Question

Using cromer's rule,solve the following system of linear equations:
2x+3y=10
x+6y=4​

Answer 1

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

Given system of equations are

2x + 3y = 10

x + 6y = 4

The matrix form of the equation is

$\rm :\longmapsto\:A = \: \begin{bmatrix} 2 & 3\\ 1 & 6\end{bmatrix}$

$\rm :\longmapsto\:\begin{gathered}\sf B=\left[\begin{array}{c}10\\4\end{array}\right]\end{gathered}$

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

So that, .

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

Now,

$\rm :\longmapsto\: |A| = \: \begin{array}{|cc|}\sf 2 &\sf 3 \\ \sf 1 &\sf 6 \\\end{array}$

$\rm :\longmapsto\: |A| = 12 - 3$

$\rm :\longmapsto\: |A| = 9$

$\rm :\longmapsto\:As \: |A| \ne \: 0$

$\rm :\implies\:Equations \: has \: unique \: solution$

Consider,

$\rm :\longmapsto\:D_1 = \: \begin{array}{|cc|}\sf 10 &\sf 3 \\ \sf 4 &\sf 6 \\\end{array}$

$\rm :\longmapsto\:D_1 = 60 - 12$

$\rm :\longmapsto\:D_1 = 48$

Consider,

$\rm :\longmapsto\:D_2 = \: \begin{array}{|cc|}\sf 2 &\sf 10 \\ \sf 1 &\sf 4 \\\end{array}$

$\rm :\longmapsto\:D_2 = 8 - 10$

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

Therefore,

$\rm :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{48}{9} = \dfrac{16}{3}$

$\rm :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 2}{9} = - \dfrac{2}{9}$