Question

Solve by using Cramers rule. 3x + 2y = 1 , x - y = 2
.

Answer 1

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

Given pair of linear equations is

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

and

$\rm :\longmapsto\:x - y = 2- - - (2)$

The matrix form of the above system of equations is

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

where,

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

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

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{c}1\\2\end{array}\right]\end{gathered}$

Now, Consider

$\rm :\longmapsto\: |A| = \begin{array}{|cc|}\sf 3 &\sf 2 \\ \sf 1 &\sf - 1 \\\end{array} = - 3 - 2 = - 5 \ne \: 0$

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

So, Consider

$\rm :\longmapsto\: D_1 = \begin{array}{|cc|}\sf 1 &\sf 2 \\ \sf 2 &\sf - 1 \\\end{array} = - 1 - 4 = - 5$

$\rm :\longmapsto\: D_2 = \begin{array}{|cc|}\sf 3 &\sf 1 \\ \sf 1 &\sf 2 \\\end{array} = 6 - 1 = 5$

So,

$\bf\implies \:x = \dfrac{D_1}{ |A| } = \dfrac{ - 5}{ - 5} = 1$

and

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

VERIFICATION

Consider first equation

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

On substituting the value of x and y, we get

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

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

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

Hence, Verified