Question

Find $ {{D}_{x}} $ and D for the equation $ 2x+3y=4 $ ; $ 7x-5y=2 $ .

Answer 1

Hint: We know that Cramer’s rule Cramer’s rule for expressions $ {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} $ and $ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} $ , can be given as, $ D=\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]=\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right) $ and $ {{D}_{x}}=\left[ \begin{matrix}
   {{c}_{1}} & {{b}_{1}} \\
   {{c}_{2}} & {{b}_{2}} \\
\end{matrix} \right]=\left( {{c}_{1}}{{b}_{2}}-{{c}_{2}}{{b}_{1}} \right) $ . So, we ill compare our expressions with the general expressions and then by substituting the values in expression we will find the value of Dx and D for the equation $ 2x+3y=4 $ ; $ 7x-5y=2 $ .

Complete step-by-step answer:
In question equations $ 2x+3y=4 $ and $ 7x-5y=2 $ are given and we are asked to find Dx and D, where D is determinant of two equations and Dx is determinant of equations with respect to x. Now, to find the value of determinant Cramer’s rule for expressions $ {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} $ and $ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} $ , can be given as,
 $ D=\left[ \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right]=\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right) $ ………….(i)
Now, comparing expressions $ 2x+3y=4 $ and $ 7x-5y=2 $ , with $ {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} $ and $ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} $ , we will get,
 $ {{a}_{1}}=2 $ , $ {{b}_{1}}=3 $ , $ {{a}_{2}}=7 $ , $ {{b}_{2}}=-5 $ , $ {{c}_{1}}=4 $ and $ {{c}_{2}}=2 $
On substituting these values in expression (i), we will get,
 $ D=\left[ \begin{matrix}
   2 & 3 \\
   7 & -5 \\
\end{matrix} \right]=\left( 2\left( -5 \right)-3\left( 7 \right) \right) $
 $ \Rightarrow D=\left( 2\left( -5 \right)-3\left( 7 \right) \right)=-10-21=-31 $
Now, using Cramer’s rule Dx can be given as,
 $ {{D}_{x}}=\left[ \begin{matrix}
   {{c}_{1}} & {{b}_{1}} \\
   {{c}_{2}} & {{b}_{2}} \\
\end{matrix} \right]=\left( {{c}_{1}}{{b}_{2}}-{{c}_{2}}{{b}_{1}} \right) $
As we are considering for x, we will replace the value of x with values of constants. Now, on substituting the values in expression we will get,
 $ {{D}_{x}}=\left[ \begin{matrix}
   4 & 3 \\
   2 & -5 \\
\end{matrix} \right]=\left( 4\left( -5 \right)-2\left( 3 \right) \right) $
 $ \Rightarrow {{D}_{x}}=\left( 4\left( -5 \right)-2\left( 3 \right) \right)=-20-6=-26 $
Thus, values of Dx and D are $ -26 $ and $ -31 $ respectively.

Note: Here, we were asked to find the value of Dx so, we replaced the value of constants of x with the constant terms in expression. If, we were asked to find the value of $ {{D}_{y}} $ , then we have to replace the value of constant of y with constant such as, $ {{D}_{y}}=\left[ \begin{matrix}
   {{a}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{c}_{2}} \\
\end{matrix} \right]=\left( {{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}} \right) $ . If students used this to find $ {{D}_{x}} $ , then the answer will be wrong. So, be careful while solving this type of problem.