Chapter SH, Problem 3.20EP

To determine

To solve: the given system of equation by using Cramer’s rule.

Answer to Problem 3.20EP

  $x=-1,y=4$ .

Explanation of Solution

Given:

The given system of equation is $\begin{array}{l}-3x-2y=-5\\ -4x+6y=28\end{array}$ .

Concept used:

Cramer’s rule: For a $2\times 2$ matrix, $x=\frac{D_x}{D}=\frac{\left|\begin{array}{l}{c}_1{b}_1\\ c_2{b}_2\end{array}\right|}{\left|\begin{array}{l}{a}_1{b}_1\\ a_2{b}_2\end{array}\right|};y=\frac{D_y}{D}=\frac{\left|\begin{array}{l}{a}_1{c}_1\\ a_2{c}_2\end{array}\right|}{\left|\begin{array}{l}{a}_1{b}_1\\ a_2{b}_3\end{array}\right|}$

Calculation:

The givensystem of equation is $\begin{array}{l}-3x-2y=-5\\ -4x+6y=28\end{array}$ .

Now, change the equation in matrix form, then $\left|\begin{array}{l}-3-2|-5\\ -46|28\end{array}\right|$

Therefore, the determinant is

  $\begin{array}{l}x=\frac{D_x}{D}=\frac{\left|\begin{array}{l}{c}_1{b}_1\\ c_2{b}_2\end{array}\right|}{\left|\begin{array}{l}{a}_1{b}_1\\ a_2{b}_2\end{array}\right|};y=\frac{D_y}{D}=\frac{\left|\begin{array}{l}{a}_1{c}_1\\ a_2{c}_2\end{array}\right|}{\left|\begin{array}{l}{a}_1{b}_1\\ a_2{b}_3\end{array}\right|}\\ x=\frac{D_x}{D}=\frac{\left|\begin{array}{l}-5-2\\ 286\end{array}\right|}{\left|\begin{array}{l}-3-2\\ -46\end{array}\right|};y=\frac{D_y}{D}=\frac{\left|\begin{array}{l}-3-5\\ -428\end{array}\right|}{\left|\begin{array}{l}-3-2\\ -46\end{array}\right|}\end{array}$

Now,

  $\begin{array}{l}x=\frac{D_x}{D}=\frac{\left|\begin{array}{l}-5-2\\ 286\end{array}\right|}{\left|\begin{array}{l}-3-2\\ -46\end{array}\right|}\\ =\frac{-5\times 6-28\times \left(-2\right)}{-3\times 6-\left(-4\right)\times \left(-2\right)}\\ =\frac{-30+56}{-18-8}=\frac{26}{-26}=-1\end{array}$

And,

  $\begin{array}{l}y=\frac{D_y}{D}=\frac{\left|\begin{array}{l}-3-5\\ -428\end{array}\right|}{\left|\begin{array}{l}-3-2\\ -46\end{array}\right|}\\ =\frac{-3\times 28-\left(-5\right)\times \left(-4\right)}{-3\times 6-\left(-4\right)\times \left(-2\right)}\\ =\frac{-84-20}{-18-8}=\frac{-104}{-26}=4\end{array}$

Hence, from Cramer’s rule, the solution of the given system of equation is $x=-1,y=4$ .