Chapter 11.3, Problem 61AYU

To determine

To prove: The Cramer’s Rule for system of two linear equation with two variable.

Explanation of Solution

Given information:

The Cramer’s Rule.

Proof:

Consider the system of equations $\{\begin{array}{c}ax+by=s\\ cx+dy=t\end{array}$ , the system is solved with help of Cramer’s Rule as first write the system in the form of augmented matrix,

  $\left[\begin{array}{cc}a& b\\ c& d\end{array}|\begin{array}{c}s\\ t\end{array}\right]$

The value of variables are $x=\frac{{△}_1}{\Delta }$ and $y=\frac{{△}_2}{\Delta }$ .

Here $\Delta$ is the determinant of the matrix and ${△}_1$ is the determinant of the terms excluding the coefficient of y and ${△}_2$ is the determinant of the terms excluding the coefficient of x .

So, $\Delta =\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|,{\Delta }_1=\left|\begin{array}{cc}s& b\\ t& d\end{array}\right|\text{ and }{\Delta }_2=\left|\begin{array}{cc}a& s\\ c& t\end{array}\right|$ .

Therefore, solution is $x=\frac{\left|\begin{array}{cc}s& b\\ t& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$ and $y=\frac{\left|\begin{array}{cc}a& s\\ c& t\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$ . The solution is valid only when $ad-bc\ne 0$ .

Case 1: When $a=0$ , then $b\ne 0\text{ and }c\ne 0$ , so $\Delta =-bc\ne 0$ .

Therefore, solution is $x=\frac{sd-bt}{-bc}$ and $y=\frac{-sc}{-bc}=\frac{s}{b}$ .

Case 2: When $d=0$ , then $b\ne 0\text{ and }c\ne 0$ , so $\Delta =-bc\ne 0$ .

Therefore, solution is $x=\frac{-bt}{-bc}=\frac{t}{c}$ and $y=\frac{at-sc}{-bc}$ .

Case 3: When $b=0$ , then $a\ne 0\text{ and }d\ne 0$ , so $\Delta =ad\ne 0$ .

Therefore, solution is $x=\frac{sd}{ad}=\frac{s}{a}$ and $y=\frac{at-sc}{ad}$ .

Case 4: When $c=0$ , then $a\ne 0\text{ and }d\ne 0$ , so $\Delta =ad\ne 0$ .

Therefore, solution is $x=\frac{sd-bt}{ad}$ and $y=\frac{at}{ad}=\frac{t}{d}$ .

Hence, theCramer’s Rule for system of two linear equation with two variable is proved under all conditions.