Question

Solve the following simultaneous equations using Cramer’s rule.3x - y = 2 ; 2x - y = 3​

Answer 1

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

Given system of linear equations is

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

and

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

Now, matrix form of the above equation is

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

where,

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

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

$\rm \: B = \left[\begin{array}{c}2\\3\end{array}\right]$

Now, Consider

$\rm \: |A|$

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

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

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

$\rm \: = \: - 1$

$\rm\implies \: |A| = - 1 \: \ne \: 0$

It means, system of linear equation is consistent having unique solution.

Now, Consider

$\rm \: D_1$

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

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

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

$\rm \: = \: 1$

Now, Consider

$\rm \: D_2$

$\rm \: = \: \begin{array}{|cc|}\sf 3 &\sf 2 \\ \sf 2 &\sf3 \\\end{array}$

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

$\rm \: = \: 5$

Now,

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

and

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

Verification :-

Consider first equation

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

On substituting the values of x and y, we get

$\rm \: 3( - 1) - ( - 5) = 2$

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

$\rm \: 2 = 2$

Hence, Verified

So,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm\implies \:Solution \: is \: \begin{cases} &\sf{x \: = \: - 1 \: } \\ \\ &\sf{y \: = \: - \: 5 \: } \end{cases}\end{gathered}\end{gathered}}$