Question

Using matrices, solve the following system of equations:
$3x+4y+7z=4$, $2x-y+3z=-3$ and $x+2y-3z=8$.

Answer 1

In order to solve this type of questions, we have to represent the above given three equations in matrix form and then using the cramer’s rule we can get the value of $x,y$ and $z$ which will be out solution of the question. According to cramer's rule if we have given a system of equation like ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}}$; ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}}$ and ${{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}$ then we can directly calculate the value of $x,y$ and $z$ as $x=\dfrac{\left| \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}$, $y=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}$ and $z=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}$.

Complete step by step answer:
As we know that it is given three simultaneous equations and we have to solve it by using matrices. So, let us represent all these three equations in the form of a matrix.
Now ,we have $3x+4y+7z=4$, $2x-y+3z=-3$ and $x+2y-3z=8$.
It can also be written in the matrix form as,
$\therefore \left( \begin{matrix} 3 & 4 & 7 \\\ 2 & -1 & 3 \\\ 1 & 2 & -3 \\\ \end{matrix} \right)\left( \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right)=\left( \begin{matrix} 4 \\\ -3 \\\ 8 \\\ \end{matrix} \right)$
After representing it into matrix form we can now use cramer’s rule to calculate the value of $x,y$ and $z$
According to cramer's rule if we have given a system of equation:
${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z={{d}_{1}}$
${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z={{d}_{2}}$
${{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z={{d}_{3}}$
then $x=\dfrac{\left| \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}$, $y=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}$ and $z=\dfrac{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\\ \end{matrix} \right|}{\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|}$.

Extending the above concept in order to solve this question, we get

4 & 4 & 7 \\\ -3 & -1 & 3 \\\ 8 & 2 & -3 \\\ \end{matrix} \right|}{\left| \begin{matrix} 3 & 4 & 7 \\\ 2 & -1 & 3 \\\ 1 & 2 & -3 \\\ \end{matrix} \right|}$$ Expanding the determinants in the above equation, we get $\Rightarrow x=\dfrac{4(3-6)-4(9-24)+7(-6+8)}{3(3-6)-4(-6-3)+7(4+1)}$ $\Rightarrow x=\dfrac{-12+60+14}{-9+36+35}=\dfrac{62}{62}=1$ Similarly, $$y=\dfrac{\left| \begin{matrix} 3 & 4 & 7 \\\ 2 & -3 & 3 \\\ 1 & 8 & -3 \\\ \end{matrix} \right|}{\left| \begin{matrix} 3 & 4 & 7 \\\ 2 & -1 & 3 \\\ 1 & 2 & -3 \\\ \end{matrix} \right|}$$ Expanding the determinants in the above equation, we get $\Rightarrow y=\dfrac{3(9-24)-4(-6-3)+7(16+3)}{3(3-6)-4(-6-3)+7(4+1)}$ $\Rightarrow y=\dfrac{-45+36+133}{-9+36+35}=\dfrac{124}{62}=2$ And, $$z=\dfrac{\left| \begin{matrix} 3 & 4 & 4 \\\ 2 & -1 & -3 \\\ 1 & 2 & 8 \\\ \end{matrix} \right|}{\left| \begin{matrix} 3 & 4 & 7 \\\ 2 & -1 & 3 \\\ 1 & 2 & -3 \\\ \end{matrix} \right|}$$ Expanding the determinants in the above equation, we get $\Rightarrow z=\dfrac{3(-8+6)-4(16+3)+4(4+1)}{3(3-6)-4(-6-3)+7(4+1)}$ $\Rightarrow z=\dfrac{-6-76+20}{-9+36+35}=\dfrac{-62}{62}=-1$ . **Hence, the solution of the above simultaneous equations are $x=1,y=2$ and $z=-1$.** **Note:** These types of questions where there is a system of linear equations are given, it is wise to involve matrices and determinants to solve them. In order to solve these questions, students should carefully substitute values and take care of signs while expanding determinants as these are places where they can make silly mistakes and get the answer wrong. So, take care of these mistakes while solving and compute the correct answer.