Question

solve by cramers rule 2x+3y= 5 3x-2y=1​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:2x + 3y = 5$

and

$\rm :\longmapsto\:3x - 2y = 1$

The above equation in matrix form can be represented as

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}2&3 \\ 3& - 2 \end{matrix} \bigg]$

$\rm :\longmapsto\:\begin{gathered}\sf B=\left[\begin{array}{c}5\\1\end{array}\right]\end{gathered}$

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

So, that AX = B

Now,

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

$\rm \: = \: \: - 2 \times 2 - 3 \times 3$

$\rm \: = \: \: - 4 - 9$

$\rm \: = \: \: - 13$

$\bf\implies \: |A| = - 13$

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

Now, Consider

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf 5 &\sf 3 \\ \sf 1 &\sf - 2 \\\end{array}$

$\rm \: = \: \: - 2 \times 5 - 1 \times 3$

$\rm \: = \: \: - 10 - 3$

$\rm \: = \: \: - 13$

$\bf\implies \:D_1 = - 13$

Now, Consider

$\rm :\longmapsto\: |D_2| = \begin{array}{|cc|}\sf 2 &\sf 3 \\ \sf 5 &\sf 1 \\\end{array}$

$\rm \: = \: \: 1 \times 2 - 5 \times 3$

$\rm \: = \: \: 2 - 15$

$\rm \: = \: \: - 13$

$\bf\implies \:D_2 = - 13$

Hence,

$\rm :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 13}{ - 13} = 1$

$\rm :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 13}{ - 13} = 1$

Verification :-

Consider the first line

$\rm :\longmapsto\:2x + 3y = 5$

On substituting the values of x and y, we get

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

$\rm :\longmapsto\:2 + 3 = 5$

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

Hence, Verified

Consider the second line :

$\rm :\longmapsto\:3x - 2y = 1$

On substituting the values of x and y, we get

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

$\rm :\longmapsto\:3 - 2 = 1$

$\rm :\longmapsto\:1= 1$

Hence, Verified