Question

solve system of linear equation using Cramer's rule x+3y=4 2x-y=2​

Answer 1

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

Given pair of linear equations are

x + 3y = 4

2x - y = 2

The matrix representation of the equations are

$\rm :\longmapsto\:A = \: \begin{bmatrix} 1 & 3\\ 2 & - 1\end{bmatrix}$

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c}4\\2\end{array}\right]\end{gathered}$

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

So that,

$\rm :\longmapsto\:AX = B$

Consider,

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

$\rm :\longmapsto\: |A| = - 1 - 6$

$\rm :\longmapsto\: |A| = - 7$

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

$\bf\implies \:Equations \: has \: unique \: solution.$

Consider,

$\rm :\longmapsto\: |D_1| \: = \: \begin{array}{|cc|}\sf 4 &\sf 3 \\ \sf 2 &\sf - 1 \\\end{array}$

$\rm :\longmapsto\: |D_1| = - 4 - 6$

$\rm :\longmapsto\: |D_1| = - 10$

Consider,

$\rm :\longmapsto\: |D_2| \: = \: \begin{array}{|cc|}\sf 1 &\sf 4 \\ \sf 2 &\sf 2 \\\end{array}$

$\rm :\longmapsto\: |D_2| = 2 - 8$

$\rm :\longmapsto\: |D_2| = - 6$

Therefore,

$\rm :\longmapsto\:x = \dfrac{ |D_1| }{ |A| } = \dfrac{ - 10}{ - 7} = \dfrac{10}{7}$

$\rm :\longmapsto\:y = \dfrac{ |D_2| }{ |A| } = \dfrac{ - 6}{ - 7} = \dfrac{6}{7}$