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.