Question

Solve the equations 2x-y=-2;x+2y=3 with the help of Cramer's rule​

Answer 1

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

Given pair of linear equations are

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

and

$\rm :\longmapsto\:x + 2y = 3$

The above equations in matrix form can be represented as

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

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

$\rm :\longmapsto\:B \: = \: \begin{gathered}\sf \left[\begin{array}{c} - 2\\ 3\end{array}\right]\end{gathered}$

So that, AX = B

Now, Consider

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

$\rm \: = \: \: 2 \times 2 - 1 \times ( - 1)$

$\rm \: = \: \: 4 + 1$

$\rm \: = \: \: 5$

$\bf\implies \: |A| = 5$

This implies, system of equations is consistent having unique solution.

Now, Consider

$\rm :\longmapsto\: D_1= \begin{array}{|cc|}\sf - 2 &\sf - 1 \\ \sf 3 &\sf 2 \\\end{array}$

$\rm \: = \: \: - 2 \times 2 - 3 \times ( - 1)$

$\rm \: = \: \: - 4 + 3$

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

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

$\rm :\longmapsto\: D_2= \begin{array}{|cc|}\sf 2 &\sf - 2 \\ \sf 1 &\sf 3 \\\end{array}$

$\rm \: = \: \: 3 \times 2 - 1 \times ( - 2)$

$\rm \: = \: \: 6 + 2$

$\rm \: = \: \: 8$

$\bf\implies \:D_2 = 8$

Hence,

$\rm :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 1}{5}$

and

$\rm :\longmapsto\:y= \dfrac{D_2}{ |A| } = \dfrac{8}{5}$

Verification :-

Consider

$\rm :\longmapsto\:x + 2y = 3$

On substituting the values of x and y, we get

$\rm :\longmapsto\:\dfrac{ - 1}{5} +2 \times \dfrac{8}{5} = 3$

$\rm :\longmapsto\:\dfrac{ - 1}{5} +\dfrac{16}{5} = 3$

$\rm :\longmapsto\:\dfrac{ - 1 + 16}{5} = 3$

$\rm :\longmapsto\:\dfrac{15}{5} = 3$

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

Hence,Verified