Question

3x+4y=8,X-2y =5
Cramer's rule​

Answer 1

Solve the following equations using Cramer's Rule :-

• 3x + 4y = 8

• x - 2y = 5

Solution :-

• The matrix form of the above equations are

$\rm :\longmapsto\:\: \begin{bmatrix} 3 & 4\\ 1 & - 2\end{bmatrix}\begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}8\\5\end{array}\right]\end{gathered}$

Here,

$\rm :\longmapsto\:A = \: \begin{bmatrix} 3 & 4\\ 1 & - 2\end{bmatrix}$

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

$\rm :\longmapsto\:\begin{gathered}\sf B=\left[\begin{array}{c}8\\5\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\longmapsto\:D = |A| = \: \begin{array}{|cc|}\sf 3 &\sf 4 \\ \sf 1 &\sf - 2 \\\end{array} = - 6 - 4 = - 10$

Now,

Consider,

$\rm :\longmapsto\:D_1 = \: \begin{array}{|cc|}\sf 8 &\sf 4 \\ \sf 5 &\sf - 2 \\\end{array} = - 16 - 20 = - 36$

Again,

Consider,

$\rm :\longmapsto\:D_2 = \: \begin{array}{|cc|}\sf 3 &\sf 8 \\ \sf 1 &\sf 5 \\\end{array} = 15 - 8 = 7$

Therefore,

Now,

$\rm :\longmapsto\:x = \dfrac{D_1}{D} = \dfrac{ - 36}{ - 10} = \dfrac{18}{5}$

$\rm :\longmapsto\:y = \dfrac{D_1}{D} = \dfrac{7}{ - 10} = - \dfrac{7}{10}$