Question

With the help of determinants , solve the following equation :
2x + 3y = 9 ;
3x -2y = 7​

Answer 1

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

Given pair of linear equations is

$\rm :\longmapsto\:2x + 3y = 9$

and

$\rm :\longmapsto\:3x - 2y = 7$

So, the above equations in matrix form can be represented as

$\rm :\longmapsto\:\bigg[ \begin{matrix}2&3 \\ 3& - 2 \end{matrix} \bigg]\begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}9\\7\end{array}\right]\end{gathered}$

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

Where,

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

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c}9\\7\end{array}\right]\end{gathered}$

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

Now, Consider,

$\red{\rm :\longmapsto\: |A|}$

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

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

$\rm \: = \: - 13$

$\bf\implies \:\boxed{ \tt{ \: |A| = - \: 13 \: }}$

So, system of equations have unique solution.

Now, Consider

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf 9 &\sf 3 \\ \sf 7 &\sf - 2 \\\end{array}$

$\rm \: = \: - 18 - 21$

$\rm \: = \: - \: 39$

$\bf\implies \:\boxed{ \tt{ \: D_1 = - 39 \: }}$

Now, Consider

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 2 &\sf 9 \\ \sf 3 &\sf 7 \\\end{array}$

$\rm \: = \:14 - 27$

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

$\bf\implies \:\boxed{ \tt{ \: D_2 = - 13 \: }}$

Now, Using Cramer's Rule, we have

$\bf\implies \:x = \dfrac{D_1}{ |A| } = \dfrac{ - 39}{ - 13} = 3$

and

$\bf\implies \:y = \dfrac{D_2}{ |A| } = \dfrac{ - 13}{ - 13} = 1$

Verification :-

Consider,

$\rm :\longmapsto\:2x + 3y = 9$

On substituting the values of x and y, we get

$\rm :\longmapsto\:2(3) + 3(1) = 9$

$\rm :\longmapsto\:6 + 3= 9$

$\bf\implies \:9 = 9$

Hence, Verified

Answer 2

$\large{\underbrace{\underline{\color{magenta}{\bf{Solution:-}}}}}$

Given equations

2x+3y=9,3x−2y=7

here,

$x = \frac{ ∆_{1} }{∆} = \frac{ - 39}{ - 13} = 3$

$and \: \: y = \frac{ ∆_{2}}{∆} = \frac{ - 13}{ - 13} = 1$

solution of equation

$\Huge{\color{magenta}{\fbox{\textsf{\textbf{X =3 }}}}}$

$\Huge{\color{magenta}{\fbox{\textsf{\textbf{Y= 1}}}}}$