Question

solve by cremer' s rule 3 x- 4 y=10 ,4 x+3 y=5.​

Answer 1

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

Given pair of linear equations are

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

and

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

Its matrix form is

$\rm :\longmapsto\:\bigg[ \begin{matrix}3& - 4 \\ 4&3 \end{matrix} \bigg] \: \begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}10\\5\end{array}\right]\end{gathered}$

So,

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

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

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

So, that

$\rm :\longmapsto\:\boxed{ \tt{ \: AX = B \: \: }}$

Now, Consider

$\rm :\longmapsto\: |A|$

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

$\rm \:  =  \:9 - ( - 16)$

$\rm \:  =  \:9 + 16$

$\rm \:  =  \:25$

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

So,

Consider,

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

$\rm \:  =  \:30 - ( - 20)$

$\rm \:  =  \:30 + 20$

$\rm \:  =  \:50$

Consider,

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

$\rm \:  =  \:15 - 40$

$\rm \:  =  \: - 25$

So, we have

$\rm :\longmapsto\: |A| = 25$

$\rm :\longmapsto\:D_1 = 50$

$\rm :\longmapsto\:D_2 = - 25$

So,

$\bf\implies \:x = \dfrac{D_1}{ |A| } = \dfrac{50}{25} = 2$

and

$\bf\implies \:x = \dfrac{D_2}{ |A| } = - \dfrac{25}{25} = - \: 1$

Hence, Solution is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x \: = \: 2} \\ &\sf{y \: = \: - \: 1} \end{cases}\end{gathered}\end{gathered}$