Question

To solve x+y=3,3x-2y-4=0 by deteminant method find D

Answer 1

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

Given pair of linear equations is

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

and

$\rm :\longmapsto\:3x - 2y = 4$

can be represented in matrix form as

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

So, that,

$\bf\implies \:\boxed{ \tt{ \: AX = B \: \: }}$

where,

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}1&1 \\ 3& - 2 \end{matrix} \bigg]$

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c}3\\4\end{array}\right]\end{gathered}$

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

Now, Consider

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

$\red{\rm \: = \: - 2 - 3}$

$\red{\rm \: = \: - 5}$

Since,

$\red{\rm :\longmapsto\: |A| \: \ne \: 0}$

It implies, system of equations have unique solution.

So,

$\purple{\rm :\longmapsto\:D_1 = \begin{array}{|cc|}\sf 3 &\sf 1 \\ \sf 4 &\sf - 2 \\\end{array}}$

$\purple{\rm \: = \: - 6 - 4}$

$\purple{\rm \: = \: - 10}$

Now,

$\purple{\rm :\longmapsto\:D_2 = \begin{array}{|cc|}\sf 1 &\sf 3 \\ \sf 3 &\sf 4 \\\end{array}}$

$\purple{\rm \: = \: 4 - 9}$

$\purple{\rm \: = \: - 5}$

So, By Determinant method,

$\green{\bf :\longmapsto\:x = \dfrac{D_1}{ |A| } = \dfrac{ - 10}{ - 5} = 2 \: }$

and

$\green{\bf :\longmapsto\:y = \dfrac{D_2}{ |A| } = \dfrac{ - 5}{ - 5} = 1 \: }$