Question

cramers rule 3x - 4y = 10 ; 4x + 3y = 5​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:3x - 4y = 10$

and

$\rm :\longmapsto\:4x + 3y = 5$

The above equation in matrix form can be represented as

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}3& - 4 \\ 4&3 \end{matrix} \bigg]$

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

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c}10\\5\end{array}\right]\end{gathered}$

So that, AX = B

Now, Consider

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

$\rm \: = \: \: 3 \times 3 - 4 \times ( - 4)$

$\rm \: = \: \: 9 + 16$

$\rm \: = \: \: 25$

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

Now, Consider

$\rm :\longmapsto\:D_1 = \begin{array}{|cc|}\sf 10 &\sf - 4 \\ \sf 5 &\sf 3 \\\end{array}$

$\rm \: = \: \: 10 \times 3 - 5 \times ( - 4)$

$\rm \: = \: \: 30 + 20$

$\rm \: = \: \: 50$

Now, Consider

$\rm :\longmapsto\: |D_2| = \begin{array}{|cc|}\sf 3 &\sf 10 \\ \sf 4 &\sf 5 \\\end{array}$

$\rm \: = \: \: 3 \times 5 - 4 \times 10$

$\rm \: = \: \: 15 - 40$

$\rm \: = \: \: - 25$

Now,

$\rm :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{50}{25} = 2$

and

$\rm :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 25}{25} = - \: 1$

Verification

Consider the first equation

$\rm :\longmapsto\:3x - 4y = 10$

On substituting the values of x and y, we get

$\rm :\longmapsto\:3(2) - 4( - 1)= 10$

$\rm :\longmapsto\:6 + 4= 10$

$\rm :\longmapsto\:10= 10$

Hence, Verified

Consider the second equation

$\rm :\longmapsto\:4x + 3y = 5$

On substituting the values of x and y, we get

$\rm :\longmapsto\:4(2) + 3( - 1) = 5$

$\rm :\longmapsto\:8 - 3= 5$

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

Hence, Verified