Question

2x=5y-1 ;x=y-1 solve the crammers rule​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given equations are

$\rm :\longmapsto\:2x = 5y - 1$

and

$\rm :\longmapsto\:x = y - 1$

can be rewritten as

$\rm :\longmapsto\:2x - 5y = - 1$

and

$\rm :\longmapsto\:x - y = - 1$

The above equation in matrix form can be written as

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}2& - 5 \\ 1& - 1 \end{matrix} \bigg]$

$\rm :\longmapsto\:X = \begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c} - 1\\ - 1\end{array}\right]\end{gathered}$

So,

$\rm :\implies\:AX = B$

Consider,

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 2 &\sf - 5 \\ \sf 1 &\sf - 1 \\\end{array}$

$\rm \: = \: \: - 2 + 5$

$\rm \: = \: \: 3$

$\rm :\implies\: |A| = 3 \: \ne \: 0$

➢ System of equations is consistent having unique solution.

Now,

Consider,

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf - 1 &\sf - 5 \\ \sf - 1 &\sf - 1 \\\end{array}$

$\rm \: = \: \: 1 - 5$

$\rm \: = \: \: - 4$

$\bf\implies \:D_1 = - 4$

Consider,

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 2 &\sf - 1 \\ \sf 1 &\sf - 1 \\\end{array}$

$\rm \: = \: \: - 2 + 1$

$\rm \: = \: \: -1$

$\bf\implies \:D_1 = - 1$

So,

By Cramer's Rule

$\bf :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 4}{3} = - \dfrac{4}{3}$

and

$\bf :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 1}{3} = - \dfrac{1}{3}$

Verification

Consider first equation :-

$\rm :\longmapsto\:2x - 5y = - 1$

On substituting the values of x and y, we get

$\rm :\longmapsto\:2\bigg( - \dfrac{4}{3} \bigg) - 5\bigg( - \dfrac{1}{3} \bigg) = - 1$

$\rm :\longmapsto\:\bigg( - \dfrac{8}{3} \bigg) + \bigg(\dfrac{5}{3} \bigg) = - 1$

$\rm :\longmapsto\:\dfrac{ - 8 + 5}{3} = - 1$

$\rm :\longmapsto\:\dfrac{ - 3}{3} = - 1$

$\rm :\implies\: - 1 = - 1$

Hence, Verified

Consider second equation

$\rm :\longmapsto\:x - y = - 1$

On substituting the values of x and y, we get

$\rm :\longmapsto\:\bigg( - \dfrac{4}{3} \bigg) - \bigg( - \dfrac{1}{3} \bigg) = - 1$

$\rm :\longmapsto\:\bigg( - \dfrac{4}{3} \bigg) + \bigg(\dfrac{1}{3} \bigg) = - 1$

$\rm :\longmapsto\:\dfrac{ - 4 + 1}{3} = - 1$

$\rm :\longmapsto\:\dfrac{ - 3}{3} = - 1$

$\rm :\implies\: - 1 = - 1$

Hence, Verified