Question

What is Cramer’s rule?

Answer 1

This question is from the topic of matrix and determinant. Cramer’s rule is a handy way to solve for just one of the variables without having to solve the whole system of equations. In this question, we will first understand the definition of Cramer’s rule. After that, we will see some application of Cramer’s rule.

Complete step-by-step solution:
Let us solve this question.
Cramer’s rule is an explicit formula for the solution of a system of linear equations with as many equations as is unknown, valid whenever the system has a unique solution.
Or in a simple way, we can say that if the numbers of equations are n and those equations are having n variables, then we can use Cramer’s rule to find the value of variables.
Now, let us understand this rule from the following:
Consider the linear system:
${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$
${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$
We can write the above in matrix form as

{{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ \end{matrix} \right]=\left[ \begin{matrix} {{c}_{1}} \\\ {{c}_{2}} \\\ \end{matrix} \right]$$ Here, we can see that there are two equations. In those two equations, there are two variables that is, x and y and the rest are constants. So, we can say that the matrix having constant is of the order $$2\times 2$$. Remember that $$D=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|={{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}$$ should not be zero. Hence, according to Cramer’s rule, the value of x and y can be written as $$x=\dfrac{\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\\ {{c}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|}=\dfrac{{{c}_{1}}{{b}_{2}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}}$$, $$y=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|}=\dfrac{{{a}_{1}}{{c}_{2}}-{{c}_{1}}{{a}_{2}}}{{{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}}}$$ Now, similarly if there are three equations and three variables. We can write the equations as $${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}}$$ $${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}}$$ $${{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}$$ The matrix form can be written using above equations as $$\left[ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} {{d}_{1}} \\\ {{d}_{2}} \\\ {{d}_{3}} \\\ \end{matrix} \right]$$ The matrix containing only constants i.e., $$\left[ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right]$$ is of the order of $$3\times 3$$. Remember that $$D=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|$$ should not be zero. Hence, using the Cramer’s rule, the value of x, y and z will be $$x=\dfrac{\left| \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}=\dfrac{{{d}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{d}_{2}}{{c}_{3}}-{{c}_{2}}{{d}_{3}} \right)+{{c}_{1}}\left( {{d}_{2}}{{b}_{3}}-{{b}_{2}}{{d}_{3}} \right)}{{{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{a}_{2}}{{c}_{3}}-{{c}_{2}}{{a}_{3}} \right)+{{c}_{1}}\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)}$$ $$y=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}=\dfrac{{{a}_{1}}\left( {{d}_{2}}{{c}_{3}}-{{c}_{2}}{{d}_{3}} \right)-{{d}_{1}}\left( {{a}_{2}}{{c}_{3}}-{{c}_{2}}{{a}_{3}} \right)+{{c}_{1}}\left( {{a}_{2}}{{d}_{3}}-{{d}_{2}}{{a}_{3}} \right)}{{{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{a}_{2}}{{c}_{3}}-{{c}_{2}}{{a}_{3}} \right)+{{c}_{1}}\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)}$$ $$z=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}=\dfrac{{{a}_{1}}\left( {{b}_{2}}{{d}_{3}}-{{d}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{a}_{2}}{{d}_{3}}-{{d}_{2}}{{a}_{3}} \right)+{{d}_{1}}\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)}{{{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{a}_{2}}{{c}_{3}}-{{c}_{2}}{{a}_{3}} \right)+{{c}_{1}}\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)}$$ **Now, we have seen how we can use Cramer’s rule. This can be used for finding any variables from the equations. Just remember that the number of variables should be equal to the number of equations.** **Note:** We should have a better knowledge in the topic of matrices and determinants. Remember that the value of D that is the determinant of the constant matrix should not be zero. If it is zero, then we can say that the value of variables cannot be unique. The solution can be infinite or zero. And, we should not use Cramer’s rule, if the value of D is zero.