:
$\Rightarrow$ ${D_x} = \left| {\begin{array}{*{20}{c}}
{{b_1}}&{{a_{12}}} \\
{{b_2}}&{{a_{22}}}
\end{array}} \right|$
$\Rightarrow$ $ {D_x} = \left| {\begin{array}{*{20}{c}}
{10}&{ - 4} \\
5&3
\end{array}} \right| = (10 \times 3) - ( - 4 \times 5) = 30 + 20 = 50$
Similarly, to find the value of ${D_y}$, eliminate the coefficient of $y$ variable from the matrix ‘$A$’, and we get a new matrix:
$\Rightarrow$ ${D_y} = \left| {\begin{array}{*{20}{c}}
{{a_{11}}}&{{b_1}} \\
{{a_{21}}}&{{b_2}}
\end{array}} \right|$
$\Rightarrow$ $ {D_y} = \left| {\begin{array}{*{20}{c}}
3&{10} \\
4&5
\end{array}} \right| = (3 \times 5) - (10 \times 4) = 15 - 40 = - 25$
Now, by using Cramer’s Rule, the value of $x$ and $y$ are:
$\Rightarrow$ $x = \dfrac{{50}}{{25}} = 2$
$\Rightarrow$ $y = \dfrac{{ - 25}}{{25}} = - 1$
Note:
Whenever in the question, it was asked to find the values of ${D_x}$ and ${D_y}$, or to value of $x$ and $y$ for any system of equations or number of variables, always remember to check for the similar number of system of linear equations and the number of unknown variables. Then, apply Cramer’s Rule to find out the result.
$D = \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right) = (a \cdot d) - (b \cdot c)$
$x = \dfrac{{{D_x}}}{D}$
$y = \dfrac{{{D_y}}}{D}$
Try to remember the above formulas for solving the simultaneous equations by Cramer’s Rule.
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